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Theft, mutilation, and underlining of books are reasons | for disciplinary action and may result in dismissal from the University. To renew call Telephone Center, 333-8400 UNIVERSITY OF ILLINOIS LIBRARY AT URBANA-CHAMPAIGN L161—O-1096 i St iN ) hy % HR 3 Y , y & v4 : 4 : qi ae ea cate wt uv BOGS r Ln 4 Ree Y 5 ee SS ee eee 4 a ALGEBRAICAL PROBLEMS, PRODUCING SIMPLE AND QUADRATIC EQUATIONS, WiTH THEIR SOLUTIONS ; DESIGNED AS ‘ AN INTRODUCTION TO THE HIGHER BRANCHES OF ANALYTICS: TO WHICH IS ADDED, AN APPENDIX, CONTAINING A COLLECTION OF PROBLEMS ON THE NATURE AND SOLUTION OF EQUATIONS OF HIGHER DIMENSIONS. BY MILES BLAND, D.D. FERS. & FS.A. LATE FELLCW AND TUTOR OF ST. JOHN’S COLLEGE, CAMBRIDGE, NINTH EDITION, WITH CONSIDERABLE ADDITIONS. > i LONDON: WHITTAKER AND CO., AVE MARIA LANE. 1849. + > 4 q , i ® , Mua lak ‘ " mh ' ae ¥ \ ¥ » j 1 ‘Wlatile by . * .! ee ie i + ¢ ve t « A x ‘ rw 3 a wl ; a a re, pi ie, ao 8. aie» a en by p ; ah P re M a sh ea pean ( sy ay i re a 0 4) he » : oy ‘wee P| me A § ee ‘, «4 : i Pegi E TES: RIES : i, a BO ea ik, ; a a C pis bit a : x ; | iy ; ue ies t a Ay" ’ “8 " 1 ay OA putea tia jd! etl ani inne ie ieee ig + eS, a My: . ib MEY eas Kae pe Ey: ro ai re ‘won i i’ 7 . ; ig 7 > % ee “A ; of y ; Pony A lg BA AR Re wd CM aie ean bs he Oh ne x ’ ‘kb oma re ’ Phd ee ao 62) Fae ake ‘vsihiagae, hes) thie Le cot in -s { ® . ; ia bh RG tt ae es if Pi neem Vere + SRS es pet wit Vere) - a _ i, a. a ; | ) the SUP 6 OM a " 7 ale ae Aa | | * me “ ee Ce * y Pps re i J © * ' : qj < ha i pak i of poate tt Bayh. oes, 188 1 ai eg @'s oe a art : el A a hy - ah ADVERTISEMENT. Tue following pages, of which. ezght large editions have been favourably received by the public, contain a col- lection of Algebraical Problems, designed to point out the various methods employed by Analysts in the So- lution of Equations. They were originally drawn up for the use of the younger Students of St. John’s College, in performance of the duties attached to the office of Sadlerian Algebra Lecturer; and were printed with the approbation of the late Very Rev. Dr. Wood, who had himself at one time designed a similar publication, but from more important occupations had not found leisure to collect materials for the work. The Examples are arranged in the usual manner: 1. Simple Equations; 2. Pure Quadratics, and others which may be solved without completing the square; and 3. Adfected Quad- ratics. Utility being the sole object of this Publication, wherever a proper Example occurred, it has been taken without hesitation, or altered to suit the purpose. Many have been selected from the questions which for a very long period have been proposed annually to the Freshmen in the College Examinations at St. John’s: and several successive Editions of the Work have been benefited by vl ADVERTISEMENT. the contributions of friends. At the head of each Sec- tion are given the common Rules; and the whole con- cludes with a Collection of Problems without Solutions, for the Kpercise of the Learner. To the Sixth Edition was added an Appendix, con- taining a Collection of Problems in Arithmetical, Geo- metrical, and Harmonical Progressions; and another on the nature of Equations, and the solution of those of higher dimensions. | And the Ninth Edition has been increased by an additional Section on the Solution of Indeterminate Equations and Problems, with a corresponding portion of the Praxis. ome ALGEBRAICAL PROBLEMS. LONDON: GILBERT & RIVINGTON, PRINTERS, ST. JOHN’S SQUARE. CONTENTS. Definitions, &c. Solution of Simple Equations, involving only one unknown Quantity Solution of Simple Equations, involving two unknown Quan- tities Solution of Simple Equations, involving three or more unknown Quantities . Solution of Pure Quadratics, &c. . . . .. Solution of Adfected Quadratics, involving only one unknown COATES Ne oR Sa PR «ta aah) ("Lhe felts is uaeee oda. Solution of Adfected Quadratics, involving two unknown Quan- TIS. "pS PRED Se pas tee aise Solution of Problems producing Simple Equations, involving Ori vOne MU KTOWI QUAN reels cot. clu oy ARE a ne Solution of Problems producing Simple Equations, involving two unknown Quantities: (2... 4. Solution of Problems producing Pure Equations Solution of Problems producing Adfected Quadratics . Solution of Problems in Arithmetical and Geometrical progres- SIONS Mt | tee ye se ee ed ae Solution of Indeterminate Equations and Problems . . EA ReRRCOMMME) Gh) 9. Also, if v— a+ b6=c— 32, then, by subtracting —a+b—34x from each side, we have x + 3v =a—d+e. et Cor. 1. Hence, if the signs of all the terms on each side of an equation be changed, the two sides still remain equal; because in this change every term is transposed. Cor. 2. Hence, when the known and unknown quantities are connected in an equation by the signs + or —, they may be separated by transposing the known quantities to one side, and the unknown to the other. Cor. 3. Hence also, if any quantity be found on both sides of an equation, it may be taken away from each; thus, ife+y=5+y, then w=5. If r—b=c+d—ZJ, then e=c+d. (18.) If every term on each side of an equation be mul- tiplied by the same quantity, the results will be equal: Because in multiplying every term on each side by any quantity, the value of the whole side is multiplied by that quantity; and (13) if equals be multiplied by the same quantity, the products will be equal. Thus, if v=5+a, then 6#=30+ 6a, by multiplying every term by 6. Cor. 1. Hence an equation, of which any part is frac- tional, may be reduced to an equation expressed in integers, by multiplying every term by the denominator of the fraction. If there be more fractions than one in the given equation, it may be so reduced by multiplying every term either by the product of the denominators, or by a common multiple of them; and if the least common multiple be used, the equa- tion will be in its lowest terms. B2 4. Reduction of Equations. Thus, if ~ +o 455 13; if every term be multiplied by 12, which is the least common multiple of 2, 3, 4; 6% + 40+ 32 = 156. Cor. 2. Hence also, if every term on both sides have a common multiplier or divisor, that common multiplier or divisor may be taken away ; » thus, if az’ +abe=cdex; each term being divided by the common multiplier, 7, aw + ab = cd. Also, if a then also 5% + 2+ 6 =40 +75 So Ur CO. tony 5 ae c Also, if cats ee ear then, multiplying by 7 e+b=d+42. Also, if (a? + x)? = 32”. (a? + 2’)3, then dividing by (a? + #)?, a + 2? = 34’. Cor. 3. Also, if each member of the equation have a common divisor, the equation may be reduced by dividing both sides by that common divisor ; Thus, if ax? — a’x = abe —a’b, each side is divisible by ax — a’, whence x = b. Cor. 4. Hence also any term of an equation may be made a square, by multiplying all the terms of the equation by the quantities necessary; as, if ax’? + bew = cd’, the first term may be made a square by multiplying each term by a, and ax’? + abcx = acd’. (19.) If each side of an equation be raised to the same power, the results are equal ; Thus, if 2 =6, 2? = 36; if +a=y— 8, then 2’ + 202 +@=y—2by+86'; And .if the same roots be extracted on each side, the results are equal: Reduction of Equations. 5 US se Wey oe 40 7s ifs a= 2) 8; then) #@= ab; if @+e2er74+1=y'’—y+i, thn v+1=y—141, and if e—s4ar+4a7=y' + 6by + 90’, then v—2a=y + 30. For (13 and 14) when equal quantities on each side of an equation are multiplied or divided by equal quantities, the results will be equal. Cor. Hence, if that side of the equation which contains the unknown quantity be a perfect square, cube, or other power, by extracting the square root, cube root, &c. of both sides, the equation will be reduced to one of lower dimen- sions : Thus, if 2 + sv + 16 = 36, thenv +4= 6, if v? + 34° + 37 +1= 27, then + 1 =3, if 2 + 27° + x’ = 100, then x2? + #7 = 10. (20.) Any equation may be cleared of a single radical quantity by transposing all the other terms to the contrary side, and raising each side to the power denominated by the surd. If there are more than one surd, the operation must be repeated. Thus, if = ./ax + 6°, by squaring each side 2? = aw + BD’, which is free from surds. Also, if 2° +7+2=7, then (17) by transposition, 4/2? +7=7—2; and (19) by squaring each side, #? + 7 = 49 — 14@ + 2’, which is free from surds. Also, ifv + aes 8. Given 3z7 + ae =5+ os to find the value of 2. (1s. Cor. 1.) multiplying by 10, the least common multiple of 2 and 5, 300 + 4” + 12 = 50 + 55% —1855 .. (17) by transposition, 12 — 50 + 185 = 55” — 30” — 42, 2 SMe 0) 6a Hip Ss Corse.) a =7= 2. 6” — 4 18 — 4@ 9. Given TER aa: +2, to find the value Oller. ” (1s. Cor. 1.) multiplying every term by 3, 60 —4—6 = 18 — 4@ + 3X5 and .*. (17) by transposition, 62 + 4v — 37 = 18 + 6 + 4, Oriya 95 Pye Gib 07a) ty - = 4, 10. Given 21 + wn nah spose it we (to find the 8 value of z. Since 16 contains 8 and 2, a certain number of times exactly, it will be the least common multiple of 16, 8, and 2; and therefore (18. Cor. 1.) multiplying both sides of the equation by 16, 336 + 30 — 11 = 10% — 10 + 776 — 5625 involving one unknown Quantity. 9 .. (17) by transposition, 37 — 1ov + 56” = 11 — 10 + 776 — 336, or 49” = 4413 Spach dis comb ergy a ea poe 49 e a meee 5 eased 1]. Given # + = pia, as ne to find the value of x. (1s. Cor. 1.) multiplying both sides of the equation by 6, the product of 2 and 3, 60 + OV — 15 = 72 — 42 + 835 (17) by transposition, 62 + 97 + 4” =72+8 + 15, or 19% = 95; egtlasn Olane ded a ; _ 5 l 12. Given cy eae i iste “=A, to find the value of 2. Since 12 is a multiple of 3 and 4, it is the least common multiple of 3, 4, and 12; therefore (18. Cor. 1.) multiplying both sides of the equation by 12, 362 — 3@ + 12 — 48 = 202 + 56 — 13 *. (17) by transposition, 36% — 37 — 207 = 56 + 48 — 1 — 12, One lse—wd le it (18. Cor. 2.) av = ls == Tie | Res’ Rn Gren = et = mk Deas 7— oe to find the value of x (is. Cor. 1.) multiplying both sides of the equation by 4X 5X 7= 140, 20” — 20+ 644 — 287 = 980 — 140 — 352; .. (17) by trans", 20” — 28v + 35” = 980 — 140 + 20 — 644, or 27% = 216; 10 Solution of Simple Equations 16 ee (18. Cor. 2.) aw = ek ent 8. oT 14. Given value of 2. pees +o +" to find the (is. Cor. 1.) multiplying both sides by 2 x 3 X 5 = 30, 70” + 50 — 96 — 24” + 180 = 45” + 135; .. (17) by transposition, 70% — 244” — 45” = 135 + 96 — 50— 180, or 7 = 1. 3@ + 4 Ge eet eee 15. Given oe a=, to find the value of 2. (is. Cor. 1.) multiplying by 20, the least common multiple of 2, 4, and 5, 12% + 16 — 702 + 30 = 5” — 805 .. (17) by transposition, 16 + 30 + 80 = 5% + 70% — 12a, or 126 = 637; 126 ave 8. CG ° = — aah A (18. Cor. 2.) 2 2 1,32 4” + 2 10+ 14 16. Given =5 — 67 + ae to find the value of x. (is. Cor. 1.) multiplying both sides of the equation by 3X 5=15, 51 — 9% — 207 — 10 = 75 — 90” + 35” + 705 . (17) by trans", 907 — 357 — 20% — 9v = 75 + 70 + 10 — 51, or 26% = 1043 a Be Oteeo,) epee IS tue 26 fia 20 — @ 62 — 8 4¥ — 4 -+4= ~ -- 17. Given 2 — : 5 2 7 5 to find the value of z. ) involving one unknown Quantity. ik (1s. Cor. 1.) multiplying both sides of the equation by 2x5x7=70, 70% — 42¥ + 42 + 280 = 700 — 35@ — 60x + 80 + 56x — 56; “. (17) by transposition, 70” + 352 + 60% — 424” — 56H = 700 + 80 — 56 — 280 — 42, Or 67#@ = 40235 402 ae (1s. Cor. 2.) v = 6. 67 : 4” — 21 57 — 3L 5X2 — 96 18. Given aaa 33 4 i= 241 eee ee a to find the value of 2. (is. Cor. 1.) multiplying by 36, the least common multiple of 4, 9, and 12, 16@ — 84 + 135 + 513 — 27H = 8676 — 15¥ + 288 — 3962; “. (17) by transposition, 162 + 15% + 396% — 27H = 8676 + 288 + 84 — 135 — 513, or 400% = 8400; 8400 a AS Or. 22) — 1, 400 : 6x2 + 18 ll — 3” 13 — @ Iola Given, ee = 450 — = 6 — 48 — — 13 36 12 ool ae to find the value of 2. (is. Cor. 1.) multiplying by 36 x 13, the least common multiple of the denominators, 216% + 648 — 2262 — 143 + 394 = 2340% — 22464 — 507 + 39% — 546 + 522; .. (17) by transposition, 648 + 22464 + 507 + 546 — 2262 — 143 = 23402 + 39” + 52” — 216% — 392, OY 21760 = 217623 21760 _ “. (18. Cor. 2.) —— = 10 = @, 2176 12 Solution of Simple Equations Soe $4 = i 20. Given Ee ly v+4a 20 4 to find the value of 2. =) (1s. Cor. 1.) multiplying by 4a, the least common multiple of the denominators, 40x — 40? + i2bv~ — 40°70? = 44abe + 1204 — 10d — abe — 40’; *, (17. Cor. 3.) 40a —40°b? = 3ab4 — 100’; by transposition, (4a° — 3a6) .w = 4a°b’ — 100’; (is. Cor. 2.) (4a — 3b) .w = 4a’? — loa; __ 4ab’— 10a — 4a—36 ‘ v+ 16 e+ 8 21: Givene — au 21 ait Multiplying both sides of the equation by 21, 21x” + 168 7H +16 — —— = : 4”%— 11 21 168 Miiy Wer abe ees 4% — 11 *, (18. Cor. 1.) 644 — 176 = 214 + 168; *, (17) by transposition, 64” — 214 = 168 + 176, OY 432 = 344; sal Be COrne sae spats ty 43 92: Given Game elas Sas er AE 6% + 3 of x. Multiplying both sides of the equation by 9, 21” — 39 6x ee SIGs 11213 ailing ab + 12; = =, to find the value of 2. 5 to find the value involving one unknown Quantity. 13 21¥ — 39 i! . Cor. a. ae Ow atone ; (17 Se ragt ee *, (18. Cor. 1.) 21@ — 39 = 10% + 5; (17) by transposition, 217 — 107 = 39 + 5, Olsia i 4a 44 oe) (184Corse2:) Pies are el : 47-3 L—» se +1 23. Given Masta Na lain 9 == oe Bd 9 5 — 12 18 , to find the value of x. Multiplying both sides of the equation by 18, 8 + 6 + cre CLL Aa het 19 bv —12 ; 1262 — 522 = o/5a, = 523 and (19) m1 cO +) (382°C 0F, 12.) 2 = x. 20 84, Given ./ (4a +2) =24/(b+ 2) —(/2, to find the value of 2. (19) squaring both sides of the equation, 4a+u=4.(b+27)—4/(br42")4+2; (17. Cor. 3.) 4a + 44/ (ba + w) = 4. (642); (is. Cor. 2.)a+/ (b2 + 2’) =b4+2; (17) by transposition, ,/ (64 + v)=b—a+a4; (19) bw +a°=(b—a)’+2.(b—a).742’'; (17. Cor. 3.) (2a— 6).%7 =(b—= = 6 3 44, Given / (5 +2) + V = DET to find the value of z. (is. Cor. 1.)5+2¢7+4/(5@4+ 2") = 15; , (17) by transposition, 4/ (57 + z#*) = 15 —5—£r=10—2@, and (19) squaring both sides, 54 + # = 100 — 202 + 2"; . (17. Cor. 3.) 25” = 100; *. (18, Cor.'2.) 2 = shales 25 45, Given JV (t+ VB—o (0-V a) =57 Fe , to find the value of z. (ig. Cor. 1.) @ + / t—/ (2? — x) = ewes .. by transposition, 7 — “= = / (2 — 2), and (18. Cor. 2.) /#z—1=/(¢—1)3 (19) squaring both sides, 7 —./v+4=2—1; *, (19. Cor. 3.) fa = =; and (19) squaring both sides, 7 = =. involving one unknown Quantity. 21 = ae | 1 4 9 46. Given Z aS vale ~ wale + a) testo find the value of 2. (19) squaring both sides, a 1 2 9 and (18. Cor, 2.) = + ary (a? +2); *, (19) squaring both sides — oon += =: aS 3 4 8 (17. Cor. 3.) — = aa wae (18. Cor. 2.) ain BF Ser (ige Ole ty) 2.0 0, SECTION ILI. On the Solution of Simple Equations which involve more than one unknown Quantity. (23.) Ir the equation involve several unknown quantities, and definite values of these are required, there must neces- sarily be as many independent equations as there are unknown quantities. In which case, the values will be found by exter- minating all the unknown quantities except one; and this may be done by either of the three following methods: 1. By equalizing the coefficients of the same unknown quantity in the several equations. 2. By substitution. 3. By equating different values of the same unknown ~ quantity. 1. Of exterminating an unknown quantity by the first — method in equations where two unknown quantities are — concerned *. If the coefficient of either unknown quantity in one — equation be contained a certain number of times exactly in the coefficient of the same unknown quantity in the other, multiply the former equation by that number, then add it to, or subtract it from, the other equation, according as the signs are different or the same, and an equation arises, in which only one unknown quantity is found. * The first of these methods is also known by the name of the method by addition and subtraction, because the unknown quantities are exterminated by addition and subtraction, after the equations have been prepared in such a manner that one unknown quantity may have the same coefficient in each. ee Solution of Simple Equations, &c. 23 Thus, if 4% + y = 34 and 4y¥ + # = 16 second equation is contained 4 times exactly in the first; multiplying therefore the second equation by 4, and sub- tracting the first from it, 40 + 16Yy = 64, © and 47 -+y = 343 sy) F==30) andy = 9) ' Here the coefficient of # in the Having thus obtained a value of one of the unknown quan- tities, the other may be determined by substituting in either equation the value of the quantity found, and thus reducing the equation to one which contains only the other unknown quantity. Thus, from the second of the preceding equations, 2=16—4y=—16—8= 8. The values of 2 and y might be found in a similar manner, by multiplying the first equation by 4, and subtracting the second from it. But if neither of the coefficients be a measure of the coefficient of the same unknown quantity in the other equa- tion, multiply the first equation by the coefficient of one of the unknown quantities in the second equation, and the second equation by the coefficient of the same unknown quan- tity in the first. If the signs of the unknown quantity be alike in both, subtract one equation from the other; if unlike, add them together, and an equation arises in which only one unknown quantity is found. Thus, if 2a + 3y = 23] and 5” — 2y = 10f cients is a measure of the coefficient of the same unknown quantity in the other equation; and therefore, multiplying the first equation by 2, and the second by 3, In this case neither of the coeffi- 4v + 6y = 46, and 15” — 6y = 30; . by addition, 197 = 76, and 7 = 4; whence, as before, 3y = 23 — 24 = 23 —8 = 15, and y = 5. 24: Solution of Simple Equations The values of x and y might also be obtained, by multiplying the first equation by 5, and the second by 2, and then sub- tracting the second from the first. 2. By substitution*. Find the value of one of the unknown quantities, in terms of the other and known quantities, in the more simple of the two equations; and substitute this value instead of the quan- tity itself in the other equation; thus an equation is obtained in which there is only one unknown quantity. —— Thus in the first of the preceding examples; from the | second equation, v = 16 — 4y; substituting therefore this value of x in the first equation, 4.(16—4y) +y = 34, or 64 — 16y + y = 343 .. by transposition, (64 — 34 =) 30 = 15y, and (Lneretoreg.— Yes whence, as before, v = 8. Here a value of x might have been obtained from the second equation, and substituted for it in the first; whence an equa- tion would have arisen, involving only y; the value of which being found, that of # also might be determined, as before, by substitution. Or a value of y might be determined from either equa- tion, and substituted in the other; from which would arise an equation involving only #, the value of which might be found; and therefore the value of y also might be obtained by substitution. Again, in the second example; from the first equation is obtained 23a 24 = 23 — 3y; and therefore 7 = at, * There is an inconvenience attending the two latter methods, which the former does not offer, viz. they originate new equations with denominators, which must be got rid of. The method of substitution may be employed with advantage whenever the coefficient of one of the unknown quantities, as in the former of the preceding examples, is equal to unity. Here the inconvenience mentioned is not perceptible. But in general the first method is preferable. involving two unknown Quantities. 25 substituting therefore this value in the second equation, 23 — 3Y Soh Iai pie aan or 115 — 15y — 4y = 20; .. by transposition, 115 — 20 = 15y + 4y, or 95 = 19Y;5 . S=Y;, 93 — 3 23 — 15 8 2 2 2 Here also a value of x might be obtained from the second equation, and substituted in the first, which would give an equation involving only y; ora value of y might be obtained from either equation, which substituted in the other would give an equation involving only x; the value of which might therefore be found, and consequently that of y might also be determined. 3. By equating different values of the same unknown quantity. From each equation find the value of the same unknown quantity in terms of the other and known quantities ; then, by equating the values so found, an equation arises containing only one unknown quantity. Thus in the first of the preceding examples; from the first equation, y = 34 — 4a, and from the second equation, 4y = 16— 2; and therefore y = a, 16—w@# = 34—42; consequently, 16 — # = 136 — 1623 *, by transposition, 162 — # = 136 — 16, Or 15” ==120; '. 2£=8, and y = 34 — 4¥ = 34 — 32 = 2, as before. 26 Solution of Simple Equations In this case also, two values of w# are deducible from the two equations, which would give an equation involving y only; and the value of y being determined, that of x might also be found. Again, in the second of the preceding examples ; from the first equation, v = oe and from the second, v = = £ oy. 1O+2y 23—3Y Baty eo ale and 20 + 4y = 115 — 15y3 by transposition, 4y + 15y = 115 — 20, Ole 99205": ne/ == 5, and 4s DeLure, Here again two values of y might have been found, which would have given an equation involving only #; and from the solution of this new equation, a value of x, and therefore of y, might be found. EXAMPLES. 1. Gi ra Given 52 + 4y 8| to find the values of # and y. and 3a” + 7y = 67] Multiplying the second equation by 5, and the first by 3, 15@ + 35y = 335, and 15@ + 12y = 174; .. by subtraction, 23y = 161, and Y= Ts whence 5a@ = 58 — 4y = 58 — 28 = 30, and therefore x = 6. If the second equation had been multiplied by 4, and sub- tracted from the first when multiplied by 7, an equation involving two unknown Quantities. 27 would have arisen, involving only 2, the value of which might be determined, and thence, by substitution, the value of y. Second Method. From the second equation, 37 = 67 —7y; Oy Pee 3 Substituting this value of w in the first equation, 5. aaa + 4y = 58, and 335 — 35y + 1l2y = 174; .. by transposition, 335 — 174 = 35y — 124, or 161 == 237; ha whence, as before, the value of # may be found. In the Same manner, a value of w might be found from the first equation, which substituted in the second, would give an equation involving only y. Ora value of y might be obtained from either equation, and substituted for it in the other; whence an equation would arise involving only 2, the value of which might be found, and therefore that of y also deter- mined. Third Method. From the first equation, 57 = 58 —4y; _ 58—4y =. From the second, v7 = a 58—4y 67—7Y o° 5 an 3 >) and 174 — 12¥ = 335 — 35Y3 by transposition, 35y — 12y = 335 — 174, or 23y = 1613 . Y=75 whence, as before, 7 = 6. 28 Solution of Simple Equations In this case, two values of y might be deduced from the — two equations; and from equating these, there would arise an equation involving x only; whose value being found, that — of y also might be determined by substitution. GE ie i to find the values of x and y. ce +dy=nJ Multiplying the first equation by c, and the second by a, ace +bcy=me, ace +ady=na; .. by subtraction, (ad — bc). y =na— me, and y = ame me had hee ieee _m by _ m _nab—mbe e Sik hee We Da i ee Ole boe _mad—mbe nab—mbc tty dene “aa _mad—nab _md—nb yh R ey Ria, Or the value of x might be determined from the second — equation, v = es dy. Deere If the first equation had been multiplied by d, and sub- tracted from the second multiplied by 8, an equation would — have arisen involving only 2, the value of which might be determined; and this being substituted in either of the equa- tions, the value of y might also be found. Second Method. From the first equation, aw = m— by; _ m—by Se AG eee involving two unknown Quantities. 29 Substituting this value of 2 in the second equation, pinpdioe by a * me— bcy + ady=an, and (ad — bc) .y=an—me; +dy=n; eh toe ie OO whence, the value of # may be determined, as before. In the same manner, a value of v might be found from the second equation, which substituted in the first would give an equation involving only y, the value of which being found, that of x might also be determined. Or, a value of y might be obtained from either equation, which substituted in the other would give an equation involving only wz, the value of which, and consequently that of y, might be found. Third Method. From the first equation, 7 = = - by ; and from the second, 7 = eee _ moby. n—dy PLO Cupheuey ke Git and mc — bcy = na — ady; “. by transposition, ady — bey = na — me; _ na— me lin ari any Pe md — nb whence, as before, 7 = Py ey In this case, two values of y might be deduced from the two equations; and from equating these, there would arise another equation involving only z, the value of which being determined, that of y also might be found by substitution. 3. Given 112 + 3y = 100 } to find the values of # and y. and 47 —7y=4 30 Solution of Simple Equations Multiplying the first equation by 7, and the second by 3, 772 + 21y = 700, and 12% — 21y= 12; .. by addition, soe =712, and # = 8; whence 3y = 100 — 11¥ = 100 — 88 = 12; See to find the values of # and y. ee $8) ote and ie ae (is. Cor. 1.) clearing the equations of fractions, by multi- plying each by 6, 3x2 + 2y = 42, and 2% + 3y = 483 and as the coefficients in this case are not aliquot parts, mul- tiplying the first by 3, and the second by 2; “. 9@ + 6Yy = 126, and 4v% + 6y = 96; .. by subtraction, ba? aa 30 and #2 = '65 whence, 24 = 42 — 3% = 42 — 18 = 24, andy 212. 5. Given + 7y = 99 to find the values of # and y. and © +72 = 51 (18. Cor. 1.) multiplying each equation by 7, .@&@+49y = 693, and 49” + y = 3573 involving two unknown Quantities. ol “. by addition, 50” + 50y = 1050, CNG aoe 2 Sea but since @ + 49y = 693, subtracting the upper equation from the lower, 48Yy = 6723 7. Ys whence wv = 21— y= 21— 14 = 7. 6, Given “= + sy = a1 | Fare to find the values of # and y. and —— + 10” = 192 | Clearing the first equation of fractions, LV+2+ 24Yy = 933 .. by transposition, # + 24y = 91. Clearing the second equation of fractions, y +5 + 40” = 768; .. by transposition, 40” + y = 763. Multiplying the first equation by 40, and subtracting the second from it, 40” + 960y = 36403 40v+y si FOdss “. 959Y ah yr and y = 3; a == OF — 24 f= OL 1 7 19, Pee Given = Ly = 8] and ppiiat ee to find the values of # and y. i +16= 10| apd: By transposition, — — 4, from the first equation, and »°, 27@—y =8, 32 Solution of Simple Equations Also aus = 3, from the second equation, and .. 2y+ 7=9; which, multiplied by 2, gives 2v + 4y = 18; but22-— y=" 8; .. by subtraction, 5y = 10, and y = 2, whence v = 9 — 2y=9—4=5. 2” + 3Yy 8. Given +o= 8 to find the values of x and y. and Mee. Yat 2 Clearing the first equation of fractions, 27 + 3y + 2¥ = 48, or 47+ 3y = 48; and clearing the second of fractions, 7y — 3% — 2y = 22, or 5Y — 3x = 22. Multiplying this by 4, and the preceding one by 3, OY + 12% = 144, and 207 — 12% = 88; .; by additions?: 297) -='239, and 7/8, whence 4% = 48 — 3y = 48 — 24 = 24, and wv = 6. 9. Given 3x + ut = 292 to find the values of x and y. 22 and lly — we ee Clearing the first equation of fractions, 62 + 7y = 443 but from the second, 55y — 24 = 100. envolving two unknown Quantities. 393 Multiplying this last by 3, 165y — 6” = 300, but 7y + 62 = 44; .. by addition, 172y ae ott, AL og) Mimbo d Now 6@ = 44 —7y = 44 — 14 = 30; Soe Mer 10. Given#@ +1:y:.5:3 | to find the values of 22 5B—Yy 41 2-1 al ie PE a tata he | x2 and y. and — 3 2 12 4 From the second equation, (18. Cor. 1.) multiplied by 12, sv — 30 + 6y =41— 627 +3; *, by transposition, 142 + 6y = 74, and 727 + 3y = 37. But from the first equation, 5y = 34 + 3, or 5yY — 3% = 3. Multiplying this equation by 7, 35y — 217 = 21, and the former by 3, 9y + 212 = 111; .. by addition, 44y oles; and y = 35 o+1=-# a5, anda =, : eL—2 10 — # — 10 1 be Chg ae ee J 3 4 to find the values mig BEE. eS SSN of # and y. (18. Cor. 1.) multiplying the first equation by 60, 12@ — 24 — 200 + 207 = 15y — 1503 and by transposition, 32” — 15y = 74. D 34 Solution of Simple Equations Also (18. Cor. 1.) multiplying the second equation by 24, 16y + 32 — 6” — 3y = 62 + 783 .. by transposition, 13y — 12” = 46. Now the coefficients of w have aliquot parts; multiplying there- fore this by 8, and the preceding by 3, 104y — 96” = 368, and 964% — 45y = 222; .. by addition, 59y = 590; and y = 10; and 324 = 15y + 74 = 150 + 74.-= 224; . v=]. i AS som yes 30—2Y 12. Given 2y Fe pas 5 to find the values y 2V+1 of # and y. s— and 40 — Sa te (is. Cor. 1.) from the first equation, 40Y — 5% — 15 = 140 + 124 — 8y; .. by transposition, 48y — 17% = 155, and from the second equation, 240 — 16+ 2y = 147 — 64 — 3; .. by transposition, 30” + 2y = 160. Multiplying this by 24, 48y + 7202 = 38403 but 48y — 17% = 155; .. by subtraction, 737@ = 3685, anate=s 5. and 2y = 160 —30@ = 160 — 150 = 10; st ye. : 2 af ia 13. Given -4 — =" ay ae are 36 3 6 to find the and: #2); sy \ii47r7 | values of # and-y. involving two unknown Quantities. 35 Reducing the first equation to lower terms, Ve Ae Neary La Y 3 3 6 9 18 a. and therefore (18. Cor. 1.) multiplying by 18, BU 47 ee) 18a 2d — OY tag Sus .. by transposition, 7 = 7# — 11y. But from the second equation, 77 = 12y. | Substituting therefore this value in the preceding equation, 12y and therefore 7 = er = 12. A Beerg 14. Given v7— y Se 11 a3 to ane aondmidipwe «12 + 152. 3y ba 6 4 12 2 find the values of w and y. (18. Cor. 1.) multiplying the first equation by 33, 4y 33% — oy +6— 3H = 33 4+ 152 + 3 b] by transposition, 15% — 9y = 27 + 663 .. by subtraction, z= 12; and y = 33°— 22 = 33 — 24 = 9; alsov = 29—y—z2=29—9—12=8. In like manner, had the first equation been multiplied by _2, and subtracted from the second, an equation would have resulted, involving only w and z; and had it been multiplied by 4, and subtracted from the third when cleared of fractions, another equation would have been obtained, involving also _@ and z; whence, by the preceding rules, the values of la and z would be found, and consequently the value of ¥ ‘also, by substitution. Or if the first equation be multiplied by 3, and the second subtracted from it, an equation would arise involving only # and y; and if the first, when multiplied by 3, be subtracted from the third when cleared of fractions, another would arise involving only w and y; whence the values of 2 and y might be determined. And hence the third, that of z, might be found. Second Method. From the first equation, v = 29 — y — z; .. substituting this value of # in the second equation, 29—y— z+ 247 + 32 = 625 “. by transposition, y = 33 — 22. Also substituting, in the third equation, the value of & found from the first, | 29 — Yo 4 ip 2 yee. | 2 Se Siar ac i *, (18. Cor. 1.) 174 — 6y — 62 + 4y + 32 = 120, | and by transposition, 54 = 32 + 2y; in which, substituting the value of y found above, 54 = 32 + 66 — 423 4A, Solution of Simple Equations .. by transposition, z = 12; whence y = 33 — 22 = 33 — 24 = 9, and # = 29 —y—z2>=29—9— 12 = 8. | It may be observed, that there will be the same variety of solution, as in the last case, according as a, y, or Z, is exter- minated. | | Third Method. From the first equation, 7 = 29— y— 2, and from the second, v = 62 — 2y — 32; “29 —Y—Z= 62 — 2y — 3z, and by transposition, y = 33 — 22. Again, from the third equation, 2 = 20 — yl mS 29 -y—z=20—%—%, ¢ - 31wikia? “yy Zz and by transposition, 9 — Sean _ ; Se le 21 aie ante 32 whence 27 — og os Ss ees seed .. by transposition, sgt and, 2 == 2); whence y = 9, and w = 8, as before. The same observation applies to this solution, as did to. the last. 8. Given wv +2 oF Desde! ) + 2y + to find the values of 2, y 20 —3y +4z= 8 and z. 3su+4y—5e2=>—4 | Multiplying the first equation by 2, : 20 + 4y + 62 = 28, but 24 —3y +4z= 8; ! .. by subtraction, 7y + 22 = 20. involving three unknown Quantities. AS Again, multiplying the first equation by 3, 3u + 6y+9z7= 42, but 3v7 + 4y —5z=—4; .. by subtraction, 2y + 142 = 46, and: bye 72 == 23's "aise 2—s1 OF. but'7 y+ 9s — "20; .. by subtraction, 47z = 141, Sit Meee Ss whence y = 23 — 72 = 23 — 21 = 2, and # = 14 —2y—37=>14—4—9=1. 4. Given bz+cy=a az + cx = b> to find the values of x, y and z. ay + bx=c Multiplying the first equation by a, the second by 4, and the third by c, abz+acy=a, abz+ bex = 0’, acy + bex=c’; .. by addition, 2ab2 4+ 2acy+2bem=0°4+0'+ 0’, but 2abz + 2acy 33 Og .. by subtraction, 2bc7 = 0? + c? — a’, Cnn 2be and #« = 2 2 2 In the same way, y = a — 2 ; ome 2ab ‘ and z= ) 46 Solution of Simple Equations 5. Givens eee ae Weg smb Zirh Cirle sen Coreg Ae } 10 15 5 and 22+ 8Y¥— 22 (28 + y— 82 y+ 2 oe 12 4 1] 6 32+ 2Y a and sy +3z 27+ 3y—2 4 12 to find the values of 2, y and z. +22=>y—1+ Multiplying the first equation by 30, the least common multiple of 5, 10, and 15, 12@ + 9yY +32 —4y— 424 24 —2 = 150 + 6H — 62 — 303) *, by transposition, 8v@ + 5y + 52 = 122. Again, multiplying the second equation by 132, the least common multiple of 4, 6, 11, 12, 99" + bey a 222 — 662 — 33y + 992 = 84y + 122 + 36 + 225° . by transposition, 334% — 62y + 652 = 58. Again, multiplying the third equation by 12, the least com- mon multiple of 12, 6, 4, Bd eal LR a NE 242 = 12y—12+ 6% +4y + “ug . by transposition, 87 + 4y — 342 = — 2; but ali the first equation, 87 + 5y + 52 = 122; *. by subtraction, y + 392 = 124. Also the third equation being divided by 2, 4u+2y—-17Z=—1. Multiplying this by 33, and the second by 4, 132@-+ 66y — 561z2 = — 33, and 132@ — 248y + 2602 = 232; *, by subtraction, 314y— 8212 = — 265; but 314y + 122462 = 38936 (by multiplying the equation found above by 314) ; .. by subtraction, 130672 = 39201, a and-therefore z= 3; ee involving three unknown Quantities. 4.7 whence y = 124 — 392 = 124 —117 = 7, and 47 = 172 — 2y—1=51— 14 —-1= 36; . @=9. It is evident that any of the quantities x, y, z may be first exterminated. And the operation will be similar by the other | two methods. (25.) If there be four unknown quantities, their values may be found from four independent equations. For from the ‘four given equations, by the preceding rules, three may be ‘deduced which involve only three unknown quantities, the |values of which may be found by the last Article; and hence the fourth may be found, by substituting in any of the four , given equations, the values of the three quantities determined. But it frequently happens that in equations involving four or more unknown quantities, we do not find each equation containing all the four. And in this case the labour of solu- tion may frequently be abridged. EXAMPLES. l. Letorv —3y +27 =13 4 — 2% = 30| to find the values of v, 2, y 4y+az=14 and z. 5y + 3u = 32 | Subtracting the first equation from the third, | 7¥Y¥ — 2% =135 | multiplying the second by 3, and the fourth by 4, | 120 — 6L= 90, | 12v + 20y = 128; | .. by subtraction, 207 + 6v = 38, | but 21y — 67 = 3; .. by addition 41y = 41, EAU Gy) bam Ie Also 22 =7y—-1=>7—1=6, Ow, = 3; 48 Solution of Simple Equations and 4v = 24% + 30 = 36, . V=9; and 29z=14—4y=14—4= 10, aan « = 9d. 2. Given av + by =a’ bx — az = 0 cap ovargi te to find the values of v, x, y, 2. dy —cz=a@’ Multiplying the first equation by 8, and the second by a, abe + by=a’d, and ade —az= ab’; .. by subtraction, b’y + a’?z = ab. (a — 5); . Bdy+adz=abd.(a—)b); but from the fourth, 0’dy — U’cz = 0d’; .. by subtraction, (a@’d + 6’c).z=abd —ab’d— Pa’, _ bd. (a? — ab — bd) and 2 = apehae ee Now from the second equation, bp ea y AE — ab — bd) Cute : _b. (ad + Be — abd’) i a’'d + bc z ‘ is. eee b°c — abd’ rd avd+ be Also from the third equation, q=c¢— cr = c.(¢— 2), = ge ad.(ac + bd — a’) — b’c.(6—Cc) | a'd+ b’c __ acd. (ac + bd — a’) — Bc’. (6 — Cc) San le d. (ad + 8c) ' involving three unknown Quantities. 49 And from the fourth equation, bed. (a’* — ab — bd) ad + b’c ‘ ad’ + a’be — abc. a’d + b’c , ad’ + a’ b’ — ab’c a’d+ bc dy=@+ez=a@+ and .. y= (26.) If there be nm unknown quantities and n independent ‘equations, the values of those quantities may be found in a similar manner. For from the n given equations, nm — 1 may be deduced, involving only n — 1 unknown quantities; and from these »—1, »—2 may be obtained, involving only n— 2 unknown quantities; and so on, till only one equation remains, involving one unknown quantity; which being found, the values of all the rest may be determined by substitution. (27.) If there be more unknown quantities than inde- pendent equations, some of these quantities cannot be found except in terms of the others; and by assuming values of these others, we may obtain an infinite number of correspond- ing values of the former quantities, which will satisfy the con- ditions proposed. See Sect. XI. But if there be fewer unknown quantities than independent equations, the values of the unknown quantities may be found from the different equations; and if these values be the same, some of the equations are unnecessary; if different, the equations are incongruous. SECTION III. On the Solution of Pure Quadratics, and others which may be solved without completing the Square. of the equation, the known quantities being transposed to the other, the simple unknown quantity will be determined by extracting the root. And by the same process, any equation containing the powers of a function of the unknown quantity, or containing the powers of two unknown quantities, may frequently be reduced to lower dimensions. | EXAMPLES. 1. Given 2’ — 17 = 130 — 22’, to find the values of z. By transposition, 32? = 147; Ramer ean Os and w= 7, answer the conditions required. Every pure quadratic ,*, will have two, and onl two, roots which are equal in magnitude, but different in Algebraical sign, Solution of Pure Quadratics, &c. 2. Given xv? + ab = 52”, to find the values of z. By transposition, ab = 42’; st Jf ab= 22, fab ——_—_= &. 2 attic: 38. Given ry =a and = = 6|’ Y | From the second equation, #7 = by. Substituting this value in the first equation, Cua a 3: a: . yas \t hi . a rec extracting the square root, y = + Ne 5? eye 7 4 4a, ao) Giver a toy Ta. 52 3] and ry = 6 J Sinceai-eey td a 305 +33 OS OOO PAVE Ree ine cee me Be eee 2h)a3 Ras Qa, and yi — =". 22" SMA | and 2 = 9; herefore, extracting the square root, 7 = + 3, | 22 whence ¥ = = + 94 | eo: Given 2 +y > 2@—yii3 51 and 2° — y*® = 56 cnd y. E 2 to find the values of x and y. - Substituting this value in the second equation, 5 to find the values 51 , to find the values of # and y. of x 52 a Solution of Pure Quadratics, &c. From the first equation, (Alg. 179, 6.) 20: 24114225 ee AIG: 1705-7.) eae 2 cel, and wv — 27. Substituting this value of # in the second equation, a sy? iS a4. y° a 565 Oly 00; Pehle SG) ll, whence 4—2y:— 4, : p Pye 6. Given vy =a l, to find the values of x and y. and 2? + y’ = sf To the second equation, adding twice the first, e+oermyt+y=s' +20’; *, extracting the square root, v + y = +4/ (s’ + 2a’); and from the second, subtracting twice the first, e—esyt+y=s'—20; *, extracting the square root, 7 — y = + 4/(s’ — 2a’); but # + y= +// (s* + 2a"); *, by addition, 27 = + ,/(s? + 2a”) + 4/ (s’? — 20’), and @ = + § {4/(s° + 20°) + o/( by subtraction, 2y = + 4/(s’? + 20’) = \/(s’ — 20’), and y = + 3 {4/(s* + 20°) — / ) i: be uphe ie sepals ‘} to find the values of w and y. (By, dlgs i790, 4, atid fs) es yen oe ets os LSE) Pa ety without completing the Square. 53 Substituting this value of x in the second equation, 6y" = 384; a aa 04: whencey= 4, RT Ma mat ft en A mmmcriven e+ yi: vit 7 : 5) andxy+y?=1296 f Ue SVC A OE Ue VRE *, (21) 27 = 54, and # = ky , to find the values of # and y. Substituting this value for z in the second equation, by” 2 ter a te 2 5D, OF U2 + Ue = 126 5 | Aye y’ = 36, | and aes | / 5 ch = -# 416. 9. Given (vw + y)*: (@—y)* 3: 6421] | and wy = 63 i of 2 and y. , to find the values (Alg. 179, 9.) ety: %@—Yyii8il; rm Algo ti 0s G20 2 Ue Sacer. ants AlgeV795i 72 ie ay yD aes * (21) 77 =9y, ! and 2 = 24, 7 Substituting this for # in the second equation, iY: Bis 7 re 54 Solution of Pure Quadratics, &c. . ¥? = 49, and y= 7; 5h peed elk 7 TU eels camer 2s 1a Tod tie wales at ene and y’? + vy = 24) Adding the two equations together, v?+2ryt+y’ = 363 *, extracting the square root, v+y= +6. Now # + ay=a@.(¢+y)=+ 628; Peery i aoe YP iG ee Se and therefore:y'="-s 6/453 = -F 4. ll. Givene +y= s| and a? — y? = d?{’ Since? —y’=(a@+y).(@—y)=s.(@—y); . §.(#—y) =a’, to find the values of x and y. rg te and #—y ae bute +y=s; 2 2 2 *. by addition, 22 =s +2 =S+t©, 2 2 Sar te f.5 thin 28 2 2 2 and by subtraction, yas Cat, s? — d’ pinnae oie 12. Givene+y=s J, to find the values of # and y. and ry == without completing the Square. 55 Squaring the first equation, 2° +ery + y=s', and from the second, 4XY Se Ye *. by subtraction, 2 — 2vy + y= s’ — 4a’, and extracting the square root, # — y = +4/(s*’ — 4a’), but@e@+y=s; *. by addition, 24 =s+,/(s? — 4a? by subtraction, 2y=s7-4/(s?— 40’ » Ys 1S / (S — 4008, ameyeD @ + y rig |, to find the values of w# and y. and 2’? + 4? = @’ Squaring the first equation, a 4+-27y + y’? =s’, and doubling the second, 22° + 2y = 30 ; : .. by subtraction, 2? —2r7y + yw? = 20 — 3’, and extracting the square root, 7 —y = +4/ (2a — s’), bute +y=s; *, Dy addition,\ "2a = st ./ (20° — $*); and 7 = 4 {fs +,/(2a’— s’)t; also by subtraction, 2y=s += (/ (2a — s’); y=4isFe/ 2a’ —s°)}. | 14, Given /u + iV/y =5 | and /a—V/y =1 Adding the two equations, 24/7 = 6; \, to find the values of # and y. Bn oer and pu ecune the equations, 24/7 = 43 AY = % andy = 8. 56 Solution of Pure Quadratics, &c. , to find the values oP Tee 20° 15. Given w Gan 1 + we sp War ri of 2. (18. Cor. 1.) 2 4/ (a? + 2) + a? + 2’ = 20’; by transposition, v \/ (a + 2’) =a’ — 2’, and squaring both sides, aa’ + v = a* — 20°27" + 2; 3 dave Co, a’ and a’ = —; 3 a As v cal e J/: 2 2 16. Given / (440) (GS-8) =5, to find the values of x. : By transposition, 4/ = se a) Sh WA (iG — i) +b; os | squaring both sides, = + >= a — b+ 2b// ie — a) + 8; (a7, Cor.) pee aay G a a), b a” : ol aVv (5-2): ies squaring both sides, oe as Os ee Ue as fe and extracting the square root, + vad rs <5 TES Sas ae 5.6 Pe ho) 17. Given - + aoe = - to find the yalues of a. | i = without completing the Square. 57 ae. II S18 2 The given equation becomes “ +f (e —] by transposition, 4/ (& — 1) = = 2a b oo &] : a a’ *, squaring both sides, — — 1 = = — Lv b 2 and by transposition, = —l1= _ ; *-2ab—P?= a and extracting the square root, + 4/ (2a6 — 6’) =x. : 1, COE oe Ca | ec sy — x—y)| to find the values of # and ) 6 Y. and LY From the first equation subtracting twice the second; gf) "y LAMP ROR SR Tell 9 Nese ws (eee ea) ae Y) - : 7 (e— y) —J)Jand 4— y= 1; ae 7 — <1. ALC ye — bes .. by addition, 2 + 27y + y’ = 25, andaw+y=+5; Dita — ye 12 .. by addition, 227 == 6, 0r 4, and # = 3, or — 2, and by subtraction, 2y = 4, or — 6; “. ¥Y = 2, or — 3. 19. Given 2? — vy = 48y and vy — y? = a to find the values of # and y. 58 Solution of Pure Quadratics, &c. 48 Y Dividing the first equation by v7, v7 — y= +23 and the second by y, 7 —y = a . _ 48y _ 3a, seep tS =A 8 9) se and 16y° = 2; consequently, + 4y = w@; and first, suppose + 4y = 2; Sap eee = (8% = #4) oO os Ba (3 y=)sy = (42 = SY 123. Y= 43 ._ ©=4y = 16. But if = —4y, ~y=)—sy= (8% = *¥ =) — Ls sty, sy = (% ty oF 125 12 y= and @ = —4y=—— 20. Given a a =e <4, , to find the values of # and y. Diyiding the first bie by the second, perig 5 yt iia) y2 = 25 . Yeas : 42 = whence, from the second equation, ie =4V/r= 243 x *. SJ & = 6, and w = 36. without completing the Square. 59 2 —_— 21. Given Tt = ], to find the values of 2. e+e2t+ ai Clearing the equation of fractions, : 18 xv Bee dy Sea es .. by transposition, v7 + 24 =9 + =, or@.(e+2)=9. (142) =2.@49)5 ica ® x’ | and 2” = 9; | A eB | 99, Given / (7**) 42 (— ind the values of z. Multiplying the equation by ,/ (2 s 2), | L+a a ) x Miss RE rat a (eae oi teta/tay, & eee II 3 1) = Jt WA (z= 1) and squaring both sides, Ce ae) one ee oe pe ys JZ. ¥ (5 i) +2 Ls by transposition, 2 4/ {- : fC _ 1)} os 7 1 é wav -(E-3}- 8 \ squaring both sides, 5 ie Jno & — oe xr’ *. by transposition, + 5-1 = aR and 4ab — 40°? = 2’; *, extracting the square root, + 2,/ (ab — 0’) =a 35. Given 3a — a =y — y| to find the values of andy’? +2=4 | IE Reducing the first equation, stoi ve = and therefore (18. Cor. 2.) a = Y; late Eevee Substituting this value in the second equation, 447 = 4; Fra Rat Ci Sa ,y=t Y3. without completing the Square. 67 36. Given2z’? + ¥/aey = "t to find the values of x and y? + 7 /a2y = 18 and y. The first equation is z? x (a + y}) = 9, and the second, y? x (# + y}) = 18. Dividing the second by the first, 2 = 2. . y? = 22%, and 4 == 4a, Substituting this value in the first equation, a+ 404/427 =9, Orton =)02 {w= + 1s CONSCQUENL MAY — at) ot as + 2 37, Given caine Y, = &@ + 2xry | to find the values of x Vy > | and y. From the first equation, 2? + 2y? = 2 /y + 2xy); and x* — 2y* = 256 — 2 /y -, by transposition, 2? — x /y = 22) — 2y’, ora. («— fy) = 24). (x9 — VY) 2. t= 2y%s and x Vy = 2y"; *, adding these equals to the second equation, 2 av = 256, and therefore v = + 16; ae oy? = + 16, and yi= + 8; . y? =+ 2, and y = 4. 388. Given v’y + ry’ = a) 1 1 5%, to find the values of w and y. and —- + - = a ey IG F 2 68 Solution of Pure Quadratics, &c. From the second equation, 2 + y = “et; 30 but from the first, vy = . eV2+y Substituting this value in the other equation, 30 25 Hea wea? | Wy “Vt+y “. (2 + y)? = 25; and extracting the root, 7 + y= +5. 9 5 EE eles Let v+y=4+5; then vy = + 6; whence 2 + 2%y + y’? = 25, and 42VY Spurs .. by subtraction, v7? — 27y + y’? = 1, anda —y=+1; butz+y= 5; .. by addition, 2a = 6, or 4; and 7 = 3; or 2; by subtraction, 2y = 4, or 6; and y = 2, or 3. But if# + y =— 5, then vy =— 63 whence x + 27y + y’ = _ 25, and 4xvY = — 24; .. by subtraction, v7 —e2rvy+ y= 49; and extracting the root, v7 —y =+7; butz+y=—5; .. by addition, 27 = 2, or — 12; andw = 1, or — 6; by subtraction, 2y = — 12, or 2; and y = — 6, or 1. 39. Gi ‘ — Given f = sae , , to find the values of # and and a*y’? + ay’ = 12 | Dividing the second equation by the first, ry = 2. But 2’y + ay = («+ y).rvy = 6, “ 2.(@ + y) = 6, without completing the Square. 69 and # + y = 3; whence 2? + 27y + y’=9; but 4ry = 83 “. by subtraction, v’? — ary + y? = 13 and extracting the root, y-—y =+1; also v7 + yY =3;3 Ee by addition, 22.—-4, or 2; i OTN» and by subtraction, 2y = 2, or 4; yan Ores, 40. Given 2° —y*: (w@—y)> 3: 61: 1| Pe a ks bah | and vy = 320 J of v and y. | Since vw — y°: v& — 3x°y + 34y? — y' 33 61213 “. (Alg. 179, 4.) 3a°y — 3@y?: (vw — y)*® 2: 60:1, or 3@7y x (w— y): (@— yy) 3: 60:15 _ . (Alg. 179, 7.) 960 : (w — y)’ +: 60 : 1, dividing the first and second terms by xv — y; and 16: (# — y)? :: 1: 1, dividing the first and third terms dy 605 | . (w@— y)? = 16, and#w—y=+ 4. But since # — 27y + y’? = 16, and 4xy 250 .. by addition, 2 + 27y + y? = 1296; =e _ and extracting the root, # + y =+36; | butz7—y =+ 4; .. by addition, 27 =-+ 40, or + 32, anda ch 20for a6: 70 Solution of Pure Quadratics, &c. but by subtraction, 2y = + 32, or + 40, and Wy i—=f-201 6, OFir 20. 41. Given (2’ + y’) x (@+y)= ami to find the values and (%? — y”) x (7 — y) = 576J of # and y. From the first equation, 2° + #’y + vy’ + y° = 2336; and from the second, 2° — #’y — vy’ + y° = 576; *. by subtraction, ou’y + 2xy’ = 1760; adding this to the first equation, 2+ 3a°y + 3ay? + y® = 4096; *, extracting the cube root, 2 + y = 16, and 2r7y. (@% + y) = 1760, or 16 X 2@y = 1760, ap UE ane Now 2’ + 2ry + y’ = 256, and AY 220 *, by subtraction, 2 — 2vy + y? = 36, and therefore 7 —y =+6; bution 16's *, by addition, 24 = 22, or 10, S00) oie Ona but by subtraction, 2y = 10, or 22; UR pEsOLiLl« 42. Given a’ + y° = (« + y) y ; FAN to find the values of # and wy + vy? =4xy J and y. Dividing the second equation by vy,# + y = 4; without completing the Square. 71 24+ 3u7y + 32y + y® = 64. But from the first equation, z*° — wy —vy +y>= 0; .. by subtraction, 4a°y + 4vy’? = 64; ro) (af) ) ey 9 16; and vy = 4 But 2’ + ary + y’ = 16, and 4xy =i163 .. by subtraction, v7 —2xy + y? = 0; and extracting the root, v7 —y = 0; but chi —e .. by addition, 27 = 4; Ang was? 5 but by subtraction, 2y = 4; Y= 2 43. Given (a? + y?) x («© +y) = HoH to find the lues of x . ie aye: value and (7 —y*) x (#v’+y’)= and y. Dividing the second equation by the first, é bx @+y).@-y =. Again, dividing the first equation by this last, V+Y = 35 a—y . vty = 3@ — 3y3 .. by transposition, 4y = 22, and 2y = 2&3 5 2 whence (2? + y’) . (7 —y) =5y’ x y =, Olan iss and therefore v = 2y = 2. és ee 72 Solution of Pure Quadratics, &c. 44, Given (v7? — y’) x (rw — y) = 32y ye ar vy tHe and (wf — y') x (& values of # and y. Dividing the second equation by the first, (a? +y') x (w@+y) = sry, ore +a@ytayt+y =15ry; but from the first, 2° —a’y—ay+y= 32y;3 .. by addition, 2% + 2y° = isxvy; and # + y° = 92y; but by subtraction, 2a’°y + 2vy° = 12@y; *, dividing by 27y,7 + y =6;3 whence # + 3a°y +3ay? + y' = 216; but 2° + y= 9xry; .. by subtraction, 3a°y + 3vy =2.6—92ry, or3.(v~@+y).ry = 18xy = 216 — 92y;3 Wary = 216, ands egy =sia: Now a + 2a7y + y’ = 36, and 4xY aa vt “. by subtraction, 27 —2ry +y= 4 and extracting the root,w—y = +2 butrz+y = 6; .. by addition, 22 = 8, or 4; Mere, =| 4 Tas and by subtraction, 2y = 4, or 8; ay ae TA SECTION IV. Solution of Adfected Quadratics, involving only one unknown Quantity. _ (29.) Let the terms be arranged on one side of the equation, according to the dimensions of the unknown quantity, beginning with the highest; and (17) the known quantities be transposed to the other side; then, if the square of the unknown quantity has any coefficient, either positive or negative, let all the terms be divided by this coefficient (i3). If the square of | half the coefficient of the second term be now added (11) to © both sides of the equation*, that side which involves the un- known quantity will become a complete square; and (19) ex- ‘racting the square root on both sides of the equation, a simple 2quation will be obtained, from which the values of the unknown quantity may be determined. It may be observed, that all equations may be solved as yuadratics, by completing the squares, in which there are two erms involving the unknown quantity or any function of it, and the index of one is double that of the other. Thus, / n n S+pe=—90"—pr =r +e =a, 02 +ar=), 3n + az’ = 6, pa — px” = d, (4? + px t+ q)’ + (@ + pz + Q) ar, a. (a? + ax)? + bx. (a + ax) =d, are of the same form * This is called completing the square; and that a complete square is thus »btained may be easily proved. _ Let 2?+2axz be the proposed quantity on one side, when the terms are rranged according to the form prescribed above; and suppose y? = the quantity equisite to complete the square. Now the square of std=a?x2dx-+d?, where ‘is evident that four times the product of the extreme terms is equal to the square f the middle term; and therefore, in order that 2? + 2aa-+ y? may be a square, ay? must be equal to 4a?2?; therefore y? = a? = the square of half the coeffi- lent of the middle term. | | 74 Solution of Adfected Quadratics, as quadratics, and the value of the unknown quantity may be determined in the same manner. Many equations, also, in which more than one unknown quantity are involved, may in a similar manner be reduced to lower dimensions by completing the square, as a’y? + pry=q, (@+y)t+p.(@t+y’) =n Instances of this kind occur in the following EXAMPLES. 1. Given 2’ + 8x2 = 33, to find the values of 2. Completing the square, 2’ + 8x + 16 = 49; and extracting the root, v+4=+7; whence, by transposition, 2 = 3, or — 11*. 2. Given 2 + 642 + 4= 59, to find the values of x. By transposition, x? + 6% = 55; and completing the square, 2’ + 6v + 9 = 64; .. extracting the root,7+3=2+8; whence & = 5, or — 11. 3. Given x? — 87 + 10 = 19, to find the values of a. * A Quadratic Equation cannot have more than two distinct values of the un- known quantity which will satisfy it. For, if possible, let the equation ax? + pa + q = 0 have three distinct values of @, Viz..a, B, 73 then aa? + pa+q=0, ap? + pB+q=0, ay? +pytq=0. If the first be subtracted from the second, a.(a*— B’?)+p.(a—B)=9, wa. (a+f)+p=0; similarly a.(a?— y?)+p.(a—y)=0, and a.(a+y)+tp=—0. Subtracting this from the former, a.(6 — y) = 0; but a cannot be = 0; for then the proposed equation would not be a quadratic, .°, 83 —y=0, or B=y. A Quadratic Equation ,*, has not three distinct values of the unknown quantity, but it may have two. involving only one unknown Quantity. 75 By transposition, 2’? — sv = 9; and completing the square, 2? — 8x7 + 16 = 25; .. extracting the root, 7—4= +5, and # = 9, or — 1. 4. Given xz’ — 2px = 9g, to find the values of x. Completing the square, 2 —2pr + p?’=p’+q; extracting the root, z—p=+/(p’+4q); G2 =pt/(p +9): 5. Given 2’ + px = q, to find the values of z. p 2 Completing the square, #? + px + ean f +93 p extracting the root, x + f arly o + 7); 2 + 2 4 e=—Psy(Z ra 1) = ee 6. Given 2? — x + 3 = 45, to find the values of 2. By transposition, 2 — 7 = 42; : l Tee 16 and completing the square, x? — # + a as ; | 1 13 .. extracting the root, 7 — Ao and # = 7, or — 6. 7. Given 32° + 24” — 9 = 76, to find the values of x. 85 | a Clash: 2 By transposition and division, 2’ + a Gres me | 2 BDe rush pin250 ‘and completing the square, 2’ + set F ey fa ha, Ss j 1 | .. extracting the root, 7 + 7 i ve whence # = 5, or — ne 76 . Solution of Adfected Quadratics, ms NA AL 156 By transposition and division, 2? — — = sara 9) : 42 4 156 4 .. completing the square, 7? — — 4+ — = — + —= P 8 ‘4d A 5 uF 25 5 if 25 28 J Ou and extracting the root, 7 — piethig 26 consequently, 7 = 6, or — ra 9. Given ax’ — bx =c, to find the values of x. b c By dividing each term by a, x? — a eS 3 : b b? b? c 2 —_—_——_- p> = completing the square, aha y ane - 6? + 4ac a 402 3 : b extracting the root, g— — = + Vv (ot + 440) : 2a 2a 3 OE 7 (On 40) : ius 2a ; 10. Given 6a + Brae (is. Cor. 1.) 64° + 35 — 3% = 442; . by transposition, 62° — 474 = — 35; 4 35 and (18. Cor. 1.) 2° ae! 2=——; 6 6 therefore, completing the square, 4 47\? 2209 35 _ 136 gt eae —)= —— ahs 6 12 144 6 144 F 47 37 .. extracting the root, 7 — Pe 5 andy \=7, or a -= 44, to find the values of 2. involving only one unknown Quantity. 77 .4—— 1 Sey | Clearing the equation of fractions, ll. Given «2 — = 14, to find the values of x. 4u°+4”v7—14+0=>140 4 14; and therefore, by transposition, 42? — 9x7 = 28, and (18. Cor. 1.) 2? — =a = 7; 0 9 |? 81 529 _ .. completing the square, v7 —-#+-| = — = — ; | sare pk he Lig tks Odi gasormh Gey? | 9 23 | and extracting the root, # — iets 7 whence # = 4, or — a : 1121 — 4@ 12. Given 32 — =o ery old to find the values of z. (18. Cor. 1.) 347 — 1121 + 4% = 2235 .. by transposition, 3%” + 227 = 11215 2 1121 and (1s. Cor. 1.) 2 + a bearer © *, completing the square vst ga gee Hege ensS oa .* re puss? 3 9 3 9 Gaia : 1 58 and extracting the root, # + Pile am = 59 t= 195 O1/— : S—2 27 —11.,.£.—2 13. Given — ——_____ = — —_-,, to find the values | 2 %—3 6 of a. | Ie 12” — 66 | Multiplying by 6, 24 — 34” — oT ous 12” — 66 by transposition, 26 — 4” = ———_— | L—3° 5 72 Solution of Adfected Quadratics, | 6% — 33 | and (18. Cor. 2.) 13 — 27 = prea | “. 192 — 39 — 22° = 62 — 335 | .. changing signs, and transposing, 2x — 138% = — 6, 13 and 4° —— #7 = — 3; 2 and completing the square, ae ee ee 7 exist 60 lale | 43/49 5x38 wan] fan . 13 11 .. extracting the root, 7 — es + ae 1 es & = 6, Or -. | 2 | 34” — 3 3” — 6 14. Given 52 — TUR oma TI to find the values of x. (18. Cor. 1.) 1027 — 362 + 6 = 4x" — 124% 4+ 3x7 — 15x + 185 | .. by transposition, 32° — 9x7 = 12, and 2? — 3% = 4; : 5 .. completing the square, 2? — 3x4 + 7 =4+ > = = : : 3 5 and extracting the root, # — 7= =p | and # = 4, or — 1. - 16 100— 9x 15. Given Poe —— = 3, to find the values of z. (18. Cor. 1.) 644 — 100 + 9@ = 122’; whence, by transposition, 122° — 737 = — 100, | involving only one unknown Quantity. ind completing the square, 73 (ey 5329 100 529 ; —— ey ee a ee feed's fia\ a4 576 .. extracting the root, 7 — Zs a ayia 25 and # = 4, or —. 12 169 — 32 16. Given 3% — = 29, to find the values of 2. | Here 32” — 169 + 34 = 2927; .. by transposition, 32° — 26” = 169, sherefore, completing the square, > Se 3 Coen oo 9 ° 13 2 and extracting the root, 7 — 3 ot 5° 2x 17. Given 16 — = =" + 1, to find the values of a. Multiplying every term by - 24—-27= ns “. (17. Cor. 1.) and by transposition, ,. 6x “a+ 77 = 24 — — | 3 id completing the square, 80 Solution of Adfected Quadratics, . 3 18 .. extracting the root, # + tS + sat 21 and # = 3, or ——. | 106. 14—997 98 | 18. Given — — —— = —, to find the values ofz. x L 9 229° (is. Cor. 1) low — 144+ 27 = : 9 | 22.0" | aan (17. Cor. 1.) Lanes 12V = 14, | 54 63 and | iB Lig | a ine the square, | 2 on r) = 729 63 36 x’ eee erat 121 ty bem da Co | 27 6 | and extracting the root, 7 — oh ecard | 21 “. & = 3, or —. ll 19. Given — +1=10— ? to find the values of a. | 4 v— : : Clearing the equ:tion of fractions, | 627 —8 +27 —8 = 20% — 80— a’ + 62 — 8; | *, by transposition, 2 — 18% = — 72; | and completing the square, 2’ — 18@ + 81 = 81 —72=9; whence, extracting the root, v —-9=+3; | | and therefore v = 12, or 6. | su +4 30-28 7H — | tint | 20. Given == = “v—6 10 ne ye find the values of x. 300 — 204 xH— 6 Multiplying by 10, 6% + 8 — = 7" — 143 involving only one unknown Quantity. 81 ie 300 — 20x .*.. by transposition, 22 — # = —_____; L— 6 and 28a” — wv — 132 = 300 — 202; .. by transposition, and changing signs, a’ — 48% = — 432; completing the square, xv’ — 48% + 576 = 576 — 432 = 144; .. extracting the root, 7 — 24 =+ 12; KE x —— 30; or 12. : 3% — 10 6a” — 40 21. Given 37 — ———— = 2 + ———__,, to find the values 9— 2” 27 —1 of x. Multiplying by 22 — 1, 62? — 234 + 10 6a? — 32 — errr = 47 — 2 + 62” — 40, 2 — 238 +10 | 6 or 72 a+ 9 — 22 423 “. 632 — 142” + 6x? — 23H + 10 = 378 — 8423 by transposition, 1242 — 82° = 368, 31 and Woirne © = 46 5 ' .*, completing the square, 2 31 961 961 925 a ——— —_—_ = = — 3 | 2 16 16 16 | . 31 15 .. extracting the root, 2 — oe + azis 23 and therefore 7 = 378 4, 22. Given a ee ee = ie to find the values 25+ 7P 86565” + 112" SS I eT Multiplying by (5 + x), #@ + Fd qe se A hd multiplying by 11z — 8, } | 82 Solution of Adfected Quadratics, 329” x? — 280 11z* — sav + e aaa = 552 + 112’, 3292 + 77x" — 280 = 6325 6— 42x 470 + 11a”? — 40 or, dividing b Oe aaa Sate 6 — 4x sda “. 472 + 112? — 40 = 542 — 362’; by transposition, 47”° — 7# = 40; Ba 40 of — HS 3 47 47 and telat the square, +(2 a + 40 7569 xv — — ie = = — 5 mee 47-8336 : 7 87 and extracting the root, # — mi Ge 2 areper 40 1. C= 1, OF a7 : 90 27 90 23. Given — — —_ fi h 1 es Ee TG to find the values of x. toad a ) Dividing every term by 9, - Saas abs Toe “. (18. Cor. 1.) 10% + 30% + 20 — 3a” — 34 = 102” + 202; .. by transposition and (17. Cor. 1.) 3z7 — 7@” = 20, anid Sata ope ou 3 3 D ” 40 M20 49 289 completing the square, 2? — —# + — at P & Hl : 3 ihe 36's 36 36° and extracting the root, 7 — i= as a; 5 ara v= 4,0r — —. 3 envolving only one unknown Quantity. 83 ‘ ] 1 a 24. Given ——— + —=—— = Bed to find the values of 2. TO E44 8x Multiplying every term by a, for ig os = =; .“. (18. Cor. 1.) 8% + 32 + 87 — 24 = 92” + 94 — 1083 *, by transposition and (17. Cor. 1.) 92? — 72 = 116, 116 and 2? — Lh! = as 9 9 completing the pane : 116 49 4225 PP ham —_—_- = SO a ° ata 324 9 324 324 ” . 65 | .. extracting the root, x — oil + e ; g ) te aes xv a 4; or Se ar ae 9 ve — 1027 +141 “v’—b6e +9 (is. Cor. 1.) v — lov’ + 1 = a* — gw’ + 274% — 27; 25. Given = x — 3, to find the values of a. *, by transposition and (17. Cor. 1.) x? + 272 = 28; md completing the square, 729 841 etore + (2 TY = a5 + 228 eae 2 .. extracting the root, x + a = =, and # = 1, or — 28. 26. Given : ~ + ree 2-2, to find the values of 2. a Clearing the equation of fractions, | 20a? — 140” + 490 = 2034 — 292”; .. by transposition, 492° — 3434” = — 490, and 2? —7# =— 10; | G2 ~at 84: Solution of Adfected Quadratics, -, completing the square, v° —7# + ~ = = — 10 =} | ° re 3 and extracting the root, # — er +33 °. © = 5, Or 2. (aad a ae a 8x2 + 110 JepfaehiEs eg ee ee 0 Oe iven — ie Sis Clearing the equation of fractions, 7 — 127 =x — 8H — 1105 pore ety tre we We and completing the square, 121 = a + 4# + 45 .. extracting the root, + 11=2# + 2; | | | and therefore 7 = 9, or — 13. 28. Given «/(# +5) x «/(u + 12) = 12, to find the value of x. ) Squaring both sides, (# + 5) . (@ + 12) = 1445 or 2 + 172% + 60 = 144; .. by transposition, 2° + 17x” = 843 and completing the square, a+ ize t+ (e Panta alae 2 4 4 extracting the root, # + “ =t => and # = 4, or — 21. 29. Given ./ (x* — a’) = x — 3, to find the values of a. Cubing each side, v — a& = & — 3bu° + 30x — DB; ‘, by transposition, 3ba2? — 3b’# = @& — 8, 3 3 § a — b and 2 — b« = — 30. hi? involving only one unknown Quantity. 85 completing the square, 7 P a—-Bv WP adce—s xv? —bxet+ eae = : 3b gab 4 b (SS b° tract th CD ee ) extracting the root, # — > = + Tyr ae Magy & AN Gude U Bente | @ H i 2 ae 126 30. Given x” — mx" = p, to find the values of x. Completing the square, 2 2 2 2° — mer pp = pp ™ ald 4 4 4 2 extracting the root, 2” — ~ Seat ues ba) : » on MH of (m* + 4p) e a = = 2 7 i; aod a= (MEV (+40) 2 31. Given 4a ey 1-2 to find the values of 2. 4 + Sa Sx Clearing the equation of fractions, 2% + 2 /t=16—2; . by transposition, 37 + 24/a# = 16, 2 _ 16 anda +—-YWr#=—;3 2 3 A 2 — 1 16 1 4 completing the Square # + — Sa at ply 3 9 3 9 9 : - 1 7 and extracting the root, Sa + - Vas a Ve = 2, or — =, 64 and # = 4,.0r ak 86 Solution of Adfected Quadratics, Meet _ NA Ee to find the values of 2. a+ at Sa Sa im < 3 Clearing the equation of fractions, ax” + 6 /x=a — 2; by transposition, (a+ 1).%+6/zr= a’, 32... Given] +1? completing the square, b _ 6° a’ 6° 4a°+4a°+ 5° ty ee ag ety acl jen aeo iy a and extracting the root, = b MV ACUYaes Ura eee). Vat eens 2.(a@ +1) 3 —b+ f(s +40? 4+ Ny 2.(a + 1) : 33. Given 2° — 2/2 — 2 =0, to find the values of 2. Dividing by / a, we find #—2— fe =03 -. by transposition, 2 — Sa = 2; tics and completing the square, 7 —/x + - —~=2+ a 3 extracting the root, 4/7 — ; =i73 *. /r = 2, or— 1, ANG 4 Oral 34. Given (2° + / 2 =6 \/2, to find the values of #. Dividing by 2, 2 +a =6; 25 bie Ht *, completing the square, #* + # + [= 6 + ; and extracting the root, 7 + -=+ , and z = 2, or — 3. involving only one unknown Quantity. Ve Q OTe # 35. Given > = 223 + err to find the values of z. Multiplying by 2, and transposing, # — : a= =; ; and completing the square, 133°) 1a 400 On EE aie enter .. extracting the root, wa Forte : — fam =, — 1 and 4/z = 7, or —=; 361 “. & = 49, or —. 9 ae 1 36. Given irate Ena 0, to find the values of a. L—5 20 Clearing the equation of fractions, 12/7 —40-#+5=0; -. (17. Cor. 1.) #7 —12 fa = — 35, and completing the square, 2 — 124/z + 36 = 36 — 35 =1; . extracting the root, «/z7 —6= +1; . Va = 7, or 5, and # = 49, or 25. 37. Given v3 + x3 = 756, to find the values of z. 3025 > 4 : 1 1 Completing the square, # + #3 + Fim 756 + _= ; 1 55 and extracting the root, #3} + Ao + a3 x = 27, or — 28, XS = 3, or 4/ (— °, @ = 243, or (— 28)3. CAN 87 88 Solution of. Adfected Quadratics, 38. Given 2° — x} = 56, to find the values of a. 225 Completing the square, 2° — 23 +: - = 56 + ia = 5 1 1 *, extracting the root, 2 — Slee aah are and #7} = 8, or — 7; erie), OTe (97 13 and # = 4,or (—7)3 39. Given 323 + a3 = 3104, to find the values of z. eae, 1 3104 Dividing by 3, 23 + 3 mer and completing the square, pie ls i 1 3104 " 1 37249 3 — —_— => oe aoe ane S40 bn ae 3 36 36.7 : 1 03 .. extracting the root, v3 + pag = ayaa 97 whence #2 = 32, or — me 40. Given aa} + bx: = c, to find the values of x. Dividing by a, af + Sai = 2; and completing the square, ee ee bial c+ 4ac. 2 a 4 4a°... 4a’ 18 4a? 3 . _=+ 2 ‘, extracting the root, 73 xt + = eth WA ee sac) eC rs and xi = 2a involving only one unknown Quantity. 2 4 | 8 5 41, Given 323 — = = — 592, to find the values of x. 5X3 pn SU0) ies oy, Aas Sai — 592; 6 1184 and therefore xv} — us — : | Completing the square, | 6 9 1184 9 5929 | v— —-# 4+ —=— + — = 5 | te hs 5 25 Py tke : 3 77 *, extracting the root, vi — —- = + —; 5 5 74 - xX = 16, or — 5? 74\% and @.=s48,,0r = 75): . 42, Given #”— re 6, to find the values of w. | Completing the square, 2” — yaate a=a+; .. extracting the root, is Gi ela 120), and ae atk /(a’+ 0d); ee — Ot 4 / (a + 6)" ' 43. Given a’x’? — bx =c, to find the values of 2. The equation is the same as a’x’ — eS eae : OE een 6? b? . completing the square, a’ a’ — at iter agen” Cie ts 90 Solution of Adfected Quadratics, 2 extracting the root, aw — 2 = af: (c + aol} 2a 4a b+ (sae + 8’) | ae 2 *, AL 2a | snide ke a 2a | : = 21 | : = —_-——_,, tofi | 44, Given s/(27+1) +2V%a@ Silent) to find | values of 2. (1s. Cor. 1.) 2@ + 1+ 2/ (227 + x) = 215 | .. by transposition, 2 4/ (227 + x) = 20 — 225 | and therefore 4/ (22° + 2) = 10 — 2; | *, squaring both sides, 2v7 + x = 100 — 20% + 2°; | and transposing, 2? + 21” = 100; | completing the square, , oty 441 841 xv + 212 + (—) = 00 + — = —; 2 4 4 21 29 *, extracting the root, v + os t>3 °, @ = 4, or — 25. 74+ 52 Algae 45. Given 24/ (v7 — @) +34/ (24) = the values of x. (1s. Cor. 1.) 247 — 2a + 34/ (2a — 2axv%) = 7a + 52; by transposition, 3 4/ (2%° — 2ax) = 9a + 323 2. of (22 — 2a”) = 3a4+ 2; squaring both sides, 22° — 2a” = 9a’ + 6ax@ + 2°; by transposition, v° — sax = 9a’; completing the square, w* — 8axv + 16a° = 25a"; extracting the root, 7 — 4a =+ 5a; *, @ = 9a, or — a. envolving only one unknown Quantity. 91 46. Given ; 24/ (a + 60x? + 9x + 540) + 89 “2+ 60 = Vv ( J+ 7 (w + 9) V/ (@ + 60) + YY (a + 9) ) to find the values of zx. | Clearing the equation of fractions, 2+60+2°+9+2// (x? + 60x + 9” + 540) = 2f (v* + 6ox® + 9” + 540) + 89; .. by transposition, x? + 7 = 20; and completing the square, v7 + # + . = 20 + “ ae : l 9 *, extracting the root, # + sats, and # = 4, or — 5. 47. Given W3+ 41/2 0 S/ar+4e 2a Beer 3a . 62a) 3 — Say to find the values of zx. This equation is 41.3+%2) 4-(5fe+a) _ 20° Mmyo—r 3—/xr (Gi Dee G4, and therefore, clearing it of fractions, | 41.(9— 2%) = 4. (25% — wv’) — 22’, or 369 — 41”2 = 100% — 42” — 24’; .. by transposition, 62° — 1414 = — 369, 47 123 and # — —# = ——; 2 2 completing the square, a 47 i 1) = 2209 123 1225 : aie Ea 16 2 ioe . 4 35 .. extracting the root, 7 — “ — + 41 and # = ZF 3. 92 Solution of Adfected Quadratics, . ae f & h | 48. Given Sx Eb wi (a x x) Ne = ee (a Pd x) ra to find the values of x. Multiplying the equation by i/at+/(a—a)}.{/e-—V (a—a)}3 TAG Siem) Tins oA (Gh) ee =. (20-0), or 2a /F= j=. (0 a) * 22° = 26H — ad, and 2’ —be=—%, completing the square, 6? GP ab && —2ab LAE mrad sa, He Dace Meola dE extracting the root, 7 — 7 — + Vv (6° — 246) : — 3 2 ays b+ f/(P — 2a6) | at Jf (# Sa lhe 49. Given ~ a ras = (7 — 2)”, to find the valg® of 2. Multiplying the numerator and the denominator of th fraction by x + (/ (wv — 9), fe+V@—9P yyy, 9 3 2 — extracting the square root, Ba EE + (@ — 2); taking the positive sign, 7 + (/ (@ — 9) = 3% — 6; by transposition, 4/ (v? — 9) = 2v — 6, and squaring both sides, #7 — 9 = 44° — 24a + 36; involving only one unknown Quantity. 93 by transposition, 327 — 247 = — 45, and a — 87 = — 15; completing the square, v” — 8% + 16=1; extracting the root, v7—4=+1; eb, DES, rs x? — But if rea B34 (od taal U+ f/f (2 —9) =— 344 6, and 4/ (vw — 9) =— 4% + 6; .. Squaring both sides, x’? — 9 = 16x” — 48v + 36, and by transposition, 152° — 48% = — 45; 16 of — =. = — 33 9 and completing the square, ar 1G 64 64 — 11 V—-—.4#4+—>— — : 5 To pe 25 : ag extracting the root, #7 — ao Va tee) ; gtk Sie py ee ay bolt) 11) 5 50. Given w + 5 = 4/(w + 5) + 6, to find the values of z. L - By transposition, (@ + 5) — \/ (vw + 5) = 6; | _and therefore, completing the square, (t+ 5)— Vf (@ + 5) + extracting the root, 4/ (#@ + 5) — 5 1 5 = + 2 2 "hia /e (Ge te 5). = 85.08 2s and squaring both sides, # + 5 = 9, or 4; whence #2 = 4, or — 1. 94, Solution of Adfected Quadratics, 51. Given # + 16—74/ (e@ + 16) = 10—44/(# + 16), to find the values of x. By transposition, (v + 16) — 34/ (@ + 16) = 10; and completing the square, (w + 16) —34/(w + 16) += = 10 + *, extracting the root, 4/ (# + 16) — and «/ (#@ + 16) = 5, or — 25 whence # + 16 = 25, or 43 and # = 9, or — 12. 52. Given \/ (w@ +12) + 4/ (@ + 12) = 6, to find the values Ole, Completing the square, Y(@#t+i2)+Y%(e4+ 12) +-=64+ 3 4 *, extracting the root, ¥/ (wv + 12) + ~ = °, and V/ (@ + 12) = 2, or — 35 whence # + 12 = 16, or 81; and # = 4, or 69. 53. Given 2? — 2% + 64/(x* — 27 + 5) = 11, to find the values of @. Adding 5 to each side of the equation, (vz? — 22 + 5) + 64/ (2? — 24 + 5) = 16; *, completing the square, (7? —2”7 + 5) + 64/ (2? —24¥+5)+9=164+9= 25; and extracting the root, \/ (v7? — 2% +5) +3=2+5, and 4/ (v” — 24 + 5) = 2, or — 8; *, squaring both sides, v7? — 2x + 5 = 4, or 64; 2 whence 2? — 2¥ + 1 = 0, or 60; involving only one unknown Quantity. 95 and extracting the root, v — 1 = 0, or + /60;* “ @=1,0rl142V/ 15. 54. Given 2a? + 3% — 54/ (2” + 32 + 9) + 3 =0, to find the values of z. | Adding 6 to each side, 22 + 37 + 9 — 5 // (24° + 34 + 9) =6; and completing the square, as aS © Qa" + 3049) — 5 (20° +32 +9) +— =64 . extracting the root, 4 (22? + 32 + 9) — : = + and 4/ (22° + 32 + 9) = 6, or — 1, suppose the value to be 6, then 22’ + 3a + 9 = 36, 3 27 and 2? + —-7 = —-; 2 ag = —— / | .. completing the square, | 27 9 225 3 9 ae’ —W Se —_—- = — * a TEST 5 i 161? | and extracting the root, @ +- ~ = 9 se v2 = 3, or — -. 2 But if — 1 be taken, 4/ (27 + 34 + 9) =—13 "20 43a - O'='T, 3 and a + >@=— 4; completing the square, 9 — 55. 3 9 PD Pl Neh Sere ey ent 4 So eae Sir lancteny, 16 : 1604 | | : * In this example, if 0 be the value, the two roots of the equation re+land+1, Solution of Adfected Quadratics, L) highs) ees 96 and extracting the root, # + = ENS SI ene 4 > eo Gv@rero— 2) J (2 aay PONG AS ii 9's J (e+ @ + 6) 55. Given to find the values opie #. (1s. Cor. 1.) (@? + 4+ 6) = 54—4)/(e? +746) +6; by transposition, (# + # + 6) + 44/ (2? + #2 + 6) = 60; and completing the square, (wt tat 6) +4 (2° +746) +4= 64; extracting the root, \/ (#” +a7+6)+2=>+8, and 4/ (x? + # + 6) = 6, or — 10; suppose the former, then, squaring both sides, # + x2 + 6 = 36; and by transposition, z? + # = 30; a completing the square, 2 + x +- tigghe N ake a 1 11 extracting the root, # + Acs Said and # = 5, or — 6. But taking 4/ (wv + # + 6) =—10, then # +2+4+6= 100; and by transposition, 2? + # = 94; : 377 completing the square, 2 + a + 7 cet eee + extracting the root, v + ' = ee —1+/ (377) and #7 = 2 a involving only one unknown Quantity. 97 56. Given {(7 — 2)? — x? — (v—2)’ = 83 — (w—2), to find the values of z. By transposition, {(# — 2)? — x}? — {(v and completing the square, | 2 1 i(w — 2)’ ~ x}? — {(w—2)'— a} + = 9 +-= =, — 2)’ — wt = 90; extracting the root, (# — 2)? se it 2 —_— v&—-- = | and (vw — 2)? —x# = 10, or — 9; ‘whence, adding 2 to each side, ( — 2)’ — (v — 2) = 12, or — 7; supposing the former, and completing the square, (@—2)*— (w— 2) +—= : 1 extracting the root, # — 2 — = +4 whence wv = — Lian 1% or —1; amare and in the second case, where (v — 2)? — (x — 2) = —7; completing the square, ae eae (@ — 2)’ —(w—2) +2 = 1-72 =, extracting the root, 2 — 2 — : = See 3 : oe A Seo) _ 57. Given (w + 6)? + 2x3. (w + 6) = 138 + x, to find the alues of x. Completing the square, (e+ 6)? +2”). (7 +6) +e7= 138+ 2+ 2); val 8 ¥ f : extracting the root, # + 6 + #2 = +4/(138 + @+2)3_ 98 Solution of Adfected Quadratics, and squaring both sides, (vw + #2)? + ive (v + #2) +36=138 +2423; . by transposition, (v + v2)? + 11. (@ + a2) = 1023 - completing the square, 121 121 529 (w + wt)? + 11. (@ + wt) + ——= 102 + = ae : i 23 extracting the root, 7 + #2 +—=+-—, 2 andw+a2= 6, or — 17; supposing the former, and completing the square, 2 + #2 + = 6+7= 73m extracting the root, #2 + and a2 = 2, or — 3; \. @= 4, Or 9. But if@ +a =—1%, i — 67 4 4 + / (= 67) and extracting the root, #2 + ~ oe and ot === VK), — 33> Vv (—67) 2 completing the square, v + # + “ = whence 7= 58. Givenag —1=2+ ta to find the values of z. 2? Since v — 1 = (#2 + 1) x (#3 — 1); i Ras esahae es en vss ere 1 he} involving only one unknown Quantity. er 2 and therefore, dividing by #3 + 1,43 -1=—; L2 whence v — #3 = 2; : 1 1 1 9 and completing the square, # — #2 + het es : 1 1 3 .. extracting the root, #2 — a aa > 1 and #2 = 2, or — 1; whence # = 4, or 1. 59. Given 2* — 22° + # = 132, to find the values of x. (15) Adding and subtracting x’, there results vi —20 + x — (@ — xv) = 132; and completing the square, (@? — x)’ — (x — 2) + == 132 +—= . 1 extracting the root, 2° — x — z= whence 2 — # = 12, or — 11; supposing the former ; then, completing the square, v? — # + =12 + - = - : 1 (s and extracting the root, v — cg eae and # = 4, or — 3. But if # — v7 =— 11, 1b eyed 43 completing the square, x? — v + iar ye oe . bgt — 43 extracting the root, # — as avatar} and # = 1 (— 43) Aber) H 2 9 99 100 Solution of Adfected Quadratics, 60. Given (x? + 2x). (vw + 4) = 2— (w+ 4), to find the values of 2. By subtraction, (a + 2#) .(v + 4) = — (@ + 2)3 *, dividing by x +2,27.(%¢+4)=—1; Nera i) gt Pn and completing the square, 27 + 47 +4=4—1=3; extracting the root, 7 +2=2+ V3; ~@=—2t+ V3. 24 33 25 V/(5— 2) 34 6l. Given J (5a — 2") J; ies Tee to find the values of 2. Multiplying every term by 25 4/ (52° — 2"); ", 849 +5 — a” = 8504/ (5 — 2’); | and by transposition, (5 — x?) — 8504/ (5 — a”) = — 849; _ completing the square, | (5 — x) — 8504/ (5 — x”) + (425)? = 180625 — 849 = 1797763 | and extracting the root, 4/ (5 — a#’) — 425 = + 424; whence 4/ (5 — 2’) = 849, or 1, and 5 — 2 = 720801, or 1, and # = — 720796, or 4; *. @ = 4/ (— 720796), or +2. 62. Given , to find the values of 2. a—wx Lv b +. — ee PRL Fs a—2 Multiplying every term by ——; A (=*) b a—x + 1 ——. 3 a Cue involving only one unknown Quantity. 101 Ae — 27's Tb a— | by transpositio (Ce a ee : bi Heme Nats Mega 1; completing the square, =) ba—«e 8 . B# b? — 4c” mee ie GON Tae Cee ae 2 Le ge 4G 9 ac 4€ 1a pede) pore: extracting the root, Coie erg + VAC & 26 2c _ 2¢+b+/ (8? — 4c’). b oa 2a¢ ~ 2¢+644/ (0 — 40) _ 63. Given 9% + 4/ (16x? + 362°) = 152%” — 4, to find the values of a. By transposition, 97 + 4 + 27 4/ (9@ + 4) = 152"; and completing the square, : (97 + 4) + 2@4/ (97 + 4) + 2 = 162"; and extracting the root, \/ (9v + 4) + v= 42; “. «/ (92 + 4) = 3a, or — 52, and 97 + 4 = 92’, or 252; supposing the former ; then, by transposition, 92° — 97 = 4; | | 9 9 : | and completing the square, 92° — 9” + = 4+ Piaget : 3 3 | .. extracting the root, 37 — pi +33 | and 34 = 4, or — 13 4 1 whence # = -, or — -. 3 3 102 Solution of Adfected Quadratics, But if 9@ + 4 = 252’, then, by transposition, 252° — 9” = 4; and completing the square, and extracting the root, 5% — 10 10 ee 2 Ae) 10 alt ahd Yee v a5), 50 ; 12 + 8x2 64. Given 7 = eee to find the values of x. (1s. Cor. 1.) v — 5% = 12 + 822, or a2’ —47=12+ sri 4a; and completing the square, 2 —47 +4=16 + 8#34 2; extracting the root, 2 —2= + (4 + #2), and first taking the positive value ; then, by transposition, z — v3 = 6; completing the square, a — v3 + - =6+ 7 a =; extracting the root, 2? — ; = ; Ry Se 3, OF — 2, and w@ =)9,,0r/4- But if the negative value be used, 7 — 2 = — 4 — a; .. by transposition, v + #2 = — 2; and completing the square, @ + gr + ee Wa tad 2 = —}3 ; 4-4 4 involving only one unknown Quantity. 103 a ase extracting the root, #3 + : = ee , —-1tY(— papi Saal eA Me 2 arid ee 2 65. Given a + I —49=9 +, to find the values of x. Adding 4 to each side, in order to complete the square, 49 2 4 then : 49 z l extracting the root, 2 — mb (: at x), and first taking the positive value ; ) nie ries 8 .. by transposition, om Heed 3 62 16 and therefore, 2’ — ia completing the square, G2 0 ue tos 9 121 OT! at eae Sy 7 49 7 49 49 ” ; 3 1] | extracting the root, v — x + ot 8 and w = 2, or — 7. "2 7 te But if the negative value be used, Lae p ee ote 104 Solution of Adfected Quadratics, ae 6 .. by transposition, re +3=-, 62x 12 and x + Baers. 4) *, completing the square, 9 93 —— —_— “a+ 2 GP ame rake am vt rs 7 Gee 3 and extracting the root, 2 + - = v. te 3 _2= 2 ss (93) 66. Given & a se — 17x = 8, to find the values of 2. 3 Multiplying by 2, #* + a2 — 34” = 16, ee Var *, by transposition, a* + ae mR Wan eet hc and completing the square, — + (2) = (=) FB et extracting the root, 2’ + we = ct CS + a) first, let the positive value be taken ; then, by transposition, x’ = 4, Vath 7 iS sh P But if the negative value be taken, involving only one unknown Quantity. 105 and by transposition, # + a =— 4; ., completing the square, 172 @ . 289 z+ — —) =— -4= : ay 2 4 16 T6ec and extracting the root, # + 1 ne Apia Oak OL es oa 2 | 4 1 232 1 67. Given 272° =, “ aestiergeet ee oem n furs to find the values of 2. Multiplying every term by 3, Ree cea 232 1 81v— = 17 = —_ - = - 153 v x We ae 1 841 232 .. by transposition, 812’ + 17 + a om ge 3S era Adding unity to each side, in order to complete the square ; 1 841 239 + 165 2 and extracting the root, 97 + . == ( ) Let the positive value be taken ; then, by transposition, 97 — 4 = i and therefore 927 — 4% = 28; 404 256 completing the square, 9”? — 4x¥ + Gams De omen ONE 2 16 extracting the root, 37 — a Sieh 106 Solution of Adfected Quadratics, 14 1 and # = 2, or ——~. But if the negative value be taken, 9x? + 4” = — 30; and completing the square, : she x extracting the root, 3a + ag gos A aoa 3 3 ops ees 2 -+4/ (— 266) 3 any ee ee LR) © 68. Given (# oe ay (« 2a values of 2. ar 4\ 4 L By transposition, (2 aa 7) whe and squaring both sides, . involving only one unknown Quantity. 107 4 4 .. by transposition, 7 = aaa and av = 4/ (a — a’). Squaring both sides, a’a’? = a — a'; .. by transposition, z* — a’?a2’? =a‘; and completing the square, x’ a’? x ae a" ph. a’ | a el 5a ° 4 4 ay 2 aie 708 : a Pete /A00, 0 extracting the root, 2? —— = zeal — peta aaa r ah anda = ct a. / (12), Pe a A a SECTION V. Solution of Adfected Quadratics, involving two unknown Quantities. (30.) Ir the equations involve two unknown quantities, they may, by the preceding rules, be reduced to one con- taining only one of the unknown quantities, the values of which may be found by Art. 29; whence, by substitution, the values of the other may also be determined. In many cases, however, it may be convenient to solve the equations first, considering one of the quantities as known; when the rules for exterminating unknown quantities (23) may be more easily applied. EXAMPLES. 1. Given # — y = 15} ‘og zy i } to find the values of 2 and y. From the second equation, 7 = 27’; Substituting this in the first, 2y° — y = 15; and completing the square, y—sytaest maa 2 1 2 16 tees Solution of Adfected Quadratics, &c. *, extracting the root, y — - — + — 5 and y = 3, or — — 2? 2 25 whence # ='247 = 18, or ris 10%+y 2. Given Y = 3 and 9y — 9” = 18 From the second equation, y — 7 = 2; RK ll 4° r to find the values of # and y. and therefore y = @ + 2; but from the first, 107 + y = 32y. Substituting in this the value of y found above, 107 +2+2= 32. (@ + 2), orllv@7+2= 32? + 62; .. by transposition, 32° — 54% = 2, 5 2 d (18 Pei pet ey and (18) # Plog ~. completing the square, ° 5 “V—-.e4+-—= 3 : 5 .. extracting the root, v7 — aia + 25 36 25 2 36483 1 and v = 2, th oe a whence y = # + 2= 4, or =. b) 109 3. Given’ +y:x#%—y:: 13:5] to find the values of and y’? + v7 = 25 fi and y¥. 110 Solution of Adfected Quadratics, (AIG ONG.) Oar 212 toe reneis Sh Be Ha hog Wd ean Oe) eae anaes Poo . (21) 42 = 9y, and 7 = ot, Substituting this value in the second equation, 9 y+ ma = 9253 completing the square, iby easly een esy ioe 7 ae Scan ae aie * 9 41 *, extracting the root, y + amoare: 25 whence y = 4, or — ret * ie ae i fi i ia to find the values of # and y. From the first equation, w’y’? + 4vy = 96; ., completing the square, v’y’ + 4@y + 4 = 100; and extracting the root, vy +2 = + 10; ad) tak Now, squaring the second equation, a + ery + y? = 365 but 4ay = 32, or — 48; .. by subtraction, v7 —2a7y + y? = 4,or 84; whence, extracting the root, #— y =+ 20r+/ 84; but ¢ + y= 76; .. by addition, 27 = 8, or 4, or 6427 213 Bree. - he involving two unknown Quantities. 111 whence # = 4, or 2, or3+ 4/21; and by subtraction, 2y = 4, or 8, or 6 24/21; eer 92 OF 4,01. 3.42) \% 21. 5. Given 2 + y” = 2a | to find the values of x and y. and ry =C = From the second equation, y = ; and substituting this value in the first equation, 2n n Ps TN rece N x + a” a 2 a 9 hy an + c2n — 2a" xe" s by transposition, 2°” — 2a” 2” = — ¢"; completing the square, 7?” — 2a"x” + a?” =a" — ¢"; Be extracting the square root, 7” — a” =+ / (a” sane oo) : anda? —= af - andy. By transposition, # + y’ + 27+ y= 183 and from the second equation, 2Ly —= 125. .. by addition, 2? + 2avy+y+2+y=30; and completing the square, =30+-=—; 1 Baw + et y).+ 4 Lom Spr RT ey 112 Solution of Adfected Quadratics, 1 11 .. extracting the root, v + y + = + — 2 2 and # + y = 5, or — 6; whence, from the first equation, #* + y? = 13, or 24; buti2ey.—=19= .. by subtraction, v7? — 2vy + y’ = 1, or 12; .e@—yeztilorteV3. Nowa +y =5,or— 6; .. by addition, 27 = 6, or 4, or —6 42/3; 2 ” = 3,0r2,0or—3+ V3; and by subtraction, 2y = 4, or 6, or — 6 = 2 \/ 3: “y= 2, or 3,0or—3 7 V3. 7. Given # + 2vy + y’ + 2y = 120 — 24 and vy — y = 8 \, to find the values of # and y. By transposition, (w + y)? +2. (@ + y) = 120; *, completing the square, (v7 + y)? +2.(@+y) +1=1213, .. extracting the root, (v+y)+1=#N1, and # + y = 10, or — 12; and first let 7 + y = 10; from the second equation, 7 — y = * .. by subtraction, 2y = 10— 7p oie y= 5y — 43 and by transposition, y* — 5y = — 4; completing the square, y’ — 5y +- = = = cae eae 5 and extracting the root, y — : = =; envolving two unknown Quantities. 113 oe Y=4,0rl1; and # = 10 — y = 6, or 9. But ife~#+y=—12 3 andw—y ==; then 2y = — 12 i and y’? + 6y =— 4; *. completing the square, y? + 6y¥+9=9—4=5; extracting the root, y +3=+05; eg ee, Ss V5, and #=— 12—-y=—9¢ V5. | 8 Givene’?+y7—2-—y= st to find the values of x 3 and vy + 47+ y = 39 and y. Since 2? + y? — (@ + y) = 783 and from the second, 2vy +2. (%+y) =783 .. by addition, # +27y+y?+xe2+y = 156; ind completing the square, 2 1 — (@+y+@ty) +7 cae 1 25 extracting the root, # + y + te mire * & + y = 12, or — 13; supposing the former ; then vy = 39 — (v@ + y) = 39 — 12 = 27, and 2 + y?=78 + (v+ y) =78 + 12 = 90; but 22y = 543 .. by subtraction, v— wy t+y = 36; i 114 Solution of Adfected Quadratics, and extracting the root, 7 —y =+ 6; but@+ y= 12; .. by addition, e27= 18, or 6, and iss peo, Olde. and by subtraction, 2y= 6, or 18, | and Y= 923; Oreo. } But if ¢ + y= — 13, then vy, —39 +13 = 52; and 2 + y? =78 —13 = 65; but 22y = A0As .. by subtraction, z? — 2a7y + y? = — 39; and extracting the root, 7 — y = + 4/(—39); , but #7 + y=— 13; | | “. by addition, 27 =—13+44/(— 39); gud OS ee but by subtraction, 2y = — 13 = 4/ (— 39); : _ — 13 Ff (— 39). | er 9 > | 9. Given w’y* — 7xy’ — 945 = 765) to find the values of and wy — y = 12 gz and y. From the first equation, by transposition, x’y' — 7xy’ = 1710; and completing the square, 6889 2 49 49 x’ ee — —_ = y Tay + : 1710 + ' rine 3 extracting the root, vy’? — i no = . @y = 45, or — 39. involving two unknown Quantities. 115 Multiplying the second equation by y, vy? — y? = 12y,. Substituting in this the value of xy’ found above, 45 — y’ = 12y, in one case; and by transposition, y? + 12y = 45; completing the square, y’ + 12y + 36=45 + 36=81; extracting the root, y + 6= +9, and y = 3, or — 153 whence # = al = 5, or; y 5 and in the other case, — 38 — y? = 124; whence y’ + 12y = — 38; completing the square, y? + 12y + 36 = 36 — 38s = — 2; extracting the root, y +6=-+ (/ a .yY¥=—64+/-2, ee 8 — 38 S19 Veet 12. /* olen 64, 10. Given 7 —2V/ay+y—Sfat yaaa: Andihe and /2+/y=5 values of v and y. Completing the square in the first equation, = = aS (oot | (V2 — Vy) — (Va - Vy) +7 = 33 / and extracting the root, Ao Vy — . = +o; Va-V/y= 1, or 03 but from the second equation, «/” + \/y = 5; | by addition, 2 «/z = 6, or 5, | - 5 and «/a = 3, or = 5 | ! 25 | ". @ = 9 or ——5 12 116 Solution of Adfected Quadratics, but by subtraction, 2 «/y = 4, or 5, or \/y = 2, or >; 25 and y= 4, or wae . x’ 40 85 11. Given y * iyi a =| té“find the values of # and y. and 7 —y=2 Completing the square in the first equation, a A ato Spee: ease 9 11 and extracting the root, kes +2=+ wees and 7 = “s or 4, supposing the former ; | J 5 then, from the second equation, =2 —y=2, | or 247 == 6, and y= 3; 5 pa oe 3 And if the second value be taken, — ae —y=2, or — 204 = 6 = ie and # = — ak a +y to find the 128 .Gi (2 7) + (ZY )= hrs J L+Y “/ values of and vy — (w+y) = 54 | xv and y. aro E+, involving two unknown Quantities. 117 (is. Cor. 1.) and by transposition, 3 — 24/ (32) .6/(e + y) + (ety) =05 ", extracting the root, \/ (37) — 4/ (@# + y) =0; by transposition, 4/ (32) = «/ (vw + y); and squaring both sides, 37 = w + y, BNO kh 27 ==.7/; substituting this value in the second equation, 22” — 3” = 54, | 3 ore — aS oy: whence completing the square, 9 44] x’ = — 2 — = w) —_—_=—=as'= | eae tae cat rae ave . 21 extracting the root, 7 ——= + oe and « = 6, or — =; whence y = 24 = 12, or — 9. 13. Given v* — 2a’°y + y? = 49, | and v° — 22°y? + yt — #’ + y? = 20 values of # and y. |, tau fidethe Completing the square in the second equation, 2 ane 2 2 ee 1 _-. 81; ee eat Sy") be 20: | | x ] 9 | extracting the root, a — y’? — = + = ) “ @ —y’ = 5, or — 4; ‘but extracting the root] ” ° ays =< c , } of the first equation, J J | oes OD, 118 Solution of Adfected Quadratics, *, by subtraction, y’° — y = 2, or — 12, or 11, or — 3, Taking the first value, and completing the square, VE ee ee hed Meal PRB ene : 1 3 extracting the root, y — mf chs and y = 2, or — 1; 6 @=t/(7 + y) =+3, ort fo. Taking the second value, y’? — y = — 12; eerinlenne the square, y? — y + - = ‘i —i2= aa, extracting the root, y — - = sei ee) aude’ ce Nowe 7, whence # = /y—7)=+ / (EY C#__) 235 Uf(S eee 2) Taking the third value, y’? — y = 11; completing the square, y? — y + - =11+4 -= =; : a Hak. extracting the root, y — == fw ety Ae a y = 1t3V 5 5 2 and vatVcty=+/ (74 Oy 5) ie Jf (eae involving two unknown Quantities. 119 Taking the fourth value, y? — y = — 3; completing the square, y? — y + +o= _— Pup — ; extracting the root, y — — = = AS Ly andy = an es ie = tV/y—7) BY (229 )na/ (Ente) 3 or 14, Given vy + vy’ = =p to find the values of # and y. andy + vy? = 18 From the first equation, 2 = cn pehises, y-Q+y) and from the second, 7 = Bikes Ng 12 whence ——__——. d dividing by ——— my) 7p ividing by - naar pee a Ds Gale ot Rares . 2y°—2y74+2=3Y, and by eas. ay? — by = — 2, oy —-.y=— 2 eh a ect gr Sy completing the square, y’ — - FEM ey ater aan e rs ° 5 3 extracting the root, y — ae? 1 ae YY = 2, Spe 12 iis | hence 7 = ~ == 9, or 16, 120 Solution of Adfected Quadratics, 15. Givenw—#i=3—y ik to find the values of x and y. and 4—w@ =y—ys Adding the two equations together, 4 — r= 3— y; .. by transposition, yr = 2 — 1, and y = (# — 1)’. Substituting this value in the first equation, - @—evt=3—a@7 +20? — 15 fa *, by transposition, 27 — 3%? = 2, 3 and @— =. a= 1; leting th Sap ee pe eee completing e square, &@ Bie 1 = 16 = 16? Ge 5 extracting the root, #2 —--=+-, 4 4° 1 and #? = 2, or — -, 9 1 . : 9 and y = (#? — 1)’ =1, or -. 16:>- Given (27) 1))) ay a to find the values and («? +1).y=a°y?— 744)’ of wand y. Since quantities which are equal to the same, are equal to each other, uy? — 744 = LY + 1265 .. by transposition, 2’y? — vy = 870; completing the square, 2” A a leit cea na Y y or , ae involving two unknown Quantities. 121 : 1 5 extracting the root, vy — i — =, and #y = 30, or — 29; let the former value be taken, then from the first equation (a + 1). = = 156; APRS 1562 262 — Bah tan Bae? nd : 262 and by transposition, 7? — ae cats , 26 169 169 144 mereerine the square, 2?,.—— .¢ + ——— = —— — |] = =: P 8 4 : 5 zn 95 25 Di 2 13 1a I+ =| extracting the root, 7 — whence # = 5, or = S0:An aN aes | 7 5 OF 150. In the second case, (# + 1) x — = = — 29 + 126 = 97, 97 @ and 27 ++ 1= — ——_- a art b Rnait: 2, 972 y transposition, 2° + Foca G: | 97 97\? 9409 6045 complet h iat a) Sia a eh Al gs aah § et P aera g Sduare, 2", + sor i 58 3364 33647 | 97 / (6045) € t — = oe Ye | xtracting the root, # + Ty ool eee — 97 + / (6045) 4 62 5 58 29 1682 gat Sry one (0048). 122 Solution of Adfected Quadratics, 17. Givneg+y+/(e¢t+y) = 12) to find the values and #* + y° = 189) ofw#andy. | Completing the square in the first equation, 1 1 49 Pema gt AM Pea Bs wpe Bea ete extracting the root, 4/ (v + y) + 5 2 / (@ + y) =3, or — 4, and # + y = 9, or 16; w+ 3a’y + 3a2y’? + y? = 729, or 4096; 3 but 2 + ty = 189 .. by subtraction, 3z°y + 3ay° = 540, or 3907 ; 390 “ (2 +y)-xvy = 180, or ae 3907 . .. 9”@Y = 180 in one case, and l67y = Tae the other, whence in the first case vy = 20. Now 2 + 27y + y’? = 81; but AxY = 80; “. by subtraction, 2? — 2vy + y? = 15 and extracting the root, 7 —y=+1; buta+y= 93 . by addition, 24 = 10, or 8, Oil a os Oe but by subtraction, 2y = 8, or 10; ANG yed/ = 4 Ol eas : 390 Now in the second case, 167y = ar; . 42 —— 3907 ° ° Y ed 12 5 = involving two unknown Quantities. 1238 and since # + 2vy + y’? = 256, 3907 d 4 hh Sa ase and 427y ar : 2 ; 835 .. by subtraction, # — 2vy + y’ =— ae and extracting the root, 7 —y=+ oh ( = = ), butaz+y= 16; 94° — 835 .. by addition, 27 = 16 + yA ( a ), Oe 2 ae ot al aoe ; 4 3 but by subtraction, 2y = 16 += ah ( =e ), 1 — 835 and y = 8 = ~/ (==). 18. Given 2’ + y’ + #—y = +132] to find the values of and (x + y*).(v¢—y) = 1220)’ wandy. From the first equation, v’ + y’ = 132 — (vw — y); ; ee b320 and from the second, 2° + y* = ey? 1220 whence 132 — (ev — y) = poy? and, .*. 132.(7 — y) — (@ — y)’ = 1220; and (17. Cor. 1.) (e — y)? — 132. (# — y) = — 1220; completing the square, (27 — y)’ — 132. (@ — y) + (66)? = 4356 — 1220 = 3136; | extracting the root, v — y — 66 = + 56; 124 Solution of Adfected Quadratics, and # — y = 10, or 122; supposing the former, aay a "Nadie ey ea but vw — 2x2y + vy’ = 100; .. by subtraction, v? + 2vy + y’ = 144, and extracting the root, 7 + y ==+ 123 but v7— wy s==)105 .. by addition, 22 = 22, or — 2, and #=11,0r — 1; by subtraction, 2y = 2, or — 225 ANC ae == 1, OF —-aia. But if 2 — y = 122, then x + y’ = 10, and 2 — 2vy + y*? = (122)’5 Duta foo) 207 .. by subtraction, v? + 2vy + y® = 20 — (122)? and extracting the root, v + y ==+ 4/ {20 — (122)? but 7 — y = 122; .. by addition, 2v = 122 + 24/ {5 — (61) andw= 61+ 4/ (— 3716); — and by subtraction, 2y = — 122 + 24/ (— 3716)$ y= 61+ Y(— 3716). teen bi pip a , to find the values of # and y. From the first equation, 73 = 243; 1 and .. 5 = yh; involving two unknown Quantities. 125 substituting this value in the second equation, 8x — : a 14, 2 and 1643 — #7} = 28; or (17. Cor. 1.) #3 — 1623 = — 28; completing the square, v3 — 1623 + 64 = 64 — 28 = 36; and extracting the root, 7} -3s =+6; at @si—4 14, OF 2, and # = (14)* or 8; 1 1 2 but yi = — #3} = 98, or 2; 2 Beye cay. OF 44 90, Gi n xv a — eel ive | ae bs I to find the values of x and y. and #i+ y3= @ Squaring the second equation, .. v + 2a:y3 + y3= 2’; but #3 + Yi = 323 .. by subtraction, 2 — #3 + 2Viyt = a2 — 32; but from the second equation, y} = # — #3. Let this value be substituted in the preceding equation, then # — #3 4+ 2@3 —2v7 = 2’ — 32; and by transposition, 27 = 2” — #3; and dividing by 7,2 =w# — a; completing the square, # — #2 + Toe aan , ase 3 extracting the root, 7: —--=+-, 2 2 Sid ah 9, OFe— 1; 126 Solution of Adfected Quadratics, *’., ©=4, orl, and ¥3 = #@ — ‘vi = 23 Sey ee 21. Givenev + a3= yryr? + al, to find the values of [ thea pa ered xv and y. From the first equation, #3 + 7—4%) =y’+y+42, and from the second, w = y + 3. Substituting this value for x in the former, etyt3—-1m=yt+yt2, and by transposition, v3 —47i = y? — 1. But sinceev =y +33 ..e%—-4=y —1, by which equation let the preceding one be divided ; Set wl Ss squaring both sides of this equation, 7 = y’? + 2y + 1. Kquating therefore the two values of z. yt 2ytl=yt 3; .. by transposition, y? + y = 23 completing the square, y? + y + . —=2+ “ = ; | . 1 3 extracting the root, y + al + s ve Uf =o 1,.0O0 — 2° | whence #3 = y + 1 = 2, or — 1. and therefore x = 4, or 1. : x? 2 22. Given ie a i at: ; = 6= oe ¥; to find the values of | it ya) ey | x and y. YY : Sey: 2 By transposition, — + 2; 5 an 2 =; involving two unknown Quantities. 127 | : : : “. adding 2 to each side, — + 2+ es Ssh a eet y yey ue | completing the square, ¢ + uy 2 (< a i) ae extracting the root, e+ 4¥4seH Ue ek 2 now from the second equation squared, vty? = 22y + 45 521 124 Pa 2LY tan, or — “4 ‘whence by multiplication and transposition, 8 vy = 8,0r— — 3 and since a — 27y + ¥” 4, 32 and 4ny 32, Or — —3 11 ° . 2 9 12 .. by addition, 2? + 27y + y’ = 36, OTe; | | : and extracting the root, wv + y = + 6, or + : | | buta—y= 3; | he 2 | .. by addition, 27 = 8, or — 4, or 2+ and # =: 4, 0r — 2, or] 128 Solution of Adfected Quadratics, : AE *. by subtraction, 27 = 4, or — 8, or — 2+ Vv 11 3 *. Y=. 2, Or — 4)-0F ee Y ’ ’ ae 23. nae i Si a eae to find the on + fy 2x — fy ‘5 values of # and + —— on —/y 15 20+ Sy j andy. Adding 4 to each side of the first equation, 2aw~+y+t4=30—-7/(27+y4 4)3 by transposition, 2v +y +4+74/ (2% + y + 4) = 305 completing the square, 4 49 169 erty ts +7/ erty +4) + —=3904+—=—; : ‘6 13 extracting the root, 4/ (24 + y + 4) ee * 4/ (2@ + y + 4) = 3, or — 10, and2v +y+4 =9, or 100; 2 +y = 5, or 96. Multiplying every term of the second equation by ety, Van (Ea 16 else) = #13 22 — /y on — Sy *, by transposition, pee vay rancid AP on + VY, = 1; sa — fy) 18 a2 — Sy completing the square, a SERA Waid penn Ae __. 289% 20 — Sy 15. Gop ce 225 225 225° extracting the root, ge", 20 +t VY eae = aA oe. 15 15 : ia 3 e === oy Ol 2u— Vy 3 2 a “i a involving two unknown Quantities. 129 Let the former value be taken, then | 6v+3V/%y=l0r—5Vy, and by transposition, 8 V/y = 4k, and 2\/y = x; on + SY he Cah *. love +5 /y =— 6x + 3S; and 167 =— 2/7; or 8x =— V/y. Now 22 + y = 5, or 96; supposing the former, but if the second value be taken, = 5 and taking the first value of 27 = 4 VY; yts/y=5;3 completing the square, y + 4/y +4=93 ~ extracting the root, Wy +2=+3; 2 Vy =1, or — 5, and y = 1, or 25; but # = 2\/y = 2, or — 10. Again, taking the value, 27 = — ~V Ys i: as y— TV y= 5; _ completing the square, NW / rae l 321 y PA aE roy oe iby gee per : = l ao 21 .. extracting the root, /y— ae ve ) he oe and Vy = EMM, | Lo lect 321 | Br a diet ACD) | ee 32 130 Solution of Adfected Quadratics, Now taking the equation 27 + y = 96, and the first value 20 =4 Vy; then y + 4+/y = 96; completing the square, y + 44/y + 4 = 100, and «/y +2=+ 10; ee /y = 8, or — 12'5 and. )t = 64, 0r.144; whence # = 24/y = 16, or — 24. Again, taking the value, 27 = — AE then y — ~Vy = 963 completing the square, : l1 ,- 1 1 6145 rg Ve ee Ome Bae te 3 : _ ue extracting the root, V4 y — “ as SHI) peat Lot 6145 and /y = ae ) ; __ 3073 of (6145) | act 32 d and @ = — 1 (/y = EV (4s) 64 24. Given /y + fei Vy — Vai Ye+221 s Hohe te cuppa 2 wih VAY ST 2 aoe se he meen A ald Va VY to find the values of x and y. From the first equation, Sy Van Vara: V@4+i3 Say t Sy =uxt+3V/e; involving two unknown Quantities. 131 and from the second, yt2Vy—SYay=37+ fet Vy; . by transposition, y + /y— /avy =32 + Ja; but Vy t+ Vay=2+3V/a; .. by addition, y + 2Sfy =—47+4 a; completing the square, y + 2./y +1=47+4/@ 41, and extracting the root, /y +1=+ (2/z# + 1); and .*. if the positive value be taken, vy = 24/2; B+ 3Yfn = 2/2 + 22, and v = 4/2; /e=1,and2=1; whence \/y = 24/u = 2, and y = 4. But if the negative value be taken, Vy ti=z—2fe—1; by transposition, «/y = — 26/7 — 2; and if this value be substituted in the first equation, 2 Fy 0 ey il Sa Ar cea ea CA a Oana ie WA as ik) Sa Ge tae Ue ee eg Ie and since the first term in this proportion is equal to the | third, the second will be equal to the fourth; ats 34/ x +-2=— 13 by transposition, 3 «/” = — 1; Ja =— = and 2 = 9 oOl=— whence (/y =— 2.(/%a@ + l)=—=, | 16 c d =— ——. and y =— K 2 132 Solution of Adfected Quadratics, 25. Given y + JY ws ‘@ {| to find the values of x and i Ps y- | a/y Y¥ Completing the square in the first equation, Y ot Pg e423 mane WY ytJby tansy = x’ and es + PE a Re eas and extracting the root, —— ; g VY Ste eae 7 we TIVES A/G ee OE itt Nase aaa aaa vase fea ae” Again, from the second equation, 4 pees Tangs a+ 2/y 9 completing the square, 2 32 05 Pv uil62. 9 OO ees a ly Y ett pal 16y ” and extracting the root, # + oat Senne tfy 4V¥ 12 — 27 ae or 6 7 But /y SG hm VAR 2 ey whence, 7 = ———, or = 6 RCT eg = 2 ¢/ gor Nae * we Fes: , ve involving two unknown Quantities. 138 ~ — 12 —9 27 A 2, Or, OF ——, OF 7 4 ? 14 alee 144 a: 81 nee 729 way o0h a) a1 G2 1967 d 49 784 49 X 196 and y = pr org ees ere 799 ee 8e@ + yEr y? + x 26. Given —= + YR CY andw+s=4y to find the values of # and y. From the first equation, by transposition, xr’ yt Fat y te MEE Fi completing the square, = 205 xv —\e se = py ee (G+ Vu) + (f+ V9) +5 = 045 = extracting the root, 5 +Vyt+ . ot Vy= us Teneo, 2 Zz 4, or — 5, and # + yi = 4y, or — 5y. Let the former be taken. Now from the second equation, x + 8 meek: .. by subtraction, yi — 8 = 0, or 7i = 8; OAL Ege oF and# =4y—8=16—8 =8. 134 Solution of Adfected Quadratics, But ifv + yi=— 5y, andw# +8 = 4Y3 .. by subtraction,s —yi= 9Y3 *, by transposition, 8 — 8y = y + ¥, ors.(1—y)=y.(1 + y2). Dividing by (1 + 2), .. 8 (l — y2) = Y3 .. by transposition, y + 8y? = 8; completing the square, y + 8y3 + 16 = 24, and extracting the root, y3 + 4=+ 2 / 6; o ys =—44+276, and y = 40 =F 16/6; .@=m4y—8 = 1527 646. : ve 3 3 27. Given sv + 23y = 24° + 2y to “fnd ae and 34y + 6x7 — 5y? = 1347y + 24 values of #@ and y. From the second equation, 6a? — 13%y = 5y’ — 34y + 24, 13 5 34 and 2* — — .wy=—y — — 43 6 I~ G4 gut 3 completing the square, ; 1 extracting the root, 7 — a y=u (2 Y- 2) and first, taking the positive value, 7 = oy — 2. involving two unknown Quantities. 135 Let this value of x be substituted in the first equation, 125y° 2 ; “. 20y — 16 + 23y = —= — 75y* + 60Y — 16 + 24's Sm 13a *, by transposition, = —75y + 17y=0, and dividing b a ge a + LO 0 Bod a Pes [aa e183 300 68 by transposition, y’ — pet inal epeny completing the eae Q y+ (22)= 22500 68 __—«13456 | to = 133 (aati e 133 we Gsa)r : 150 116 extracting the root, y — ——- = + —; 8 2 J ~ 433 Va 7 aor Least cp va 5 — 171 A a 2 = 3, or é 2 133 But if the negative value be taken, =o = — ao" Let this value be substituted in the first equation, 3 é 16 — ~y + 23y = 16 — sy + eae oa + 2y°; 3 *, by transposition, a 3 4 8h Rca aio 2 and dividing by = on us y? + =y = a ? completing the square, sare 9 ee 2) = (2 ) 4 __ 10026 Ceis! 26 ~ (26)? ? : + 4/ (10026 extracting the root, y + =. — ge 3 136 Solution of Adfected Quadratics, . —9 #4 (10026) —9+34/(1114) i se 26 ny 26 : and #=2—2y = SEY (Ns), 28. Given ~ —s J (2 — 9x’) = 9y — l6axy to find the Y , values of and 54 = 4 + 25y” x and y. From the first equation, x — 8xry/ (v — Oy’) = oy’ — 16xy’; .. by transposition, (~ — gy’) — sahy / (v — oy’) + 16xy’ = 0; extracting the root, 4/ (vw — 9y’) —4#3y =0, and 4/ (vw — 9y’) = 4x2y, ora” — 97° = 16zy'; and # = (16% + 9).y’; e Ae xv ct ag 16v+9 But from the second equation, y’? = a ; _ be —4 Ff ; 25 Thiel ai-10 2 and 80x? — 19% — 36 = 252; *, by transposition, sov* — 44% = 36, 22 18 and 2? — —.#%# =—; 40 * 40 completing the square, yw 2 o+(4 2)) 121 _ 841 40° 40) — 1600 * 40 ~ i600° 11 29 extracting the root, x —— =x —; 40 40 9 peg av so l, or a weer 20 involving two unknown Quantities. 137 ee Pee fh havo) § Dae eee 4? l 1 ata Y = ca a ) or ots N/m 1. 5 2 29. Given 162 — y? = 6y?x? v 12 y+, to find the values of 2 and y. Bia oe } From the first equation by transposition, 16” = y? + 6y?x?; completing the square, 25a = y? + 6y?x? + 94; extracting the root, + 522 = y? + 343; °. 202, or — 8x22 = y?, and therefore 4%, or 64v = 4. 4 Now from the second equation, mF a i saa Vy band XL 1 12 1 20% Vy 48 a 4a 4a? 4 completing the square, = _ : £ 1 7 SS at ne « extracting the root, Vue ie eR are °° Sy b) ’ x maar ace Now /y = 42, or 642; which, substituted in the last equation, gives, a or le ane 3 4206p ra HE z x ome AP or -=+2,0rt/ — 3; 138 Solution of Adfected Quadratics, whence wv = + 4, or + 16, or +2/ — 3,0r ks — 35 ’*, or — 192, or — 3 xX (64)*. and y = 256, or 256 1 oe ates rast ce? to find the values of x and y. and y — 4 = 2y?x? From the first equation, by transposition, yo — 3a! y = 64; completing the square, y? — 8a?y + 16” = 16” + 64; extracting the root, y — 4a? =+4/(# + 4), and y = 4@2 +44/ (@ + 4). Also from the second equation, y — 23a = 4; completing the square, y — 2y2v2 +v=@7 +4; extracting the root, y2 — v3 =+ 4/ (@ + 4); *. 4y? = 4739 +44/ (uv + 4) = y, from the last equation; °. 4=y%, and 16 =y; Wp ee —- —_ — — 2y2 8 3 And from the second equation, x m 9 oe C= 4 81. Given 4/ (5a + 5V/y) + Vy =10- V2) to find and fa + fy = 275f the values of a and y. From the first equation, VERS Y +S 5 (VE + VY) = 103 completing the square, (Ve + Vx) + M5. A (Va +79) +2 = 1042 =>; extracting the root, «/ (\/a@ + /y) + oe ae involving two unknown Quantities. 139 and \/ (\/a + /y) = V5, or —25; /@ + /y = 5, or 20, supposing the former, .. by involution, wh + 52°y? + 10x8y + lory? + 5a2y’? + yi = 3125; but #3 Si yer .. by subtraction, bay? + 10xdy + 10xy) + 5x2y? = 2850; or 5x2y? (wi + 27y? + 2x2y + 3) I bo @ Or io) ) and x2y? (v3 + 2vy? + 2vty + yi) = 570; but vy? (#8 + avy? + 3xty + yi) = 125x2y?; ”. by subtraction, z2y? x (wy? + wty) = 125x2y? — 570; or vty? x wy? x (a? + y?) = 125x”3y2 — 570; or 5@y = 125x2y? — 570; | fy — 25e2y? = — 114; completing the square, Tor 1: 25 BY — 25x27 y? + 7 ° 1 25 13 extracting the root, #2y3 — arte Eire *. vty? = 19, or 63 but 2 + 2v2y? + y = 25, and 4x2y2 = 24, or 763 ”. by subtraction, 2 — 2v2y3 + y = 1, or — 51; extracting the root, #3? — y? = + 1, or +4/(— 51); but v2? + y3 =5; . by addition, 273 = 6, or 4, or 5+ 4/(— 51); 140 Solution of Adfected Quadratics, &c. uve 3, UL 2, Or see — 13 + ht | and we = 9, or 4) or = = Yee al by subtraction, 2y3 = 4, or 6, or 5 = 4/ (— 51); ee 3 5 = — 51 Ao yt = 2, or 3, or EY I 5Y PN eerne. toe MIE aa teea * The other case, where Wy 4 »/y = 20, is solved in the same manner. a. SECTION VI. On the Solution of Problems which involve Simple Equations. (31.) Tue solution of a problem, or method of discover- ing by analysis quantities which will answer its several con- ditions, is performed by assuming algebraic symbols to repre- sent the quantities sought, and by deducing equations from the application of these, in the same manner as if they were known quantities, to the conditions of the problem. The independent equations derived from this process, if the con- ditions be properly limited, will equal in number the unknown quantities assumed; and from the solution of these several equations by the rules already given (23. 28, 29, 30), the values of the algebraic symbols will be determined. Whether these values are correct, may be determined synthetically, by apply- ing them instead of their respective symbols to the several conditions of the problem. If the conditions of the problem are not properly limited, that is, are not sufficient in number, or not sufficiently inde- pendent of each other, the resulting equations will either exceed in number the unknown quantities; and will therefore some of them be identical or inconsistent, or will be fewer in number than the unknown quantities, and consequently will admit of an indefinite number of answers; see Section XI. In many cases, instead of assuming a symbol to repre- sent each of the required quantities, it is convenient to assume one only, and from the conditions of the problem to deduce expressions for the others in terms of that one and known quantities. And as the number of conditions ought to be one more than the number of quantities thus expressed, there will remain one to be stated in an equation; from which the value of the unknown quantity may be determined (22. 28, 29): and this being substituted in the other expressions, their value also may be discovered. 142 Examples of the Solution of Problems Examples of the Solution of Problems producing Simple Equations involving only one unknown quantity. 1, What number is that, to the double of which if 18 be added the sum will be s2° Let x = the number required. Then by the problem, 27 + 18 = 82; .. by transposition, 27 = 64, and # == 32. 2. What number is that, to the double of which if 44 be added the sum is equal to four times the required number? Let # = the number. Then 24” + 44 = 42, by supposition ; .. by transposition, 44 = 22, and 22 = @. 3. What number is that, the double of which exceeds its half by 6? Let #« = the number. Then by the problem, 22 — = 6, "4% —XH= 12, or 3% = 12, AY iy 4. From two towns which are 187 miles distant, two travellers set out at the same time with an intention of meeting. One of them goes 8 miles, and the other 9 miles a day. In how many days will they meet? Let # = the number of days required ; then sv = the number of miles one travelled, and 9z = the number the other travelled ; and since they meet, they must together have travelled the whole distance, consequently sv + 947 = 187, or 17.2 = 187, oo C=11. producing Simple Equations. 143 5. A Gentleman meeting 4 poor persons distributed five shil- lings amongst them: to the second he gave twice, to the third thrice, and to the fourth four times as much as to the first. What did he give to each? Let vw = the pence he gave to the first, *, 2v = the pence given to the second, and 3~=---------- to the third, 40 =---------- to the fourth. B+ +32 +40 = 60, or 102 = 60, oe C= 6, and .*. he gave 6, 12, 18, 24 pence respectively to them. 6. A Bookseller sold 10 books at a certain price; and after- wards 15 more at the same rate. Now at the latter time he received 35 shillings more than at the former. What did he receive for each book ? Let # = the price of a book, in shillings, Then 10% = price of the first set, and 152 = price of the second set. But by the problem 15z = lov + 35; .. by transposition, 5% = 35, and # = 7. 7. A Gentleman dying bequeathed a legacy of £140 to three servants. 4 was to have twice as much as B; and B three times as much as C. What were their respective shares ? Let # = C’s share, *, 3” = B’s share, and 64 = A’s share; whence (67 + 3% + # =) 10% = 140, or A .. received £84; B, £42; and C £14. | 8. Four Merchants entered into a speculation, for which they subscribed £4755; of which B paid three times as much | as 4; C paid as much as 4 and B; and D paid as much | as Cand B. What did each pay? 144 Examples of the Solution of Problems Let # = number of pounds 4 paid; *, 37 = number B paid, 4¢@ = number C paid, and 7@ = number D paid; . (w+ 304+4@74+ 72 =) 152 = 4755, and .°. @ = 317. *, they contributed 317, 951, 1268, and 2219 pounds respect- ively. 9. A Draper bought three pieces of cloth, which together measured 159 yards. The second piece was 15 yards longer than the first, and the third 24 yards longer than the second. What was the length of each? Let « = the number of yards in the first piece, *, e + 15 = the number in the second, and # + 39 = the number in the third. .e@+t+er+1s+xe+ 39 = 159, and by transposition, 37 = 105, *, @ = 35, .. the lengths are 35, 50, and 74 yards respectively. 10. A cask which held 146 gallons, was filled with a mixture of brandy, wine, and water. In it there were 15 gallons of wine more than there were of brandy, and within 4 gal- lons as much water as both wine and brandy. What quantity was there of each ? Let # = the number of gallons of brandy, *, @ + 15 = number of gallons of wine, and 227 + 11 = number of gallons of water. - &@+e+15 +274 11 = 146, .. by transposition, 4# = 120, and # = 320. .. there were 30, 45, and 71 gallons respectively of brandy, wine, and water. 11. A person employed 4 workmen; to the first of whom he gave 2 shillings more than to the second; to the second producing Simple Equations. 145 3 shillings more than to the third; and to the third 4 shil- lings more than to the fourth. Their wages amounted to 32 shillings. What did each receive? Let # = the sum received by the fourth, DO mr nh me Ces enh wth! es) ee’ wn fe third, eA Te hah erty ahs yp es ie second, Be G8 ol oa a pan SS first .@betstot+7+e+9=32, and by transposition, 4@” = 12, . 723. .. they received 12, 10, 7, and 3 shillings respectively. 12. A Father taking his 4 sons to school, divided a certain sum amongst them. Now the third had 9 shillings more than the youngest; the second 12 shillings more than the third; and the eldest 1s shillings more than the second; and the whole sum was 6 shillings more than 7 times the sum which the youngest received. How much had each ? Suppose the youngest received 2 shillings, then the third received xv +9 --- - the second ----#+21 ---- and the eldest - - -- 7+39 ---- .@+rtot+rt+21+%74+ 39=70%4 6, .. by transposition, 63 = 32, Andel 7, consequently they received 21, 30, 42, and 60 shillings respect- _Ively. 13. A sum of money was to be divided amongst six poor per- | sons; the second received 1od. the third 14d. the fourth o5d. the fifth asd. and the sixth 33d. less than the first. ’ Now the sum distributed was 10d. more than the treble of what the first received. What did each receive? Yi 146 Examples of: the Solution of Problems Let 2 = what the first received, .- @2—10= - --- second - = - f— 14 eae) J sthird se 2— 25 - - - - fourth --— - 2—28= ---- fifth --- Pm 33a: ---- sixth ~ - - The sum of which = 6% — 110 = 3w + 10 by supposition ; .. by transposition, 37 = 120, S00 0; .. they received 40, 30, 26, 15, 12, 7 pence respectively. 14. It is required to divide the number g9 into five such parts, that the first may exceed the second by 3; be less than the third by 10; greater than the fourth by 9; and less than the fifth by 16. Let # = the first part, AYP Beet 8 om second, e+10= third, L— 9= fourth, YU+16= fifth. ~ &©+H7—-34+74+104+%—9+274+16= 99, or 5% + 14= 99, *, by transposition, 52 = 85, and # = 17% *, the parts are 17, 14, 27, 8, and 33. 15. What two numbers are those whose sum is 59, and differ- ence 17? Let x7 = the less, -. & + 17 = the greater, and .. @+27+417=59, by transposition, 247 = 42. and w# = 21 the less, *, the greater = 38. 16. What number is that, the treble of which increased by 12, shall as much exceed 54 as that treble is below 144? 17. Then vw — 14 producing Simple Equations. 147 Let v = the number. *, 3@ + 12 — 54 = 144 — 3a by supposition ; .. by transposition, 6% = 186, and # = 31. Two persons began to play with equal sums of money: the first lost 14 shillings, the other won 24 shillings; and then the second had twice as many shillings as the first. What sum had each at first ? Let # = the sum; = th ine’; Me. aa the sums each had after playing; .. by the problem 2a” — 28 = x + 24; "¢ 405s 52 is. At a certain election 943 men voted, and the candidate 19. chosen had a majority of 65. How many voted for each? Let x =the number of votes the unsuccessful candidate had ; . & + 65 = the number the successful one had. . v2ta+t 65 = 943; by transposition, 27 = 878, and # = 439. .. the numbers were 439 and 504. Two Robbers after plundering a house found that they had 35 guineas between them; and that if one of them had had 4 guineas more, he should have had twice as many as the other. How many had each? Let 2 = the number one had, . 35 — # = the number the other had, and 35 —v7+4= 22. by transposition, 39 = 32, and 13 = a. .. they had 13 and 22 guineas respectively. L2 148 Examples of the Solution of Problems 20. A Mercer having cut 19 yards from each of three equal pieces of silk, and 17 from another of the same length, found that the remnants taken together were 142 yards. What was the length of each piece? Let # = the length, . @ — 19 = the length of each of the 3 equal remnants, and # — 17 = the length of the other. then 3. (#— 19) + @ — 17 = 142, or 3% — 57 +. #@— 17 = 142. by transposition, 47 = 216; . &= 54. 21. A Farmer has two flocks of sheep, each containing the same number. From one of these he sells 39, and from the other 93; and finds just twice as many remaining in one as in the other. How many did each flock originally contain ? | Let v = the number required. Then x — 39, and x# — 93, are the numbers remaining ; ". & — 39 = 22 — 1865 and by transposition, 147 = a. 22. Bought 12 yards of cloth for £10. 14s. For part of it I gave 19 shillings a yard, and for the rest 17 shillings a yard. How many yards of each were bought? Let # = the number of yards at 19s. per yard; 12 — # =the number at 17s. and 19” = the price of the former, and 17.(12—w) = the price of the latter. *, 19% + 204 — 17” = 214. and by transposition, 22” = 10, vale We dta—a op, . there were 5 yards at 19 shillings, and 7 at 17 shillings. 23. Divide the number 197 into two such parts, that four times the greater may exceed five times the less by 50. producing Simple Equations. 149 Let x = the less, and .*. 197 — x = the greater. Then 788 — 442 = 54 + 50; and by transposition, 738 = 92, and; 82. == 273 ~~ the greater 115; 24. A Courier, who travels 60 miles a day, had been dispatched 5 days, when a second was sent to overtake him; in order to which, he must go 75 miles a day: In what time will he overtake the former ? Let « = the number of days the second courier travels ; then @ + 5 = the number the first travels ; .. 75a = the number of miles the second travels, and 60.(#+5) = the number the first travels. But by the supposition they both travelled the same number of miles ; , 752 = 602+ 300, by transposition, 152 = 300, ance == 20. 25. After A had lost 10 guineas to B, he wanted only 8 guineas in order to have as much money as B; and together they had 60 guineas. What money had each at first ? Let v = the number of guineas 4 had ; “. 60— 2 =the number - --- B had. Then after playing A had w — 10, and B had 70 — 2; . @—-10+8=70—2, by transposition, 22 = 72, and # = 36. .. they had 36 and 24 guineas respectively. 26. A and B began trade with equal stocks. In the first year A tripled his stock, and had £27 to spare; B doubled his stock, and had £153 to spare. Now the amount of both their gains was five times the stock of either. What was that stock ? 150 Examples of the Solution of Problems Let x = the stock ; then 3x + 27 = A’s stock at the end of the year, °. 22 + 27 = his gain, and 2” + 153 = B’s stock at the end of the year ; *.. @ + 153 = B’s gain; . 527 = 297 + 27 + @ 4.153. by transposition, 27 = 180, and # = 90. 27. Two workmen A and B were employed together for 50 days, at 5 shillings per day each. A spent sixpence a day less than B did, and at the end of the fifty days he found he had saved twice as much as B, and the expense of two days over. What did each spend per day? Let « = what A spent per day (in pence) ; *, 60 — # = what he saved per day, and 54 — # = what B saved per day. and .*. 3000 — 50” = 5400 — 100a” + 22. by transposition, 482 = 2400, and @ = 50; .. A spent 50 pence, and B 56 pence a day. 28. A and B began to trade with equal sums of money. In the first year A gained 40 pounds and B lost 40; but in the second 4 lost one-third of what he then had, and B gained a sum less by 40 pounds than twice the sum that A had lost; when it appeared that B had twice as much money as 4. What money did each begin with? Let # = the number of pounds each had at first. then # + 40 = the sum 4 had after the first year, and #@ — 40 = the sum B had, also 2. (v + 40) = the sum 4 had after the 2° year, and # — 40 + 3(# + 40) — 40 = the sum B had; .4.(%+ 40) =#—40+ 2. (w+ 40) — 40, and 2. (% + 40) = x — 80; *, 22 + 80 = 34 — 240, and by transposition, 320 = #. 31. 30. producing Simple Equations. 15] 29. Divide the number 6s into two such parts, that the differ- ence between the greater and 84 may equal three times the difference between the less and 40. Let # = the less, then 68 — # = the greater; .. 84 — (68 — 7) = 3. (40— 2), or 16+ 2% = 1200— 32. by transposition, 47 = 104, and # = 26; and .*, the greater = 42. A and B being at play severally cut packs of cards so as to take off more than they left. Now it happened that 4 cut off twice as many as B left, and B cut off seven times as many as A left. How were the cards cut by each? Suppose 4 cut off 22 cards, then 52 — 2% = the number he left, and # = the number B left; . 52 — # = the number he cut off; whence 52 — # = 364 — 1423 by transposition, 13@ = 312, and 2 = 24; .. A cut off 48, and B cut off 28 cards. What number is that whose one-third part exceeds its one- fourth part by 16? Let * 124 = the number; °. 42 — 32 = 16, Ole —iG. and .*, the number = 12 x 16 = 192. 32. Upon measuring the corn produced by a field, being 48 quarters; it appeared that it yielded only one-third part more than was sown. How much was that? * 12 being the least common multiple of 3 and 4, 152 Examples of the Solution of Problems Let 32 = the number of quarters sown, then 32 + #7 = 43, or 4” = 48, and w= 133 *, the quantity sown was 36 quarters. 33. A Farmer sold 96 loads of hay to two persons. To the first one-half, and to the second one-fourth of what his stack contained. How many loads did that stack contain ? Let 44 = the number of loads, then 24 = the number the first bought, and # = the number the second had. pipe Daanl eee tei Oe and 2 == 39; whence, the stack contained 128 loads. 34. A Gentleman bequeathed £210 to two servants; to one he left half as much as to the other. What did he leave to — each 2 Let 27 = the sum one received ; *, # = the sum left to the other. ea ye fae e210, anda == 70: *, they had 140 and 70 pounds respectively. 35. A prize of £2329 was divided between two persons 4 and | B, whose shares therein were in proportion of 5 to 12. — What was the share of each? Let 54 = A’s share, then 122 = B’s share; *. (5a@°+ 12% =) 17H = 2320, AHR T= 37 *. their shares were 685 and 1644 pounds respectively. 36. A sum of money is to be shared between two persons 4 and B, so that as often as 4 receives 9 pounds, B takes 4. Now it happens that 4 receives 15 pounds more than B. What are their respective shares? 37. 39. producing Simple Equations. 153 Since for every £9 that 4 receives, B receives £4, Let 9% = the whole sum JA receives ; . 4v = the whole sum B receives; OV = 4x 415; and by transposition, 52 = 15; &G=33 .. A receives £27, and B £12. A Gentleman gave to 3 persons 98 pounds. The second received five-eighths of the sum given to the first, and the third one-fifth of what the second had. What did each receive ? Let sv = the number of pounds the first received ; 0 SUS - - - ee Ke second - - - Sat gy ke ee cms Pe fol third -- - .. (82+ 527+ #7 =) 1427 = 98, and. 4 7s .. they received 56, 35, and 7 pounds, respectively. A person bought two casks of beer, one of which held exactly three times as much as the other. From each of these he drew four gallons, and then found that there were four times as many gallons remaining in the larger, as in the other. How many were there in each at first ? Let 3 = the number of gallons in the larger ; and .*. # =the number in the smaller ; .°4.(7—4)=>37—4; by transposition, # = 12; *, they held 36 and 12 gallons, respectively. A man at a party at cards betted three shillings to two upon every deal. After twenty deals he won five shil- lings. How many deals did he win? Let « = the number of deals he won; *, 20 — x = the number he lost; also 2” = the money won, and 3. (20 — 2) = the money lost; 154 40. 41. 42. Examples of the Solution of Problems whence 22 — 3. (20— 4) = 5; .. by transposition, 5a = 65, and # = 13. What two numbers are as 2 to 3; to each of which if 4 be added, the sums will be as 5 to 7? Let 2x and 32 be the numbers; 20 +4530 +4:5°5°7; and (21) 142 + 28 = 15% + 20; by transposition, 8 = @; and .*. the numbers are 16 and 24. A sum of money was divided between two persons 4 and B, so that the share of A was to that of B as 5 to 3, and exceeded five-ninths of the whole sum by 50 pounds. What was the share of each person? Let 5x2 = A’s share; * aa == DS. share, and sv = the whole sum; 7. 50 = 2.82 + 50, or v7 =3.4% + 10, and 9% = 8@ + 903 .. by transposition, 2 = 90, and the sums were 450 and 270 pounds. Being sent to market to buy a certain quantity of meat, I found that if I bought beef, which was then 4 pence a pound, I should lay out all the money I was entrusted with; but if I bought mutton which was then threepence halfpenny a pound, I should have two shillings left. How much meat was sent for? Let 22 = the number of pounds; *, 8H = the price of 2 lbs. of beef, and 7# = the price of 2 lbs. of mutton, and 8v¥ = 7x2 + 243 \, ©= 245 whence 48 lbs. were sent for. producing Simple Equations. 155 43. A Fish was caught, whose tail weighed 9 lbs.; his head weighed as much as his tail, and half his body; and his body weighed as much as his head and tail. What did the fish weigh ? Let 2v = the number of lbs. the body weighed ; then 9 + # = the weight of the head; ODL =2IV; by transposition, 18 = #; .. the fish weighed 36 + 27 + 9 = 72 lbs. 44, The joint stock of two partners whose particular shares differed by 40 pounds was to the share of the lesser as 14 to 5. Required the shares. Suppose 142 = the joint stock ; *, 5a = the less, and 9x = the greater ; oe OV = 5H + 405 by transposition, 47 = 40, rina eas Wag .. the shares are 90 and 50 pounds, respectively. 45. A Bankrupt owed to two creditors 140 pounds; the dif- ference of the debts was to the greater as 4 to 9. What were the debts? Let 4v = the difference of the debts ; *, ge == the greater, and 52 = the less; SSG ia == | EA ae 1S0y and #7 = 10% .. the debts are 90 and 50 pounds. 46. A Gentleman employed two labourers at different times, one for 3 shillings, and the other for 5 shillings a day. Now the number of days added together was 40; and they each received the same sum. How many days was each employed ? 156 Examples of the Solution of Problems Let # = the number of days the second was employed ; *, 40 — # = the number the first was employed ; and 5a = the sum received by the second, and 3. (40 — 2) = the sum received by the first ; ‘BT S= Bi 40e ays by transposition, 87 = 120, andeayes 15: .. the second was employed 15, and the first 25 days. 47, Some persons agreed to give sixpence each to a waterman for carrying them from London to Gravesend; but with this condition, that for every other person taken in by the way, three pence should be abated in their joint fare. Now the waterman took in three more than a fourth part of the number of the first passengers, in consideration of which he took of them but five pence each. How many persons were there at first ? Let 42 = the number of passengers at first ; then x + 3 = the number taken in, and 3@ + 9 = the sum deducted from their joint fare; *, 242 — (3@ + 9) = 202; by transposition, 7 = 9; consequently there were 36 passengers. 48. In a mixture of wine and cyder, half of the whole + 25 gallons was wine, and one-third of the whole — 5 gallons was cyder. How many gallons were there of each? Let 62 = the number of gallons in all; *, 3@ + 25 = the number of gallons of wine, and 24 — 5 = the number of gallons of cyder; “. 62 = 3H +25 +2” —5; by transposition, v = 20; consequently there were 85 gallons of wine, and 35 of cyder. 49. A and B engaged in trade, A with £240, and B with £96. A lost twice as much as B; and upon settling their accounts it appeared that A had three times as much remaining as B. How much did each lose? 50. 51. producing Simple Equations. 157 Let x = what B lost; . 96 — & = what he had remaining ; then 2x = what A lost, and 240 — 24 = what he had remaining ; *, 240 — 27 =3. (96 — 2) by transposition, v = 48 ; ., A lost £96, and B lost £48. Four places are situated in the order of the four letters A, B, C, D. The distance from A to D is 34 miles, the distance from A to B: distance from C to D:: 2:3, and one-fourth of the distance from 4 to B added to half the distance from C to D is three times the distance from B to C. What are the respective distances ? Let 27 = the distance from A to B; °, 3@ = the distance from C' to D. Ts Cee 3 and (20 + 3@ +0 =)"2 a he a 7 2 and # = 6; whence AB = 12, BC =4, and CD = 18. A Field of wheat and oats which contained 20 acres was put out to a labourer to reap for six guineas, the wheat at 7 shillings an acre, and the oats at 5 shillings. Now the labourer falling ill, reaped only the wheat. How much money ought he to receive according to the bargain? Let x = the number of acres of wheat ; then 20 — x = the number of acres of oats; and 7# = the price of reaping the wheat (in shillings), and 100 — 54 = the price of reaping the oats ; 158 Examples of the Solution of Problems “2 72 + 100 — 5” = 1265 by transposition, 27 = 26, anda? — i132 .“. he ought to receive £4. 118. 52. A General having lost a battle, found that he had only half his army + 3600 men left, fit for action; one-eighth of his men + 600 being wounded, and the rest, which were one-fifth of the whole army, either slain, taken prisoners, or missing. Of how many men did his army consist ? : Since 40 is the least common multiple of 2, 8, and 5, let 40” = the number required ; . 20” + 3600 = the number fit for service; 5x2 + 600 = the number wounded, and sv = the number missing ; “. 40” = 20% + 3600 + 5H + 600 + 823 by transposition, 7” = 4200, and 2 = (600. .. his army consisted of 24000. 53. Three men, 4, B, and C, entered into partnership; 4 paid in as much as B, and one-third of C; B paid in as much as C, and one-third of 4; and C paid in £10, and one- third of A. What did each man contribute to the stock? Let 37 = the sum 4 contributed ; S110 oP See rt a a ee ee S00 10.6, 9a eee es eee $4 32 = 10 +22 ++"; wae fpga@ by transposition, ine ue an 2a =A ° £2: A: = 20. and the sums contributed were £60, £50, and £30, by A, B, C, respectively. producing Simple Equations. 159 54. It is required to divide the number 91 into two such parts that the greater being divided by their difference, the quotient may be 7. Let 2 = the greater ; “. 91— # = the less, Pia ae ey ar 2%—91 a5 oo © = 14H — 6375 by transposition, 637 = 132; GE e401. ane .. the parts are 49 and 42. 55. From each of 16 coins an artist filed the worth of half a crown, and then offered them in payment for their original value: but being detected, the pieces were found to be really worth no more than 8s guineas. What was their original value ? Let # = the number of sixpences each was worth; .“. &@ — 5 = the number each was worth after filing; “. 16. (@— 5) = 336. by transposition, 16” = 416, and # = 26 = 13 shillings. | 56. A and Bmade a joint stock of £833, which, after a suc- cessful speculation, produced a clear gain of £153. Of this B had £45 more than 4. What did each person contribute to the stock ? Let # = the sum brought in by B; 9 ° OV thet 8§33\5,0°.c1loaes B's gain = 753 Meee Lys .. A’S gain = — — 45 9 Ov and — -—- — — 45 = 153; 49 i 49 ; SE EG by transposition, Fry Kee 160 57. part was to the latter :: 3: 3, of each ? And 8@ xe 2 58. 59. Examples of the Solution of Problems 49 X 198 SRE lier spety reel CR aU BU 2 whence, B brought in £539, and A £294. Sold a quantity of tobacco for 19 shillings, part at 1 shil- ling a pound, and the rest at 15 pence. Now the first 3: 2 How much was sold Since 3 212025 9 : 8, Let 9z = the number of lbs. of the former ; “. 8a” = the number of lbs. of the latter; .. 92 = the number of shillings the first sold for, = 104% = the number of shillings the second sold for. SLO sO) oA 2a rune ie ae aagy Be .. there were 9 lbs. at 1 shilling, and s lbs. at 15 pence. A Gentleman gave in charity £46; a part thereof in equal portions to 5 poor men, and the rest in equal portions to 7 poor women. Now a man and a woman had between them £s. What was given to the men, and what to the women? Let 52 = the number of pounds the men received ; “. 46 — 5a = the number the women received ; *, 2 = the sum one man received, the sum one woman received ; *. 56 — 72 = 46 — 5235 by transposition, 27 = 10, ander =o .. the men received £25, and the women £21. and 8—2= Suppose that for every 10 sheep a farmer kept, he should plough an acre of land, and be allowed one acre of pasture for every 4 sheep. How many sheep may that person keep who farms 700 acres? producing Simple Equations. 161 Let x = the number of sheep required ; then 10: x :: 1: the number of acres ploughed = <, and 4:4 ::1: the number of acres of pasture = Ae eR See Seeder: ands (2d e ds —) fa — 205 700 § “¢ 2 = 20,.x 100° 2000- 60. A person being asked the hour, answered that it was between five and six; and the hour and minute-hands were together. What was the time? eta — the time pases ; then since the minute-hand goes 12 times round, whilst the hour-hand goes once, we have this proportion, 124) bat 5 ich dae abet CONG PSU OR G Bera) Bere Fe elie 85, nats Wi re NO A hea 61. Divide the number 49 into two such parts that the greater increased by 6 may be to the less diminished by 11 as 9 to 2. Let # = the greater; “. 49 — x = the less, andx+6:38—2#::9: 23 wea Alg. 179, 2-3.) & + 0.7 44/2) 9 3) 11, BUC GAO, 179,.8.)00 +t, 4c 9 os les “.@ + 6 = 36, and # = 30; .. the parts are 30 and 19. 62. A, B, and C make a joint stock; A puts in £60 less than B, and £68 more than C; and the sum of the shares of 4 and B is to the sum of the shares of B and C as 5 to 4. What did each put in? M 162 Example of the Solution of Problems Let # = what 4 put in; *, 2 + 60 = what B put in, and # — 68 = what C put in; then 2v + 60: 247 —8:3:5:4, and (Alg. 179, 7.)% +30: #©—4:25:43 0 A101 70, 42) (SRO ie — pe LS . 136 =v — 4, ANG wie 140.3 .. they put in £140, £200, and £72 respectively. 63. It is required to divide the number 34 into two such parts, — that the difference between the greater and 18, shall be to the difference between 18 and the less :: 2: 3. Let # = the greater; *, 34 — x = the less, and #— 18: %— 16:32: 33 sy { ALY. 91 70,925 45) ee Se IN ee eed .@&€—18 =4, and £2 = 22’; .. the parts are 22 and 12. 64. A Bookseller sells two books, one containing 100 sheets, for 10 shillings, the other containing 50 sheets, for 6 shil- lings, each being bound at the same price. What was that price ? Let # = the price; then 10 —%:6—#:: 100: 50:32:13 os (Als 17954. Ae Oe — eed tel .4=>6-—-2; by transposition, 7 = 2. 65. A man wished to inclose a piece of ground with palisades, and found that if he set them a foot asunder, he should have too few by 150; but if he set them a yard asunder, he should have too many by 70. How many had he? producing Simple Equations. 163 Let # = the number; then # — 70: %@ + 150::1: 3, ans 419, 179, 5.) © — 70 5 220 O22 eso, On — (0. TOs kbs hs *, @ — 70=110, and 7=-180. 66. A Footman, who contracted for £8 a year, and a livery suit, was turned away at the end of 7 months, and received only £2. 3s. 4d. and his livery. What was its value? Let v# = its value, in pounds; men 12°97 °% (o+s e+ 8:1 )60 +48: 6x + is PAG HFA) bee Ok 35 Ors tS: EULER S ACHR A IG ae 4 LTS :3 6S 13 = 403 by transposition, 62 = 36, and @ —6s 67. What number is that to which if 1, 5, and 13, be severally added, the first sum shall be to the second, as the second to the third? Let # = the number required ; thenvw@ +1:@2+5::@%@+5:2424+13; Beetalg..1/79, 5.) oe -- 1s U4 eh & + 5.2 8s AU Bibs Lote lea ea tet ee mee ote eae sig atch he az they 3 Wabi 9 vr+il=4, and # 3. 6s. A Landlord let his farm for £10 a year in money and a corn-rent. When corn sold at 10s. a bushel, he received at the rate of 10 shillings an acre for his land; but when it sold at 13s. 6d. a bushel, 13 shillings an acre. Of how many bushels did the corn-rent consist ? M 2 164 69. Examples of the Solution of Problems Let 2 = the number of bushels ; then 10# + 200 = the annual income (in shillings) ; and .. 2+ 20 = the number of acres; : 27x + 400 also in the second case Peper the number of acres ; 27x" + 400 : ae =x + 20, and 27x” + 400 = 26@ + 520, by transposition, 7 = 120. When the price of a bushel of barley wanted but 3d. to be to the price of a bushel of oats as s to 5, nine bushels of — oats were received as an equivalent for four bushels of barley, and 7s. 6d. in money. What was the price of a bushel of each? Let 5v = the price of a bushel of oats ; *, 8v — 3 = the price of a bushel of barley ; 452 = 32% — 12 + 903 by transposition, 13v = 78, and Waa Gs .. the price of a bushel of oats = 30d. and the price of a bushel of barley = 45d. A Countryman had two flocks of sheep, the smaller consisting entirely of ewes, each of which brought him 2 lambs. On counting them he found that the number of lambs was equal to the difference between the two flocks. If all his sheep had been ewes, and brought forth 3 lambs apiece, his stock would have been 432. Required the num- ber in each flock. Let # = the number in the less; *, 2” = the number of lambs this flock produced = the difference of the flocks, and 3% = the number in the larger flock ; , 42 eo 8 xX wd aa and 7” = 27; *, 27 and 81, are the numbers required. producing Simple Equations. 165 71. A Market-woman bought a certain number of eggs at two a penny, and as many at three a penny, and sold them out at the rate of five for two-pence; after which she found that instead of making her money again, as she expected, she lost four-pence by them. How many eggs of each sort had she? Let # = the number required ; then 2 : # :: 1: the price of # eggs at 2a penny = <5 in the same way, = the price of x eggs at 3 a penny. and 5 ; 2a :: 2: the price at which she sold all, = =; : Lane BAHL 5 ies ao <9l2 G37 and 24% + 120 = (154 + 10% =) 252: by transposition, v = 120. 72. A man and his wife did usually drink out a vessel of beer in 12 days: but when the man was out, the vessel lasted the woman 30 days. In how many days would the man alone drink it out ? Let # = the number of days required ; then # : 12 :: 1: part drunk by the man in 12 days = =, Se : ee and 30 : 12 :: 1: part by the woman in 12 days =e ee 2 de cle <<. ames: Ba ie and 60 + 2% = 52, by transposition, 60 = 32, and 20: == a 73. A cistern into which water was let by two cocks 4 and B, will be filled by them both running together in 12 hours, and by the cock 4 alone in 20 hours. In what time will it be filled by the cock B alone? 166 Examples of the Solution of Problems Let # = the number of hours ; then @ $12 : the quantity supplied by B in 12 hours = =. In the same way, the quantity supplied by 4 in 12 hours = 5 123 eee _-_=— l, oe RAED and 60 + 3% = 523 by transposition, 60 = 2a, and 30 = 2. 74. The hold of a ship contained 442 gallons of water. This was emptied out by two buckets, the greater of which, holding twice as much as the other, was emptied twice in three minutes, but the less three times in two minutes; and the whole time of emptying was 12 minutes. Required the size of each. Let x = the number of gallons the less held; 2x” = the number the greater held; and 4a = the quantity thrown out by the greater in 3 minutes; *.3312%:4: the quantity thrown out in 12 minutes = 162. In the same manner the quantity thrown out by the less in 12 minutes = 182; aise slGa i= lod hao and # = 13; *, the less held 13, and the greater 26 gallons. 75. A hare, 50 of her leaps before a greyhound, takes 4 leaps to the greyhound’s three; but two of the greyhound’s leaps are as much as three of the hare’s. How many leaps must the greyhound take to catch the hare? Let 3v = the number of leaps the greyhound must take; *, 4@ = the number the hare takes in the same time; *, 4@ + 50 = the whole number she takes, and 2:32:33”: 42 + 50; . 9V = 8x + 100; by transposition, # = 100, and the greyhound must take 300 leaps. producing Simple Equations. 167 76. If 10 apples cost a penny, and 25 pears cost two-pence, and“ I buy 100 apples and pears for nine-pence halfpenny, how many of each shall I have? Let # = the number of apples; *, 100 — x = the number of pears ; and 10: #:: 1: the price of # apples = ~~ In the same manner the price of the pears = met; %, 200-—-2% 19 wig Apanangat % and 54” + 400 — 4” = 475; by transposition, v = 75; .. the number of apples is 75, and pears 25. b) 77. A person has two sorts of wine, one worth 20 pence a quart, and the other 12 pence; from which he would mix a quart to be worth 14 pence. How much of each must he take? Let # = the quantity of the first, the whole quart being represented by unity ; *, 1— # =the quantity of the second; also 20” = the value of the first, and 12 — 1242 = the value of the second; * 2000 +12— 127 = 143 and by transposition, ¢7 = 2; es .. he must take } of the first, and 3 of the second. 7s. A person engaged to reap a field of 35 acres, consisting partly of wheat, and partly of rye. For every acre of rye he received 5 shillings; and what he received for an acre of wheat augmented by one shilling, is to what he received for an acre of rye as 7 to 3. For his whole labour he received £13. Required the number of acres of each sort. Let # = the number of acres of wheat; *, 35 — x = the number of acres of rye; and 175 — 54 = the price of reaping them. 168 Examples of the Solution of Problems Now 3:7::5:1-+ the price of reaping an acre of 35 wheat = aa .. the price of reaping an acre of wheat afi and the price of reaping all the wheat ==; 322 ed wey) + 175 — 52% = 260, and 324” + 525 — 15@ = 780; by transposition, 17v = 255, and’ v= 15; .. there were 15 acres of wheat, and 20 of rye. 79. Two pieces of cloth of equal goodness, but of different lengths, were bought, the one for £5, the other for £6. 10s. Now if the lengths of both pieces were increased by 10, the numbers resulting would be in the proportion of 5 to 6. How long was each piece, and how much did they cost a yard? Let 2 = the number of 10 shillings that each yard cost, then “ = the length of the least, and = = the length of the longest ; (Alg. 179, 5.) 10. (= 1)t es baal, and (Aly. 179, 8.) 2. (+ +1) gs ga ws adh) re Ae and 27 +4+2=3; by transposition, 27 = 1, 1 and # ==; 2 *, the price is 5s. and the lengths are 20 and 26 yards. producing Simple Equations. 169 so. A General, whose horse was 1 of his foot, after a de- feat found, that before the battle ~, — 120 of his foot, and 54, + 120 of his horse had deserted; + of his whole army was in garrison; and 2 remained, the rest being either taken prisoners or slain. Now 300 + the number slain = + the foot he had at first. Of how many did his whole army consist ? Let # = the number of horse; *, 3@ = the number of foot, and 4@ = the whole army; and # = the number in garrison ; 32 also ge the number slain ; and se = the number that remained. ee 1g Da agi kas Bat Soe A ae yy a AY 4 12 2 2 and 3% + # + 48x — 3600 = 48243 by transposition, 44 = 3600, and # = 900; .. the whole army consisted of 3600 men; viz. 900 horse, and* 2700 foot. 81. Two persons 4 and B have both the same annual income. A lays by ith of his; but B by spending £s0 per annum more than A, at the end of 4 years finds himself £220 in debt. What did each receive and expend annually ? Let 52 = their annual income; *, 4” = A’s annual expenditure, and 4@ + 80 = B’s annual expenditure ; *, (4@ + 80 — 5a =) 80 — = the debt B annually incurs; .. 320 — 4@ = 220; by transposition, 4# = 100, and a= 25 5 .. their annual income is £125; A’s annual expenditure is £100, and B’s £180. 170 Examples of the Solution of Problems s2. A person at play won twice as much as he began with, and then lost 16 shillings. After this he lost four-fifths of what remained, and then won as much as he began with, and counting his money, found he had so shillings. What sum did he begin with? Let # = the number of shillings he began with; then 37 = the sum he had, after winning 2a, and 3# — 16 = the sum remaining after the next loss. 3% — 16 . 4 e . . Now since he lost e of this, = the sum remaining ; 3x2 — 16 Ate + 2 = 80, 5 and 3”@—16 + 5@ = 400; by transposition, 827 = 416, and @ = 52. 83. Having lost one-third of my money at play, I won 3 times as much as I had left, half as much money as I began with, and £50; and then found I had as much above £100, as the sum I began with was below £100. What sum did I begin with? Let 62 = the number of pounds required ; then 4a = the sum remaining after 4 was lost, and 12@ + 3@ + 50 = the sum afterwards won; *, (122 + 32 + 50 + 4” =) 197 + 50 = the whole sum he had, and 19# + 50 — 100 = 100 — 62; by transposition, 254 = 150, and) gem G3 .. he began with 36 pounds. 84. A and B began to pay their debts. A’s money was at first two-thirds of B’s; but after 4 had paid £1 less than two- thirds of his money, and B £1 more than seven-eighths of his, it was found that B had only half as much as A had left. What sum had each at first ? producing Simple Equations. 171 Let 22 and sv = the sums 4 and B had respectively, 2 then after payment, A had — +1 remaining ; and B had “to l 20 32 .~—+1=—-—2; 3 . 8@ +12 = 94 — 24; by transposition, 36 = 2; .. A had £72, and B had £108. g5. It is required to divide the number 36 into three such parts, that one-half of the first, one-third of the second, and one- fourth of the third may be equal to each other. Let 24 = the first part ; part of the second, and 37 = the second ; = 1 part of the third, and .. 4a = the third; , (2@ +- 32 + 42 =) 92 = 36. and # a= 43 .. the parts are 8, 12, and 16. * 86. Divide the number 116 into four such parts, that if the first be increased by 5, the second diminished by 4, the third multiplied by 3, and the fourth divided by 2, the result in each case shall be the same. Let # = the third; . 3a = half the fourth, and 6x2 = the fourth; whence 3x + 4 = the second, and 32 — 5 = the first; “3@—5+3@+4+274+67=116; by transposition, 1327 = 117, ANC Qa Gis .. the parts are 22, 31, 9, and 54. 87. A Gentleman had some of his horses at grass at 3 shillings each a week, and the rest at livery stables at 10 shillings each a week. The horses in the stables cost him twice as 172 Examples of the Solution of Problems much a week as the horses at grass. But he finds that if he had sent 3 horses to grass out of the stables, the expense of the stables would have been only 6 shillings a week more than the grass. How many horses had he? Let v = the number of horses at grass ; *, 3@ = the weekly expense of these, TNC ae teh ai i et eg ean of horses in stables; , OF — their number ; 10 also 64 — 30 = the expense of the stables after 3 horses were sent out to grass, and 3” + 9 =the expense of the horses at grass, when 3 more were added ; “. 64 — 30 (=37 +94 6) =3H+4 15; by transposition, 3v = 45; Sndw ib: .. there were 15 at grass, and 9 in the stables. ss. A Silversmith received in payment for a certain weight of wrought plate, the price of which was £10, the same weight of unwrought plate, and £3. 15s. besides. At another time he exchanged 12 oz. of wrought plate of the same work- manship as before for 8 oz. of unwrought (for which he allowed the same price as before), and £2. 16s. in money. . What was the price of wrought plate per ounce, and the weight of the first sold? Let # = the number of ounces; 200 : XS ee ay the price of an oz. wrought, 125 ° and re sc the price of an oz. unwrought ; 2400 1000 + —— = — + 56; v xv se 400 by transposition, Te ae 25 and — = 1; & Ee Bem Pe whence there were 25 ounces; and the price was 8 shillings per oz. producing Simple Equations. 173 s9. In changing a bill of £85 into guineas and shillings (the number of shillings being } number of guineas) on exami- nation they all proved adulterated below the standard value, to the amount in the whole of £3. 5s. To make up the deficiency, nine more such guineas were paid; and four such shillings and three good ones returned. Required the number and average value of the guineas and shillings paid at first. £25 = 85s. x 20 =the N°. of shillings + 4 x 21 x N°. of shillings = 85 x the number; .. the number of shillings = 20, and the number of guineas = 80; Let x = the value of an adulterated guinea; *, (1700 — 165 =) 1535 = 804 + 20 x value of an adulte- rated shilling, and .*. the value of an adulterated shilling = ee ms 307 — 162 4 3 °, 165 = 9” — 3 — 307 + 162 = 254 — 310; by transposition, 475 = 252, and 19 = #; . the value of an adulterated guinea = 19s. and the value of S07 1007 4 = 9d. loo an adulterated shilling = 90. Before noon, a clock which is too fast, and points to after- noon time, is put back five hours and forty minutes ; and it is observed that the time before shown is to the true time as 29 to 105. Required the true time. Let w = the time the clock pointed to; then # 3a +b 65,53 29.2105; PAlgc 17 Oue) eran co ace Os And: ae 4 20 4s 174 Examples of the Solution of Problems 29 whence #2 = San obo if ., this be added to 6 20’, the true time is 8" 45’, or 15’ before 9. 91. The crew of a ship consisted of her complement of sailors and a number of soldiers. Now there were 22 seamen to every 3 guns and 10 over. Also the whole number of hands was 5 times the number of soldiers and guns together. But after an engagement, in which the slain were one-fourth of the survivors, there wanted 5, to be 13 men to every 2 guns. Required the number of guns, soldiers, and sailors. Let 3@ = the number of guns; then 224 + 10 = the number of seamen, and since, seamen + soldiers = 5. soldiers + 1527; *, the number of soldiers = 1 (22a + 10 — 152) = fs eae shy A ie gt 95x + 50 4 +2097 +10 = — and the complement = ; ; (es + 50 and the survivors were + ) = 19% + 10; 3x “1 +18 — 5 = 192 + 105 ee by transposition, aie and # = 30; *, the number of guns = 90; the number of seamen = 30 X 22 + 10 = 670, 7 xX 30 +10 and the number of soldiers = Ske awe 55. 92. A Shepherd, in time of war, was plundered by a party of soldiers, who took } of his flock, and } of a sheep; another party took from him 4 of what he had left, and } of a sheep; then a third party took 3 of what now remained, and 3 of a sheep. After which i had but 25 sheep left. How many had he at first ? Let # = the number he had at first ; producing Simple Equations. 175 then rit hee = the number the first party took away, _ 3@— 1 CoG bees = the number remaining. Now the second party took away 3 of these + 4 of a sheep; .. there remained (2—*) 3, 2% =! RS ae Rn aha Dee e 4 2 58.6 Ta Abii,” Shao dae then the third party took away half of these + 3 of a sheep; 2—1l », w@-—-3 = 4 .. there remained ; — 3 = 25, x whence and wv — 3 = 100; by transposition, 2 = 103. 93. A man being at play lost + of his money, and then won ) 3 shillings; after which he lost + of what he then had, and won 2 shillings; lastly he lost 1 of what he then had; this done he had but 12 shillings left. What had he at first ? Let 42 = the number of shillings required ; then after the first loss he had 32, and afterwards 3% + 3; after the second loss he had “ . (32 + 3) = 2v + 2, and afterwards 2” + 4. Having lost 4 of this, he had $.(2v + 4) = ae, 122 + 24 & f9, and .*. and 12” + 24=7 xX 123 by transposition, 124 = 5 x 12, and # = 53 .. he had at first 20 shillings. 94. A Trader maintained himself for 3 years at the expense of £50 a year;.and in each of those years augmented that part of his stock which was not so expended by + thereof. At the end of the third year his original stock was doubled. What was that stock ? 176 Examples of the Solution of Problems Let x = the number of pounds required ; : then 2 — 50 =the sum not expended; and with this h traded ; PR ete Uh MS ON Ee his gain the first year, and “ . (v — 50) = the sum he had at the end of the first year ; 40” — 200 4” — 350 50 = 3 second year; _ 4 4@—350 164 — 1400 AER 3 as 9 of the second year ; 16” — 1400 16% — 1850 " and Saas aac atie 50 = aah ieas Yael =the sum he traded with the third year ; _ 4 16” — 1850 e 3 e 9 year ; =the sum he had at the end of the third 4 162% — 1850 whence — . ——_——— = 22, 3 9 and 324” — 3700 = 272; by transposition, 5” = 3700, and w# = 740. 95. A Merchant buys a cask of brandy for £48, and sells a quantity exceeding three-fourths of the whole by 2 gallons at a profit of £25 per cent. He afterwards sells the re- mainder at such a price as to clear £60 per cent. by the whole transaction; and, had he sold the whole quantity at the latter price, he would have gained £175 per cent. Re- quired the number of gallons contained in the cask. Let 4@ = the number of gallons; then = the original price per gallon (in pounds), 12 : 15 and 100 : 125 :: ie the first price of sale = ath Mo ie =the sum he traded with the = the sum he had at the end producing Simple Equations. 177 12 ‘ and 100 : 275 :: a: : the latter price of sale = =; 15 33 36, ie (34 + 2) Seah (@ — 2) -—, — 48 = whole gain = 30 — —-3 36 hence 100 : 60:3: 48 ; 30 — Pe 6 0) es a a xv 30 “. 2425 ——; v oh 30 by transposition, ae ted ee etay and the number required = 4# = 120 gallons. 96. Water flows uniformly into a cistern, capable of containing 720 gallons, through a pipe; and at the same time is dis- charged by a pump, worked by three men, who take four strokes in a minute; but this not being sufficient, the cistern becomes full in 6 hours; they therefore now put in another pump, of such power that the quantity discharged at one stroke by this pump is to the quantity discharged at one stroke by the former :: 2: 3; but being obliged to detach one of their number to work the pump, the former pump makes only 10 strokes in 3 minutes, and the latter 5 strokes in two minutes; by which means the cistern is emptied in 12 hours. How much water was discharged by each pump at one stroke? and how much flowed in through the pipe in one minute? Let 37 = the number of gallons discharged by the first pump at one stroke ; .. 2” = the number discharged by the second, and 12@= the quantity discharged by the first in one minute, when 3 men worked; “. 6 X 60 X 12% = the quantity discharged in six hours ; . 6 X 60 X 12” + 720 = 720. (6” +1) =the quantity in- troduced through the pipe in that time. N 178 Examples of the Solution of Problems, &c. Now when the additional pump is worked, 102 = quantity discharged by the first in one minute ; and 5x2 = the quantity discharged by the second in one minute; *, 15@ X 12 x 60 = the whole quantity discharged in 12 hours; Se 152 Oe 20 — 720 X21 b ao) eos or 152 = 127 + 33 by transposition, 3% = 3, aut 1s *, the first discharged 3 gallons, and the second 2, at one stroke; and the quantity introduced by the pipe in one 720 x (6% + 1) 6 xX 60 minute = =2 x 7= 14 gallons. 97. A poor man with a wife and seven children, found during a scarcity that he could only earn sufficient to procure ; of a white loaf of bread per day for each of his family, himself included. He therefore applied to the parish- officers for assistance, by whom being allowed a daily sum = 3 his earnings, and mixed bread being made by order of Parliament, which was cheaper than white in the pro- portion of 4 to 5, he was now enabled to procure 4 of a mixed loaf per day for each of the family (himself still included) and had is. 73d. over. Required the sum allowed him by the parish. Let # = the price of a white loaf (in pence) ; & e ~ = what he earned, and “= = what the parish allowed him ; also — Hyde a of a mixed loaf; ode pad 9x =) 27.2 Te ee Regen 8 and 96a” + ab = 13523 by transposition, 780 = 392, and! 20'= ‘a; whence it appears that he earned 45 pence, and had 223d. allowed him by the parish. * SECTION VII. Examples of the Solution of Problems producing Simple Equations involving two unknown Quantities. . Arrer 4 had won four shillings of B, he had only half as many shillings as B had left. But had B won six shillings of A, then he would have had three times as many as 4 would have had left. How many had each? Let # = the number of shillings 4 had, and y = the number B had; then y —4= 244 8, andy + 6=3#— 18; *, by subtraction, 10 = &# — 26, and by transposition, 36 = 2, and y—4= 80; .. y= 84; .. 4 had 36, and B 84. . A person bought a quantity of brandy and rum for £19. 4s., and gave for the brandy 9 shillings, and rum 6 shillings per bottle. He found however that he could have bought as many bottles of rum as he now had of brandy, and as many of brandy as he now had of rum for £1. 13s. less. How much was bought ? Let 2 = the number of parties of brandy, and y = the number of bottles of rum; then 9% + 6y = 384, and 67 + 9y = 3513 *, by addition, 1547 + 15y = 735, andw + y= 49. But since 3@ + 2y = 128, and2v+ 2y= 983 *, by subtraction, # = 30, andy =49—w«w= 19; ‘, he bought 30 bottles of brandy, and 19 of rum. N 2 180 Examples of the Solution of Problems 3. What fraction is that, to the numerator of which if 4 be added, the value is one-half, but if 7 be added to the deno- minator, its value is one-fifth ? Let ~ = the fraction required ; = -, and “22 4+8=Y5 x SSO i as ea BI Mme TS TR A by subtraction, 3% —8 =7; by transposition, 3% = 15; and # == 5; ”, y= 27 +8 = 18, and the fraction is “, 4. Find two numbers, the greater of which shall be to the less as their sum to 42, and as their difference to 6. Let x and y = the numbers; then v2: yi: @ +-y : 42, and#:yii:@—y: 6. But ratios which are equal to the same ratio are equal to each other ; w LEYLA U—Y 26, altp oe ayes ee ta Gt ay Ee 5 UAT.V1995 0.) "a. 2 oe Bos and a sea 3s 4y t= — gi? and 4:3::2%:6; 3 ie peas & 4 and @ = “2 = 39; .. the numbers are 32 and 24. 5. What two numbers are those, whose difference, sum, and product, are as the numbers 2, 3, and 5, respectively ? producing Simple Equations. 181 Let # and y = the numbers; then @—y:r+y:: 2 te eta Yaka ts ORE pean Ree alsov+y: «xy 3 or6y: xy 3 raph sane ue 1 anu oa 10; # and y = = = 23 ., the numbers are 10 and 2. 6. A and B playing at bowls, says 4 to B, If you will give me a guinea, I will bet you half a crown to eighteen pence on each game, and will play 36 games together. B won his guinea back again, and £1. 17s. besides. How many games did each win? Let 2 = the number of games A won, and y = the number B won; ~~ Y+tLr= 36, and 3y + 3x2 = 108; but 5y —3% = 116; _.. by addition, sy = 224, and y = 283 . V=36—-—Yy=85 .. 4 won 8, and B 28 games. A person exchanged 12 bushels of wheat for 8 bushels of barley, and £2. 16s.; offering at the same time to sell a certain quantity of wheat for an equal quantity of barley, and £3. 15s. in money, or for £10 in money. Required the prices of the wheat and barley per bushel. Let x = the price of wheat per bushel, in shillings, and y = the price of barley ; then ae — the number of bushels in the second offer : 182 Examples of the Solution of Problems 12” = 8y + 56, 200 and spin pe aa eRe b rs and .%; 2a eG by transposition, 77 = 56, Alay. Oe y= 55 *, the prices of wheat and barley per bushel were 8 and 5 shil- lings, respectively. s. A Vintner sold at one time 20 dozen of port wine, and 30 of sherry, and for the whole received £120; and at another time sold 30 dozen of port, and 25 of sherry at the same prices as before, and for the whole received £140. What was the price of a dozen of each sort of wine? Let v = the price of a dozen of port, andy= --------- of sherry ; *, 207 + 304 = 120, or 27 + 3y = 12, and 30% + 25y = 140, or 642 + 5y = 28. Multiplying the first equation by 3, 6x + 9y = 36, but 62 + 5y = 28; *, by subtraction, 4y = 8,. ANG i= 12 whence 24 = 12 — 3y = 12 —6 = 6, andes 135s *, the prices of port and ere per dozen were £3 and “4 respectively. 9. Aand B severally cut packs of cards, so as to cut off less than they left. Now the number of cards left by 4 added to the number cut off by B make 50; also the number of cards left by both exceed the number cut off, by 64. How many did each cut off? Let v = the number cut off by 4; “. 52 — #2 =the number left by him. producing Simple Equations. 183 Let y = the number cut off by B; *, 52 — y = the number left by him; then # + y = the whole number cut off, and 104 — (vw + y) = the whole number left, whence 104 — 2. (@ + y) = 643 by transposition, 2. (@ + y) = 40, and @ + y = 20. Now 52—2%+ y= 503 .. by transposition, 7—y = 2, but 7+ y = 20; .. by addition, 2” = 22, and 2# ==,11.; by subtraction, 2y = 18, and tir Oi; .. A cut off 11, and B 9. 10. A countryman, being employed by a poulterer to drive a flock of geese and turkeys to London, in order to dis- tinguish his own from any he might meet on the road, pulled 3 feathers out of the tail of each turkey, and one out of the tail of each goose, and upon counting them, found that the number of turkeys’ feathers exceeded twice those of the geese by 15. Having bought 10 geese and sold 15 turkeys by the way, he was surprised to find, as he drove them into the poulterer’s yard, that the number of geese exceeded the number of turkeys in the proportion of 7 to 3. Required the number of each at first. Let # = the number of turkeys ; *, 3@ = the number of feathers from their tails; let y =the number of geese; and .*. of the feathers, and 3% — 2y = 15. Alsoy +10: @%—15:37: 33 * 3y + 30=72 — 105; by transposition, 77 — 3y = 135, and 14% — 6y = 270; but from the first equation, 97 — 6y = 45; 184 Examples of the Solution of Problems .. by subtraction, 54 = 225, Ania ie= 4515 * 2Y = 3@ — 15 = 135 — 15 = 120, and y = 60; .. there were 45 turkeys, and 60 geese. 11. A Farmer with 28 bushels of barley at 2s. 4d. a bushel, would mix rye at 3 shillings per bushel, and wheat at 4 shillings per bushel, so that the whole mixture may con- sist of 100 bushels, and be worth 3s. 4d. per bushel. How many bushels of rye, and how many of wheat, must he mix with the barley ? Let # = the number of bushels of rye, and y = the number of wheat ; then the value of the barley = 196 (fourpences), of the wheat = 12y, of the tye 0m. *. 196 + 9” + 12y = 100035 by transposition, 9@ + 12y = 804, and 37 + 4y = 268. Now 2 + y + 28 = 100, and by transposition, #7 + y= 72; *,.3@ + 37 = 216, but 3a + 4y = 268; .. by subtraction, y = 52, and.#. = 72.— 4 = 20. Hence he must mix 20 bushels of rye, and 52 of wheat. 12, A and B speculate with different sums; 4 gains £150, B loses £50, and now 4’s stock is to B’s as 3 to 2. But had A lost £50, and B gained £100, then A’s stock would have been to B’s as 5 to 9. What was the stock of each? Let # = A’s stock, and y = B’s; then 7+ 150: y—502:3 : 2; *, 2@ + 300 = 3y — 150, and by transposition, 3y — 2v = 450; producing Simple Equations. 185 alsov —50 : y+ 100:35 : 93 “. 9@ — 450 = 5y + 500; by transposition, 94 — 5y = 950; multiplying this equation by 3, and that found above by 5, 272 — 15y = 2850, and 15y — 10% = 2250; .. by addition, 172 = 5100, and # = 300; *, 3y = 2a” + 450 = 1050, and y = 3503 .. A’s was £300, and B’s £350. 13, A Merchant having mixed a certain number of gallons of brandy and water, found that if he had mixed 6 gallons more of each, he would have put into the mixture 7 gallons of brandy for every 6 of water; but if he had mixed 6 less of each, he would have put in 6 gallons of brandy for every 5 of water. How many of each did he mix? Let # = the number of gallons of brandy, and y = the number of gallons of water ; thenv?+6:y+6::7 : 6; » (Alg. 179,°5: arb Glee Yt Fe X15 but@#@—6:3: y—6:36: 5, and .. v—y i: v%@—6:i1: 6, and sincea? +6: ¥—yYyii:7: 13 * ex equal, v +6: 7—6::7 : 6; AN 7 01D) ny 1S, 7, OG ds aon. rls) les °, V=78;5 whence 84 : y¥+6::7 : 6, OFe 1 2p) Paths, Lak 6:5 oY t6=72, and y = 66; .. he mixed 78 gallons of brandy with 66 of water. 14, A person had a bag of money worth £93; but a servant having robbed him of one-sixth of his moidores, and three- 186 Examples of the Solution of Problems fifths of his guineas, left him only £54. 15s. How many moidores and guineas had he at first ? Let # = the number of guineas, and y = the number of moidores ; 45y | 142 then ive + Ee = 365, and .*. 75y + 28@ = 3650; also7#@ + 9Y = 620; *, 36y + 28” = 24805 but since 75y + 28a” = 3650; .. by subtraction, 39y = 1170, and 4) — "30 also 7# = 620 — 9y = 620 — 270 = 3503 Neel .. he had 50 guineas, and 30 moidores. 15. A Vintner bought 6 dozen of port wine and 3 dozen of white for 12 guineas; but the price of each afterwards falling a shilling per bottle, he had 20 bottles of port, and 3 dozen and 8 bottles of white more, for the same sum. What was the price of each at first ? Let # = the price of the port Y=-------- white then 724% + 36y = 252 = 924 + 80y — 172, and .*. by transposition, 20” + 44y = 172, or 5@ + 11y = 43. Now, since 722 + 36y = 252; 7 +y=7, and 224% + 11y = 77, but 52 + lly = 43; .. by subtraction, 177 = 34, and 7 == 93 whence y=7— 22 =7—4=3;3 .. the price of port was 2s. and of white 3s. per bottle. per bottle (in shillings), 16. A rectangular bowling-green having been measured, it was observed, that if it were 5 yards broader, and 4 yards longer, producing Simple Equations. 187 it would contain 116 yards more: but if it were 4 yards broader, and 5 yards longer, it would contain 113 yards more. Required the length and breadth. Let # = the number of yards in length, and y = the number of yards in breadth ; then (7 +4). (y+5)=vy+5%7+4y+20=116+ zy, and .°. 5@ + 4y = 96; also (vw +5). (y¥+4)=avy+4e74+ 5y+20=113+ ay; p 42 + 5y = 933 | multiplying the former equation by 4, and the latter by 5, 20” + 16y = 384, and 20” + 25y = 465; by subtraction, 9y = 81, SNC tae « 42 = 93 — 5y = 93 — 45 = 48, aNd == 412, .. the length was 12, and the breadth 9 yards. 17. Find two numbers in the proportion of 5 to 7, to which two other required numbers in the proportion of 3 to 5 being respectively added, the sums shall be in the proportion of 9 to 13; and the difference of those sums = 16. Let 5x and 72 = the two first numbers, and 3y and 5y = the others; then 5% + 3y:7@ 4+ 5y 3:9: 133 be + 3y 22% +24 129543 or 52 + :3y 2 @ + ytt92 2, and 102 + 6y = 9x + 9Y; by transposition, v7 = 34; but 22 + 2y = 16; *, (6y + 2y =) 8y = 16; ¥ = 2; and @ =='65 whence, the two first numbers are 30 and 42; the two others, 6 and 10. is. A Merchant finds that if he mixes sherry and brandy in quantities which are in the proportion of 2 to 1, he can sell 188 Examples of the Solution of Problems the mixture at 78 shillings a dozen; but if the proportion be as 7 to 2, he must sell it at 79 shillings a dozen. Re- quired the price of each liquor. Let # = the price of the sherry y = the price of the brandy then 27 + y=3 X 78 = 234, and 7@ + 2y=9 xX 79 =711; but the first equation being multiplied by 2, per dozen ; | 40 +2 = 608 .. by subtraction, 3% = 243, and 2 = 81; whence, y = 234 — 2” = 234 — 162 = 72; .. the price of the sherry was 81s., and of the brandy 72s. 19. A Corn-factor mixes wheat-flour, which costs him 10 shil- lings a bushel, with barley-flour, which costs him 4 shil- lings a bushel, in such proportion, as to gain 433 per cent., by selling the mixture at 11 shillings a bushel. Required the proportion. Let the proportion be v: y; then 102 + 4y = the cost of # + y bushels, and 1142 + 11y = the selling price; a. 2 7 ene earn = whence, LO®. by 5 ARYA 100 eh Se A007 Seno es and 5@ + 2y:@+7y::8:73 *, 352 + 14y = 8H + 56Y;3 by transposition, 27v7 = 42y, and 97 = 1443 a pega MTL 4108 and .*, he must mix 14 bushels of wheat-flour with 9 of barley. 20. A number consisting of 2 digits when divided by 4, gives a certain quotient and a remainder of 3; when divided by 9 gives another quotient and a remainder of 8s. Now the value of the digit on the left hand is equal the quotient which was got when the number was divided by 9; and producing Simple Equations. 189 the other digit is equal =th of the quotient got when the number was divided by 4. Required the number. Let # and y = the digits in order ; then 10x + y = the number, dese Sia Ba 9 . 10% + y=—9H + 8; by transposition, 7 + y = 8; LOD te Yen 5 4 107 +y =3 + 68y; by transposition, 102 — 67y = 3; but from the preceding equation, 107 + 10y = 80; .. by subtraction, 77y = 77, 8 v+-3 0:7 also + 17Y3 and y = 1; ~~ &=—s— Y = 15 and the number required is 71. 21. A man and his wife could drink a barrel of beer in 15 days. After drinking together 6 days, the woman alone drank the remainder in 30 days. In what time would either alone drink a barrel ? Let 2 = the number of days in which the man could drink it, and y = the number in which the woman could drink it; then Ls + Ee ts x and 2 + i oper Ug tall 6 36 OF =r = S Cove hence from the first equation, pire aes eb Dees 15 from the last, yal Oe ; wv RONG 190 Examples of the Solution of Problems es | 1 3 1 .. by subtraction, yi ce hipin eae OR and?.", y= 505 ] 1 1 1 ") also —- = — — —- = — — — = —; 0° 16 aoe eee and .°. 2 = 21-3; .. the man would drink it in 2143, days, and the woman in 50 days. 22. A purse holds 19 crowns and 6 guineas. Now 4 crowns and 5 guineas fill ~. of it. How many will it hold of each ? Let # = the number of crowns, and y = the number of guineas ; then # : 1%: 4 : the space occupied by 4 crowns = =. In the same way, the space occupied by 5 guineas = 4 As. Ghest 117, ee v “1 y — 63° dee x The first equation being multiplied by 6, and the second by 5, # vat § 30 a pee —=5; . ; 2471 .. by subtraction, — = 5 —— = —; oan et 7. 2 = 2; or: ] 4 5 ap ML NO EN ear . ¥Y = 635 .. the purse would hold 21 crowns, or 63 guineas. 23. 24, producing Simple Equations. 191 Some smugglers discovered a cave, which would exactly hold the cargo of their boat, viz. 13 bales of cotton, and 33 casks of rum. Whilst they were unloading, a custom- house cutter coming in sight, they sailed away with 9 casks and five bales, leaving the cave two-thirds full. How many bales or casks would it hold? Let vw = the number of bales, and y = the number of casks ; 3 then : Ha — "35 55 11] He eee @ y 3 : 16 2 .. by subtraction, — = = 7. 20 = 48, and @ = 24; 9 1 5 1 5 3 1 consequently = = — — -=>—~— —=—=-; Gea eta to 24 94 8 and y = 72; and .*. the cave would hold 24 bales, or 72 casks. Round two wheels, whose circumferences are as 5 to 3, two ropes are wrapped, whose difference exceeds the differ- ence of the circumferences by 280 yards. Now the larger rope applied to the larger wheel wraps round it a certain number of times, greater by 12 than the smaller round the smaller wheel; and if the larger wheel turns round 3 times as quick as the other, the ropes will be discharged at the same time. Required the lengths of the ropes and the circumferences of the wheels. 192 Examples of the Solution of Problems t Let 52 and 32 =the circumferences of the wheels (in yards) ; - then 22 + 280 = the difference of the ropes, and (15” : 3@::) 5:13: the length of the longer string : the length of the shorter. Let .*. 5y and y = the lengths of the ropes (in yards) ; then 4y = 2% + 280, OY ay and ~ = 3, 1 135 or by transposition, =o it 133s 32 pee Wb Fe and.) Foe — O77. 1950) by transposition, 70” = 280, . anda = = 723 | .. the circumferences of the wheels are 20 and 12 yards, and the length of the strings 360 and 72 yards. 25. Three guineas were to be raised on two estates, to be charged proportionably to their values. Of this sum, /’s estate, which wes 4 acres more than B’s, but worse by 2 shillings an acre, paid £1. 15s. But had 4 possessed 6 acres more, and B’s land been worth 3 shillings an acre less, it would have paid £2. 5s. Required the values of. the estates. Let # = the number of acres B had; then # + 4 = the number 4 had. Let y = the value of an acre of A’s land; “. y + 2 = the value of an acre of B’s, and (@ + 4).y:@.(y + 2) 33.355 28 335343 . 4vy + 1l6y = 52y + 10a, and #y = 16y — 102. Again, (@ + 10).y:@.(y —1) 2:45:18 2:5: 2; 7 2@y + 2y = 5xry — 52, and by transposition, 3v7y = 20y + 5a, whence 48y — 30% = 20y + 52; producing Simple Equations. 193 by transposition, 28y = 352, and 4y = 5a, which value substituted in the first equation, gives 26. LY = 210% — 10% = 102; EMAs ee ead Oy 42 whence # = “2 ss .. the value of 4’s estate = (v + 4). y = 120 = £6; and the value of Bs = xv. (y + 2) = 96 = £4. 16s. A coach set out from Cambridge to London with a certain number of passengers, 4 more being on the outside than within. Seven outside passengers could travel at 2 shil- lings less expense than 4 inside. The fare of the whole amounted to £9. But at the end of half the journey, it took up 3 more outside and one more inside passengers ; in consequence of which the fare of the whole became in- creased in the proportion of 17 to 15. Required the number of passengers, and the fare of the inside and outside. Let # = the number of inside “. & + 4 = the number of ey eee and y = the fare of an outside passenger ; - I Sia ile = * = the fare of an inside passenger, and ry 28 +y.(@+ 4) = 180. - + ne * = the fare of the passengers taken up half- _ ly +2, 8 1oy +2 oy ae nes a is 8 19y+2 7 - = 94, and 19y +2 = 192; 194, Examples of the Solution of Problems *, by transposition, 197 = 190, and y = 10; 702 + 22 . from the first equation, pase + 10.(# + 4) = 180; by transposition, 287 = 140, and sue 5% *, there were 5 inside, and 9 outside passengers, and the fares were 18 and 10 shillings, respectively. 27. In one of the corners of a rectangular garden there is a fish-pond, whose area is one-ninth part of the whole garden; the periphery of the garden exceeding that of the fish-pond by 200 yards. Also if the greater side be increased by 3 yards, and the other by 5 yards, the garden will be enlarged by 645 square yards. The fish-pond is a rec- tangle about the same diameter with the garden. Required the periphery of the garden, and the length of each side. Let x = the length of the lesser side, and y = the length of the greater ; ; and i = the lengths of the lesser and greater sides of the fish-pond, (Hucl. B. vi. Prop. 24.) Also 2.(#@ + y) = the periphery of the garden ; and Pete = the periphery of the fish-pond ; 2 2.(~+y)— >-(@ + y) = 200, or —. (e+ y) = 100; Se) “. &@ + Yy = 150. Also (y + 3).(v@ + 5) = avy + 645; .. by transposition, 3% + 5y = 630, but from the former equation, 3v% + 3y = 450; *, by subtraction, 2y = 180, and y = 90; “. &= 150 —y = 60; and .*, the periphery = 300 yards, and the sides are 60 and 90 yards, ) | 98. producing Simple Equations. 195 A and B each bought £300 into the stocks, 4 into the three per cents., and B into the fours. These stocks were at such a price that B received one pound interest more than A. When afterwards each of the stocks rose 10 per cent., they sold out their money, and 4 found himself £10 richer than B. Required the prices of the stocks. Let x = the price of the three per cents., and y = the price of the fours; per 300 143. 4 8 INLeres, == — : 1200 and y : 300:: 4: B’s interest = mare 900 1200 +1 eee Again, 2 : # + 10 :: 300 : what 4 received when selling m 300. (xv + 10). ae) bo co . Hi 300 . (y + 10) and in the same way B received si ; “ 300 . (# + 10) Balas (y + 10) + 10, vo y 300 300 and 30 + — = 30 + —d¢41; v y 300 300 oe Su hiss GS wee a i. | y /and — sane a (a -) ae from the first equation ; | oe 300 .. by transposition, 4 = ay and y = 75; 7 * 300 pages and consequently, 7 = 60; *, the prices of the stocks were 60 and 75 per cent. £500 was to be lent out at simple interest in two separate sums, the smaller at 2 per cent. more than the other. The interest of the greater sum was afterwards increased, and 02 196 Examples of the Solution of Problems, &c. that of the smaller sum diminished by 1 per cent. By this, the interest of the whole was augmented by one-fourth of the former value. But if the interest of the greater sum had been so increased, without any diminution of the less, the interest of the whole would have been increased one- third. What were the sums and the rate per cent. of each? Let x = the less sum; *, 500 — # = the greater; let y + 1 = the interest of the less; .. y — 1 = the interest of the greater, BY 11) SOO eae 100 100 former interest, vy | (500—x2).¥ v latch SY Gree aaa and —~ + ——____-* = 5y = the second interest ; 100 100 sy oe Gan 5) art BANE aivtas L ancik, AYE Sabah Yap by transposition, y = 5 — ae 50 Again, the third interest = Yin) (500 ew = 100 100 a e “100 + 5Y5 oe Focus -3=4. Pry a: ays ’ or ati ls she + 20 20 100 Lie 100 y a by transposition, (20 nee =) 20 en °F eee 100 0 from the former equation ; o Sa —_ 5, 20 and #2 = 100; alsoy +1 =6——=4; 50 .. the sums were 100 and 400 pounds, and the rates of interest £4 and £2, respectively. 1. 2. SECTION VIII. Examples of the Solution of Problems producing pure Equations. Wuart two numbers are those, whose sum is to the greater as 10 to 7; and whose sum multiplied by the less produces 270? Let 10” = their sum; .. 72 = the greater number, and 32 = the less; whence 302” = 270, AV a ie a0 ete te and the numbers are + 21 and + 9. There are two numbers in the proportion of 4 to 5, the dif- ference of whose squares is 81. What are those numbers? Let 47 and 54 = the numbers; iNetal2og 160) — 0a. — i f= 9, and. 7) == ct. 3; and the numbers are + 12 and +15. What two numbers are those, whose difference is to the greater as 2 to 9, and the difference of whose squares is 128° Let 2a = their difference ; . 9¢ — the oreater, and 7x” = the less; ““e(8la" — "40a =) 3227 = 198, and! 2 43 . @&@= 4 2; and the numbers are + 18 and + 14, 198 Examples of the Solution of Problems 4. A Mercer bought a piece of silk for £16. 4s.; and the num- ber of shillings which he paid for a yard was to the number of yards as 4:9. How many yards did he buy, and what was the price of a yard? Let 44 = the number of shillings he paid for a yard; .. 9%” = the number of yards, and 362° = (the price of the whole =) 324; SQ MAYS (A pe wl) consequently there were 27 yards, at 12s. per yard. 5. It is required to divide the number 1s into two such parts, that the squares of those parts may be in the proportion of 25 to 16. Let # = the greater part; then 18 — x = the less; ve Ga) cba et) ce oes and (Aig. 179,/9.) @ 3.18 —#@ 215 2 43 re We, LO Mate COME as and ys: M2 25 ea Sof FV and the parts are 10 and 8. _ ; ‘ 2 3 6. Find three numbers in the proportion of >3 and me: the sum of whose squares is 724. Reducing the fractions to a common denominator, the required numbers will evidently be in the proportion of 6, 8, and 9; let .. 62, 8%, and 92, represent the numbers; then (362° + 64”? + 814° =) 1814” = 724; CE gine and =a, and consequently, the numbers are + 12, + 16, and + 18. 7. It is required to divide the number 14 into two such parts, that the quotient of the greater part divided by the less, producing pure Equations. 199 may be to the quotient of the less divided by the greater as 16: 9. Let 2 = the greater part ; *. 14 — # = the less, & 14—2@ and : AG ee 14—w & or 2": (14— #)7 3216 : 93 ana (Alg. 179, 9.) v 2 14 aE, Xv ara 4 3 35 ania bn tens orn7s wa water 1 and # = 8; .. the parts are 8 and 6. 8. What two numbers are those, whose difference is to the less as 4 to 3; and their product multiplied by the less is equal to 504? Let 44” = the difference ; then 3x = the less, and 7x# = the greater; whence 632° = 504, OV 35 Pe iam uas and the numbers are 14 and 6. 9. What two numbers are as 5 to 4, the sum of whose cubes is 5103 2 Let 5@ and 47 = the numbers; >, (25a? - G4e* ==) 1894" == 5103, UNE ye Neamt ae and the numbers are 15 and 12. 10. A number of boys set out to rob an orchard, each carrying as many bags as there were boys in all, and each bag capable of containing 4 times as many apples as there were boys. They filled their bags, and found the number of apples was 2916. How many boys were there? 200 Examples of the Solution of Problems Let # = the number of boys; then 2 = the number of bags, and 4° = the number of apples; *\ 47 == 100 16! and x* = 729; oe V=9;5 .. there were 9 boys. 11. A person bought for one crown as many pounds of sugar as were equal to half the number of crowns he laid out. In selling the sugar he received for every 25 lbs. as many crowns as the whole had cost him; and he received on the whole 20 crowns. How many crowns did he lay out, and what did he give for a pound? Let 22 = the number of crowns he laid out; *, # =the number of lbs. for one crown. and 2x” = the number of lbs. in all, 2a. ° . and te the selling price of one lb. ; 5 eed ‘3 22 (X —— = 20, 25 anda = 125; whence 7 = 5; .. he laid out 10 crowns, and gave one shilling for a lb. 12. A detachment from an army was marching in regular column with 5 men more in depth than in front; but upon the enemy coming in sight, the front was increased by 845 men; and by this movement the detachment was drawn up in five lines. Required the number of men. Let x = the number in front; *, 2 + 5 = the number in depth, and a + 5@ = 5x + 4225, Or ap we RSI andi’ ==!-- 65% .. the number of men = 5” + 4225 = 4550, the negative value not answering the conditions of the problem. 13. 14. producing pure Equations. 201 A number of shillings were placed at equal distances on a table, so as to form the sides of an equilateral triangle; then from the middle of each side a number of shillings, equal to the square root of the number in the side, were taken, and placed upon the corner shilling opposite to that side; it then appeared that the number on each side was to the number previously upon it, as 5 to 4. Required the num- ber of shillings on one side at first. Let x2? = the number; then gee, 32 ont) 4. Vat ert Rage fabhia fake sey a oe. and a= 4; whence x’ = 16 = the number required. A certain sum of money is divided every week among the resident members of a corporation. It happened one week that the number resident was the square root of the num- ber of pounds to be divided. ‘Two men, however, coming into residence the week after, diminished the dividend of each of the former individuals £1. 6s. sd. What was the sum to be divided ? Let x’ = the number of pounds; then # = the number of men resident, and also = the sum each 15. received. 4 xr? Hence 2 ——- = : 3 GL -- 2 ora? +2e4— 2a"; 3 3 2 Ws 2 8 by transposition, Re. and i 45 .°, 2 = 16 = the sum required. Two partners 4 and B dividing their gain (£60), B took £20; A’s money continued in trade 4 months, and if the number 50 be divided by A’s money, the quotient will give 202 Examples of the Solution of Problems the number of months that B’s money, which was £100, continued in trade. What was A’s money, and how long” did B’s money continue in trade? Suppose 4’s money was # pounds; 50 : Oar the number of months B’s money was in trade, and since B gained £20, 4 gained £40; 5000 AU Pe re hae, Melts x 2500 One fs PE x “5 (ee P 25005 and a= + 50; .. A’s money was £50, and B’s money was one month in trade. 16. Two workmen 4 and B were engaged to work for a certain number of days at different rates. At the end of the time, A who had played 4 of those days, had 75 shillings to re- ceive; but B who had played 7 of those days, received only 48 shillings. Now had B only played 4 days, and 4 played 7 days, they would have received exactly alike. For how many days were they engaged; how many did each work, and what had each per day? Let =the number of days for which they were engaged ; *, # —4= the number 4 worked, and # — 7 = the number B worked, and Pa = the number of shillings A received per day; “ ae the number of shillings B received per day ; and # _ 75.(@—7) 48. (%—4) ee ne ee ee and 25. (w— 7)’ = 16. (~# — 4)’; “5. (@—7) = a4. (vw —4); ] 17 producing pure Equations. 203 and .*. they were engaged to work 19 days, and 4 worked 15, and B 12 days, and 4 received 5 shillings, and B 4 shillings per day. 17. Two Travellers A and B set out to meet each other, A leaving the town C at the same time that B left D. They travelled the direct road CD, and on meeting it appeared that 4 had travelled 18 miles more than B; and that A could have gone B’s journey in 152 days, but B would have been 28 days in performing 4’s journey. What was the distance between C and D? Let # = the number of miles 4 has travelled ; *, #2 — 18 = the number B has travelled, andx— 18 : #::152 : the number of days A travelled x 632 : 4. (#@ — 18)’ also wv : x—18:: 28 : the number of days B travelled 28 . (” — 18) , 3 L 98. (@—18) Goat ar x 4 (@ — 18)’ or 16. (# — 18)’ = 92"; 4. (@— 18) = 3a, anit 2 =—179,.0F 102; whence A travelled 72, and B 54 miles; and .*. the whole distance CD 126 miles. is. 4 and B lay out some money on speculation. A disposes of his bargain for £11, and gains as much per cent. as B lays out; B’s gain is £36, and it appears that 4 gains four times as much per cent. as B. Required the capital of each. Let 4@ = B’s capital, and .*, A’s gain per cent. ; then # = B’s gain per cent., and 100) 742135 2736; 204 Examples of the Solution of Problems Syeaie t= 736 eX 1003 BNC O91 KA1008 . &= + 30, and .*. B’s capital = 120, and 220 : 100 :: 11 : 4’s capital = ic: ~~ oe i9. The Captain of a privateer descrying a trading vessel 7 miles ahead, sailed 20 miles in direct pursuit of her, and then observing the trader steering in a direction perpen- dicular to her former course, changed his own course so as to overtake her without making another tack. On comparing their reckonings, it was found that the privateer had run at the rate of 10 knots in an hour, and the trading vessel at the rate of s knots in the same time. Required — the distance sailed by the privateer. Let A, B, be the original places of the privateer and trader, E the point of concourse, D the place where the A B D 0 | ay captain changed his course, CE being perpendicular to AC. Ae ee) — 120 Now (10: 8 ::) 5: 4:3: the velocity of the privateer : the velocity of the trader tt (ADS Oe oO ee Ds iat fea oh == = 163 oC S16 Die 1b io , and D iieG is 205 4A. Ce C Meee (9+ 2) 2B 654, Bi Der eat ees kG oh ON a ORS LG COr= bpanda she, and .. DE =5,and AD + DE = 25. producing pure Equations. 205 20. A Vintner draws a certain quantity of wine out of a full vessel that holds 256 gallons; and then filling the vessel with water, draws off the same quantity of liquor as before, and so on, for four draughts, when there were only 81 gallons of pure wine left. How much wine did he draw each time? Let # = the number of gallons drawn the first time ; .. 256 — x = the quantity of wine left, and 256 : 256 —wx#:: # : the quantity of wine drawn the x. (256 — 2) | second time = ; 256 v.(256—2@ 256 — wv)’ . 256—2— @ (256 #) = 2565 ek = the 256 256 quantity left after the second draught. In the same way, Eee 2) . 2 = the quantity drawn the third time, (256 — @)° and "6 = the quantity left, 56 — 2° 256 — x)" ye and ( : =) .x and a = the quantities drawn 256 and left the fourth time ; and 256 — v = 256|4 x 3 = 64 xX 3 = 192; .. by transposition, 64 = 2, and the quantities drawn off each time were 64, 48, 36, and 27 gallons. 21. What two numbers are those, whose difference multiplied by the greater produces 40, and by the less 15? Let z= the creater, and y = the less; 2? — ry = 40, and ay —y’ = 15; 206 Examples of the Solution of Problems l bo es “. by subtraction, 2 — 2vy + y’ andw—y =+5; .. from the first equation, + 5% = end 2g) == 86 and from the second equation, + 5y = 15, .. the numbers are + 8, and + 3. 22, What two numbers are those, whose difference multiplied by the less, produces 42, and by their sum 133? Let # = the greater, and y = the less; . (@—y) y= 42, and (7 — y).(@ + y) = 1333 .. by subtraction, (v — y) .# = 91, subtracting the first equa- tion from this, (v7 — y).(7— y) = 49; or (v — y)” = 49; 7. &@—y = Eis whence + 7y = 42, and y = + 6, and @ == 7+ -y St 43, .. the numbers are + 13, and + 6. 23. In a mixture of rum and brandy, the difference between the quantities of each is to the quantity of brandy as 100 is to the number of gallons of rum; and the same differ- ence is to the quantity of rum as 4 to the number of gallons of brandy. How many gallons are there of each? Let # = the number of gallons of rum, and y = the number of gallons of brandy ; $2) gemmed ot A) sted OO ently and. 2 2.2@—yiiy 3 43 *. CL COUGH. ae Ms 2G wee, and 2” = 254’; “. &= + 5y, the negative value not answering the conditions of the problem. =a producing pure Equations. 207 Now from the second proportion 5y : 4y :: y : 43 Sy) ai nas Wea sete Uf by. = 8, ander =.95 ; .. there are 25 gallons of rum, and 5 of brandy. 24. What two numbers are those, whose difference being mul- tiplied by the greater, and the product divided by the less, quotes 24; but if their difference be multiplied by the less, and the product divided by the greater, the quotient is 6? Let v = the greater, and y = the less; a then (7 — y) . . ae Bey ee eae: and (wv — y) Smee dividing the first equation by the second, a ie ene Nae y or #7 = + 2Y, and .*. in the first case, (v7 — y =) y = 12, and # = 24; but if# =—2y; —3y x —2=24; ey == 4, ANd 2 = — 8. 25. It is required to find two numbers such, that the pro- duct of the greater and square root of the less may be equal to 48, and the product of the less and square root of the greater may be 36. Let x’ and y’ be the two numbers ; “a yf = ee ie hoe 48 36 . ore (vy =) ay? 208 Examples of the Solution of Problems 3 32 whence = 48, and. = .64- Re Rea and consequently, y = 3; .. the numbers are 16, and 9. 26. Find two numbers such, that the square of the greater — multiplied by the less may be equal to 448, and the square of the less multiplied by the greater may be 392. Let 2 = the greater, and y = the less; then a’y = 448, and vy* = 392; 448 392 e MY) 7 OF fe 555 sane 8 _%1=—, 7? 8 3 and consequently, =< = 392, 3 or = 493 eri == 3443; Ally 27s “. C= 8; .. the numbers are 8, and 7. 27. A and B carried 100 eggs between them to market, and each received the same sum. If A had carried as many as B, he would have received 1s pence for them, and if B had taken only as many as 4, he would have received only 8 pence. How many had each? Let # = the number 4 had, and y = the number B had; then fs = the price of one egg of A’s (in pence), 8 and == the price of one of B’s; producing pure Equations. 209 18V By yo ANG OW 44/5 “. 3% = + 2y, the negative value of which will not answer the 28. . 29. conditions of the problem. Now (e+ y=)x@+ <= 100 3 *, (20 + 34 =) 5% = 200, and # = 40; AILS 4 ts == GO, What two numbers are those, which being both multiplied by 27 the first product is a square, and the second the root of that square: but being both multiplied by 3, the first product is a cube, and the second the root of that cube? Let xv and y be the numbers ; there (372 2, ey ANGs so = 27 Ys also (/ (3%) = 3y, TOM digo ees TE whence 97° = 274’, and y= 3; 8 = WY = 2435 .. the numbers are 243, and 3. It is required to find the three sides of a right-angled triangle from the following data. The number of square feet in the area is equal to the number of feet in the hypo- thenuse + the sum in the other two sides; and the square described upon the hypothenuse is less than the square described upon a line equal in length to the two sides, by half the product of the numbers representing the base and area. Let « = the number of feet in the altitude, and y = the number in the base; “. / (2? + y’) =the number in the hypothenuse, (Eucl. B. 1. Pp. 47.) iy 210 Examples of the Solution of Problems and “2 = the area; aye VfeVty)\+et+y; also + y?={(@+y)— pry =sa + ery t+y—4qry; .. by transposition, vy’ = 227y, BDCsy—— iss hence from the first equation, 47 = ,/ (w + 64) + 4+ 8, and by transposition, 3% — 8 = 4/ (wv + 64); “. 92? — 48% + 64 = 2 4 64, and 82" = 484; “ C= 63 whence the hypothenuse = 4/ (64 + 36) = 10; .. the sides are 6, 8, and 10 feet, respectively. 30. A Farmer has 2 cubical stacks of hay. The side of one is 3 yards longer than the side of the other; and the dif- ference of their contents is 117 solid yards. Required the side of each. Let x = the side of the greater, and y = the side of the less; i ee Pd and#w—y =3; cubing the latter equation, v° — 3a°y + 3vy? — y*° = 273 but 2° —y = lle “. by subtraction, 3a’°y — 3xy? = 90, and wy . (vw — y) = 30, OM Di = sie *. vy = 10. Now 2 —2%7y +7’ =9, and 4@y ==i40 § .. by addition, v’ + 27y + y’ = 49, and 2 + y= 7; but v—y=3; .. by addition, 22 = 10, or — 4; “. &@= 5, or — 2, and by subtraction, 2y = 4, or — 10; “y= 2,0r — 5, and the sides of the stacks are 5, and 2 yards, respectively. i producing pure Equations. 211 31. When a parish was enclosed, the allotment of one of the proprietors consisted of two pieces of ground; one of which was in the form of a right-angled triangle; the other was a rectangle, one of the sides of which was equal to the hypothenuse of the triangle, the other, to half the greater side; but wishing to have his land in one piece, he exchanged his allotments for a square piece of ground of equal area, one side of which equalled the greater of the sides of the triangle which contained the right angle. By this exchange he found that he had saved ten poles of railing. What are the respective areas of the triangle and rectangle; and what is the length of each of their sides ? Let 2% = the greater side of the triangle, and y = the less; “. «/ (42° + y’) = the hypothenuse; and also the greater side _ of the rectangle, and # = the less side of the rectangle ; *, wy = the area of the triangle, and x,/ (4z° + y’) = the area of the rectangle ; 47 =ay +e /(42'?+y’), or da — y= o/ (40 +9) alsosw+10=27+y+/(4a+y’) +20 + 2/ (42? +’), or 44% + 10 =y + 34/ (42° + y’); in which equation substi- tuting the value of ,/ (42? + y’) found above ; .4@+10=y+3 (47—y) = 127 — 24; .. by transposition, 2y = 8% — 10, and y= 42% — 5; .. from the first equation, 5 = 4/ {42 + (4x — 5)’, and 25 = 44° + 16x” — 40% + 25; | by transposition, 40” = 202”; | 2 a, | and y= 47 —5=3; _.. the sides of the triangle are 3, 4, and 5; the sides of the rectangle are 2, and 5; and the areas of the triangle and | rectangle are 6, and 10, respectively. | P2 SECTION IX. Examples of the Solution of Problems producing Adfected Quadratic Equations. | 1. A Mercuant sold a quantity of brandy for £39, and gained as much per cent. as the brandy cost him. What was the price of the brandy ? Let x = the price of the brandy ; : oh then J00.> #5. 7s the, cain jaa and .°. or 2 = 3900 — 10023 by transposition, 2 + 100% = 3900, completing the square, x? + 100% + (50)? = 3900 + 2500 = 6400; extracting the root, v + 50 == 80; .. @ == 30, or — 1305 .. the price was £30. 2. ‘There are two numbers whose difference is 9, and their sum multiplied by the greater produces 266. What are those numbers ? Let # = the greater; “. & — 9 = the less, and xv. (24 — 9) = 266; plat ao 266 -@—-.¢@ = —; 2 2 j 9 81 266 81 2209 completing the square, 2? — —# + — = — 4+ —=-——; P 8 4 2 GF at6 2 16 1608 ; 4 extracting the root, v — = + ~, Examples of the Solution of Problems, &c. 213 19 “ @= 14, or — 3 oF arts Bre aA ae and both values answer the conditions of the problem. 3. It is required to find two numbers, the first of which may be to the second as the second is to 16; and the sum of the squares of the numbers may be equal to 225. Let # = the first number ; *, «/ (162) ='the second, and xv + 16” = 225; completing the square, 2? + 16% + 64 = 225 + 64 = 289; extracting the root, v7 +8=+17; . © = 9, or — 25; but as this latter value of « makes the second number an impossible quantity, 9 is the only value of # which answers the conditions, and therefore the numbers are 9 and 12. 4. Bought two sorts of linen for 6 crowns. An ell of the finer cost as many shillings as there were ells of the finer. Also 28 ells of the coarser (which was the whole quantity) were at such a price that 8 ells cost as many shillings as 1 ell of the finer. How many ells were there of the finer, and what was the value of each piece? Let # = the number of ells of the finer ; *, x = the price of the finer (in shillings), e Ta and 8 : 28 :: 2: the price of the coarser = 758 x et 30; 2 : 49 49 529 complet the re, a = 330 bE = pieting Square, v + Be ar 7 + ma rie . if 23 extracting the root, # + a= + a3 15 and w= 4, or ——, 214: Examples of the Solution of Problems and .*, the price of the finer = 16 shillings, and of the coarser = 14 shillings. 5. Two partners A and B gained £18 by trade. .d’s money was in trade 12 months, and he received for his principal and gain £26. Also B’s money, which was £30, was in trade 16 months. What money did 4 put into trade? Let « = the number of pounds he put in; .. 26 — # = the number he gained, and 12% + 16 X 30: 12% 3: 18: 26—@2; .@+40:2::18 3: 26—2, and 18” = 1040 — 144 — 2”; by transposition, 2 + 324% = 10403 completing the square, x’? + 32x” + (16)” = 1040 + 256 = 12963 extracting the root, 7 + 16 = + 36; *\ & = 20, or — 52, and consequently 4 put £20 into trade. 6. A person bought some sheep for £72; and found that if he had bought 6 more for the same money, he would have paid £1 less for each. How many did he buy, and what was the price of each? Let # = the number of sheep bought; then 2 = the price of one (in pounds), 12 PBT the price of one, if he had bought 6 more ; and 72 72 .-— +1=—; L+6 x we 720 + & + 6H = 72x + 4325 pt oe = 402) completing the square, 2° + 62 + 9 = 441; extracting the root, 2 + 3 = 21, and # = 18, or — 24, 2 and .*. he bought 18, and the price of one = ae = 4 pounds. producing Adfected Quadratic Equations. 215 7. The plate of a looking-glass is 1s inches by 12, and is 8. to be framed with a frame of equal width, whose area is to be equal to that of the glass. Required the width of the frame. The area of the glass = 12 x 18 = 216. Let x = the width of the frame (in inches) ; then the area of the frame = (18 + 22). (12 + 2x”) — 216, and .*. (18 + 2x). (12 + 2”) — 216 = 216, or 42” + 60” = 216, and 2 + 15@ = 54; completing the square, 2? + 15a + te = 54 + —- = —_; : 15 aie extracting the root, 7 + —=+ “. @ = 3, or — 18, and .*, the width must be 3 inches. There are two square buildings, that are paved with stones, a foot square each. The side of one building exceeds that of the other by 12 feet, and both their pavements taken together contain 2120 stones. What are the lengths of them separately ? Let # and # + 12 =the number of feet in the sides of each ; .. 2 and (x + 12)’ = the number of stones in the squares, and 2 + x + 24x” + 144 = 2120; by transposition, 2° + 24” = 1976, OG. tals @ =—.088 « completing the square, 2* + 12” + 36 = 988 + 36 = 1024; extracting the root, 7 + 6 = + 32; *, & = 26, or — 38, whence the lengths are 26, and 38 feet, respectively. A labourer dug two trenches, one of which was 6 yards longer than the other, for £17. 16s. and the digging of each of them cost as many shillings per yard as there were yards in its length. What was the length of each? 216 Examples of the Solution of Problems Let # and 2 + 6 = the number of yards in each ; xv’ + (w + 6)’ = 356 shillings, or 22° + 12x” + 36 = 356; by transposition, 2%” + 124% = 3203 or xv? + 6x2 = 160; completing the square, x? + 6@ + 9 = 169; extracting the root, #7 + 3 ==+ 13, and # = 10, or — 16; -, the lengths were 10, and 16 yards. 10. A company at a tavern had £3. 15s. to pay; but before the bill was paid, two of them sneaked off, when those who remained had each 10 shillings more to pay. How many were in the company at first ? te # = the number; then =" = = the number of shillings each had to pay at first, and Li? a = = the number en had to pay, after two had sneaked off; oy SNe. aa —— 2 hl 7a as Or 2. —" 9a =— 35: completing the square, 2? — 2% + 1 = 36; extracting the root, 7 —1=+6, AU. a= OL consequently there were 7 at first. 11. A grazier bought as many sheep as cost him £60; out of which he reserved 15, and sold the remainder for £54, gaining 2 shillings a head by them. How many sheep did he buy, and what was the price of each? Let « = the number; i: ~ = the price of each in pounds, 0 1 and (# — 15). a+ = 545 producing Adfected Quadratic Equations. 217 or (vw — 15). (600 + #) = 5404, and a + 585% — 9000 = 5402; by transposition, v* + 452” = 9000; ; 45 |? 2025 completing the square, x? + 454” + ce amps ee aan 38025 =o 7 : : 45 195 extracting the root, # + a ra ea? and # = 75, or — 120; and .*. the number bought was 75, and the price = ~£ = 16 shillings. 12. A and B set out from two towns which were at the dis- tance of 247 miles, and travelled the direct road till they met. A went 9 miles a day; and the number of days, at the end of which they met, was greater by 3 than the num- ber of miles which B went in a day. How many miles did each go? Let 2 = the number of days they travelled ; .. 9% = the number of miles 4 went, and 247 — 9% = the number B went, 247 — 9@ and ———_— e = the number B went per day ; 247 — Ov oa ars Xv and 2° — 3x4 = 247 — 94; by transposition, x? + 6a” = 247; completing the square, 2 + 6” + 9 = 256, extracting the root, 7 +3= + i6, and 7 = 13, or — 19, and .*. A went 117, and B 130 miles. 13. A person bought two pieces of cloth of different sorts; whereof the finer cost 4 shillings a yard more than the other; for the finer he paid £18; but the coarser, which exceeded the finer in length by 2 yards, cost only £16. 218 Examples of the Solution of Problems How many yards were there in each piece, and what was the price of a yard of each? Let « = the number of yards of the finer ; *, 2 + 2 = the number of yards of the coarser, and “ = the price of a yard of the finer (in pounds) ; also p es the price of a yard of the coarser ; Loxeb i glo 1 | ep aloe ve gs and 90” + 180 = 80% + @& + 2a%3 by transposition, 2 — 8x” = 180; completing the square, 7° — 8# + 16 = 180 + 16 = 196; extracting the root, 7 —4=+14; *, #2 = 18, or — 10, consequently, there were 18 yards of the finer, and 20 of the coarser; and the prices were £1, and 16 shillings, respectively. 14. A set out from C towards D, and travelled 7 miles a day. After he had gone 32 miles, B set out from D towards C, and went every day =1,th of the whole journey; and after he had travelled as many days as he went miles in one day, he met A. Required the distance of the places C and D. Suppose the distance was # miles; : = the number of miles B travelled per day; and also = the number of days he travelled before he met A. Se aes + alee XR; * 361 10 paca in x’ 12@ by transposition — —— | = 39 y P Any 19 J ie ae mt Me 122 completing the square, Pee oT + 36 = 36 — 32 = 4; extracting the root, = == Gicmni O . Pp or 4 ee Fours 5 Sy producing Adfected Quadratic Equations. 219 and # = 152, or 76, both which values answer the conditions of the problem. The distance therefore of C from D was 152, or 76 miles. 15. A and B sold 130 ells of silk, (of which 40 ells were 4’s, and 90 B’s,) for 42 crowns. Now 4 sold for a crown one-third of an ell more than B did. How many ells did each sell for a crown? Let 2 = the number B sold , @ ++ = the number A sold { for a crown, and # ; 90 37.12 the price of oo'ells = ~, 1 : 120 and # + —: 40:: 1: the price of 40 ells = : 3 3@ +1 90 120 om 42 —_ — + ie) x 32% +1 Pe ale 15 iM 20 Pi ged Sa 1h whence, 2127 + 77 = 4542 + 15 + 202; by transposition, 212° — 5847 = 15, 58 1s and 2? ——-r=—; 21 841 L156 completing the square, x* — — +) ai: mice 29 extracting the root, 7 — Py a =+ = OG a) Sh ak . ie ae 21’ whence, B sold 3 ells, and 4 31, for a crown. 16. Three Merchants, 4, B, and C, made a joint stock, by which they gained a sum less than that stock by £80. .A’s share of the gain was £60; and his contribution to the stock was £17 more than b’s. Also B and C together con- tributed £325. How much did each contribute? 220 Examples of the Solution of Problems Let 2 = the number of pounds that 4 contributed; *, #& — 17 =the number that B contributed, and 325 — (@ — 17) = 342 — # = the number that C contributed; *, 325 + # = the whole stock, and 325 + xv — 80 = 245 + w = the whole gain; 325 -+@7i7:: 245 +2: 60, and x? + 245” = 60a + 19500; by transposition, v* + 185” = 19500; 185 \* completing the square, v* + 1852 + (=) = 19500 34225 112225 Era e « | 185 335 extracting the root, v + am fa a and # = 75, or — 260; .. the stocks of 4, B, and C were 75, 58, and 267 pounds, respectively. 17. The joint stock of 2 partners 4 and B was £416. A’s money was in trade 9 months, and B’s 6 months: when they shared stock and gain, 4 received £228, and B £252. What was each man’s stock ? Let « = A’s stock; *, 228 — x = his gain; also 416 — x = B’s stock; and # — 164 = his gain; and .*. 64 = the whole gain, and 92 + 6. (416 — x2) : 9H 3: 64: 228 —@; or 3@ + 2.(416 — xv) 1 3@ 3: 64 3 228 — a2; “. 1922 = (@ + 832) . (228 — w) = 189696 — 6044 — x’; by transposition, x’ + 796% = 189696; completing the square, 2’ + 796” + (398)? = 189696 + 158404 = 348100; extracting the root, # + 398 = + 590; and # = 192, or — 988; .. the stocks were £192, and £224. producing Adfected Quadratic Equations. 221 1s. A body of men were formed into a hollow square, three deep, when it was observed, that with the addition of 25 to their number, a solid square might be formed, of which the number of men in each side would be greater by 22 than the square root of the number of men in each side of the hollow square. Required the number of men in the hollow square. Let # =the number of men in a side of the hollow square ; “. x — (x — 6)? = the whole number of men, and a’ — (# — 6)* + 25 = (#3 + 22)’, or 12% — 36 + 25 = &@ + 4445 + 484; .. by transposition, 11a — 44”3 = 495, OY @ — 4@2 = 455 completing the square, 2 — 4#3 4- 4 = 49; extracting the root, 72 —-2=+7; oe, 0a =)\9,\0F — 5, and # = 81, or 25, and .*. the whole number = 936. 19. A Mercer bought a number of pieces of two different kinds of silk for £92. 3s. There were as many pieces bought of each kind, and as many shillings paid per yard for them, as a piece of that kind contained yards. Now 2 pieces, one of each kind, together measured 19 yards. How many yards were there in each? Let x = the number of yards in one piece; and .. = the number of pieces, and also the number of shillings per yard ; *, 19 — x = the number in the other, and 2’, and (19 — 2)* = the whole prices of each kind; 2 + (19 — xv)* = 1843, or 572° — 1083@ + 6859 = 1843; by transposition, 57x* — 1083” = — 5016; or 2? — 192 = — 88; 19 \ove A361 completing the square, v’? — 19% + (2 ae =; 3 wo cs) ri) Examples of the Solution of Problems 1 3 extracting the root, x — <= het ates lis OTS 19 — 2 ==18, or 11; both which values answer the conditions of the problem ; 20. is .. there were 11 yards in one, and s in the other. A square court-yard has a rectangular gravel-walk round it. The side of the court wants 2 yards of being 6 times the breadth of the gravel-walk; and the number of square yards in the walk exceeds the number of yards in the periphery of the court by 92. Required the area of the court. Let # = the breadth of the walk (in yards), *, 6@ — 2 = the side of the court, and 4” — 2 = the side of the interior square ; *, (6@ — 2)? — (4a — 2)’ = the area of the walk, and 20” — sx — 92 = 4 x (6H — 2); by transposition, 20%° — 3247 = 84; atts 21 Neh Oe Aeon imag 5 5 8 lO mre 16 121 completing the square, 2? — Pa es ae am etion e a P 2 ; o ORs - 25 be : 4 ib extracting the root, # — Pe t3 ‘ 7 and (62 — 2)’ = (16)’ = 256, the area required. A Merchant bought 54 gallons of Cognac brandy, and a certain quantity of British. For the former he gave half as many shillings per gallon as there were gallons of British, and for the latter 4 shillings per gallon less. He sold the mixture at 10 shillings per gallon, and _ lost £28. 16s. by his bargain. Required the price of the Cognac, and the number of gallons of British. Let 22 = the number of gallons of British ; producing Adfected Quadratic Equations. 223 .. #@ =the number of shillings one gallon of Cognac cost, and 54” = the price of all the Cognac; also 7 — 4 = the number of shillings one gallon of British cost, and 22° — sx = the price of all the British ; * 2U° — 8H + 54” = 10. (54 + 2”) + 576; by transposition, 2”? + 26a = 1116, or @ + 132 = 5583 : ; 13;,\¢ 169 completing the square, x? + 13% + = ) = 558 ue re pega i. 7 3 4 13 4 extracting the root, 7 + Sahat ame =, and # = 18, or — 31; .. he bought 36 gallons of British: the Cognac cost 18 shil- 22. lings per gallon, and .*. the whole price = £48. 12s. During the time that the shadow on a sun-dial, which shows true time, moves from one o’clock to five, a clock, which is too fast a certain number of hours and minutes, strikes a number of strokes = that number of hours and minutes, and it is observed that the number of minutes is less by 41 than the square of the number which the clock strikes at the last time of striking. The clock does not strike twelve during the time. How much is it too fast ? Let # = the number of hours too fast ; then the clock strikes (v + 2) + (vw + 3) + (@ + 4) + (v% + 5) times = 4u + 14, and the number of minutes = # + 10% + 25 —41 = xv +10” — 16; -@+n?+1or—16=4v 4+ 14; by transposition, 2 + 7x7 = 30; 224 Examples of the Solution of Problems ‘i 4 16 completing the square,.w’ + 7#@ + (Z) == 30 + “ = =; . 7 13 extracting the root, x + 2S ao ma ae Wh — 35 or ery 10, and the number of minutes = 23; .. the clock is too fast 3 hours and 23 minutes. 23. A Vintner sold 7 dozen of sherry and 12 dozen of claret for £50. He sold 3 dozen more of sherry for £10 than he did of claret for £6. Required the price of each. Let 2 = the price of a dozen of sherry (in pounds) ; “. @: 103%: 1: the number of dozens of sherry fog 10 £10, =— a? 10 10i—13 0 and = i 3= OAT es the number of dozens of claret for £6; 10 — 32 Aa uae gage rae vt, bt the price of a dozen of claret = 62 10 — 32’ 894s 4 are 76a + Sor See MCRL Sonne 50, 10 —.37 and 70v% — 21x" + 72a” = 500 — 15023 by transposition, 2922 — 21x? = 500, » 292 500 or a? — —.47 = ——; 21 21 completing the square, pom 292s 146 i 500 10816. 21s 21 aioli feos ean om 14 104 extracting the root, es ——— = + —.,, 21 21 250 and # = 2, or —; 21 .. the price of a dozen of sherry was £2, and the price of a 6x dozen of claret = ———— = £3. 10 — 3a producing Adfected Quadratic Equations. 225 24. A and B hired a pasture into which 4 put 4 horses, and B as many as cost him 18 shillings a week. After- wards B put in two additional horses, and found that he must pay 20 shillings a week. At what rate was the pasture hired ? Let #« = the number of B’s horses at first ; then - = the pay of each per week (in shillings) ; z 2 = what 4 paid, and a + 18 = the price of the pasture ; also vw + 6=the whole number of horses in the second case ; 72 a *, 20" + 120” = 18x” + 108H + 144; by transposition, 2%° + 12” = 144, Ort -+.o2.— 72. completing the square, v7 + 67 +9=72+9=81; extracting the root, 7 +3=+9, and # = 6, or — 12; pee OS 22S: pe lSi 200s. 72 182 2.202% .. B had 6 horses in the pasture at first, and 2 + 18 = 30 | shillings, er week, was the price of the pasture. | 25. An Upholsterer has two square carpets divided into square yards by the lines of the pattern. Now he observes, that if he subtracts from the number of squares in the smaller carpet the number of yards in the side of the other, the square of the remainder will exceed the difference of the number of squares in the smaller carpet, and the number of yards in its side, by ss. Also the difference of the lengths of the sides of the carpets is 6 feet. Required the size of each carpet. Let # = the number of yards in a side of the less ; *, # + 2 = the number in a side of the greater, Q 226 Examples of the Solution of Problems and (@’ —vx—2?’=2—x+s¢s=27?>—xX%7 —2+4+ 90; by transposition, (@ — x — 2)? — (2 — # — 2) = 90; completing the square, (v7 — x — 2)’ — (v’° —#7—2) +4 HAN ue HBOTN i Dee AT 1 19 extracting the root, x’ — 7 — 2 — Se and 2” — x = 12, or — 7, the former of which only will give a possible value of v; and .*. ‘ 1 1 completing the square, 7? — x + = pee = ih: 1 extracting the root, 7 — a= and # = 4, or — 3; consequently the carpets contain 16 and 36 square yards, respectively. 26. A Man playing at hazard won at the first throw as much money as he had in his pocket; at the second throw he won 5 shillings more than the square root of what he then had; at the third throw he won the square of all he then had; and then he had £112. 16s. What had he at first ? Let x = the number of shillings he had at first ; *, 24 = the number he had after the first throw, 60 + —, y 4 1 62 + 1 or = ='6 —-= aes a bar 22 and ih — y 6% +1 Again, 80% + 100y = 65, or 16% + 20¥Y = 135 oe 1627 + sbebibesd = 13, 60 — 1 and 962° + 56” = 782 + 13; by transposition, 9647 — 224 = 13, 22 134 and #7 ——.#= 96 96° eeneith pape. Nar (ss) 121 13 |) ieae completing the square, Ga + oe ee 37 extracting the root, #—-—- = + —; 6 96 1 13 and .. 7 = eau which last does not oe l answer the conditions; and y = re . the price of a pound of mace is 10 shillings, and ot a poung | of cloves is 5 shillings. 32. A and B engage to reap a field for £4. 10s.; and as-4 alone could reap it in 9 days, they promise to complete it in 5 days. They found however that they were obliged to call in C, an inferior workman, to assist them for the two last days, in consequence of which B received 3s. 9d. less producing Adfected Quadratic Equations. 231 than he otherwise would have done. In what time could B or C alone reap the field? Let « = the number of days in which B could reap the field, and y = the number in which C could reap it; ] then at 7 yt the number of shillings B would have 810 received = > 9+ 2 5 450 : and eo = the number he did receive ; 810 450 33 15 20 oe da Ap ae 4 5A S07 aie I *, 2162 — 1080 — 1208 = 94N + 2’; by transposition, x? — 877 = — 1080; 56 224 completing the square, x’ — s7v7 + me a =o" 1080 = —, extracting the root, # — a =a —, aI = 7 FOr 15. ete es then er ae Qn AB y by transposition, ~ =1l—-—-== “. y= 18 the number of days in _ which C could reap the field. The other value of a is excluded _ by the nature of the question. 33. ‘Throwing out the three court cards from a suit of spades, and placing the remainder in two heaps, | find the sum of | the pips in the smaller heap is to the sum in the greater as the number of cards in the greater heap is to the | number of cards in the smaller. But if I add the seventh card to the smaller heap, the difference of the number of pips in the two heaps is equal the square of the number of 232 Examples of the Solution of Problems cards in the smaller. Required the number of pips and cards in each. The whole number of cards = 10, and the whole number of pips = 55. If .*. 2 =the number of cards in the larger heap ; 10 — # = the number in the smaller, and if y = the number of pips in the smaller ; 55 — y = the number in the larger ; Sp 1A) pa DD By hn gee ce A aes and (Alg. 179, 3.) y ¢ 55 3.2 3 10; Or Ys 115. hoes Se, See Again, after the change, y + 7=the number of pips in the smaller heap, and 55 —y—7=48 —y =the number of pips in the larger heap, and .*, their difference = 2y — 41 = (11 — 2)’, and by substituting for 2y its value, 11v — 41 = (11 — 2)’ = 121 — 22% + 2”, and .*. by transposition, 2? — 33% = — 162; 108 44] completing the square, x’ — 33 + (|—) = —— 162 =—— es oUSle ce 33 21 extracting the root, v — Saunt and # = 6, or 27; but 27 being inconsistent with the nature of the problem, the number of cards in the larger heap = 6, and *, the number in the smaller heap = 4; and the number of pips” ; 1 in the smaller = Le = 33, and .*, the number in the greater heap = 22. 34. ‘The fore-wheel of a carriage makes 6 revolutions more than the hind-wheel in going 120 yards; but if the periphery of each wheel be increased one yard, it will make only 4 revolutions more than the hind-wheel in the same space. Required the circumference of each. producing Adfected Quadratic Equations. 233 Let # = number of yards in circumference of the larger, and y = the number in the circumference of the less; or 207 = 20% — xy; .. by transposition, vy = 202 — 204. 120 120 n = — 4, +1 ytl or 30.(y¥ + 1) = (# + 1).(29 — y), or 307 + 30 = 29% + 29 — vy — YY; by transposition, ry = 29” — 1 — 31y, and .. 29% —1— 31y = 20% — 20Y3 by transposition, 97 = lly + 1, 9 Tt eee Lae 20 20 .. by substitution, wnat = my — 20Y, or 11y’? + y = 220y + 20 — 180y = 40y + 20; ah BONES wk 2 Y fice epliy : 39 a9\"" 1591 completing the square, y? — —. = a pleting SUE geod en s (22)? 20 2401 | Ligwe (23)7? extracting the root, y — . cot ae =; 5 a y= 4, OL es errs 11 .*, the number of yards in the circumference of the less = 4, and the number in the circumference of the greater 11 eo ches Vines iti Sa 9 35. On the late jubilee, a gentleman treated his tenantry at the following rate. He allowed for each poor child a certain number of sixpences, for each poor woman sixpence more, 234: Examples of the Solution of Problems and for each poor man sixpence still in addition. The number of women was one-fourth greater than the number of men; the number of children was equal to twice the square of the difference between the numbers of men and women; and the whole expense was £8. 2s. But had each child been allowed as much as each woman, the expense on their account added to nine times the difference of what the men and women cost, would have been £4. 18s. Required the number of men, women, and children, and the allotment to each. Let 42 = the number of men; -, 5@ = the number of women, and 24? = the number of children. Let y = the number of sixpences each child had; “. ¥Y + 1 = the number each woman had; and y + 2 = the number each man had; “(Dae Uact or AOS len tty ee Le Oe also 24°y + 247 + Qxy — 27% = 196; .. by subtraction, 22° — 40v = — 128, and a — 20” = — 64; completing the square, x’ — 202 + 100 = 100 — 64 = 36; extracting the root, 7 — 10 = + 6, and # = 16, or 4, the former of which will not answer the con- ditions of the problem; .*. the number of men was 16, of women 20, and of children 32. Also 32y + 36y + 52 = 324, or 68y = 2723; = 45 *, each child had 2 shillings, each woman 2s. 6d., and each man 38. 36. A and B were going to market, the first with cucumbers, and the second with three times as many eggs; and they find that if B gave all his eggs for the cucumbers, 4 would lose 10 pence, according to the rate at which they were then selling. A therefore reserves two-fifths of his cucumbers; by which B would lose sixpence, according to producing Adfected Quadratic Equations. 235 the same rate. But B, selling the cucumbers at sixpence apiece, gains upon the whole the price of six eggs. Re- quired the number of eggs and cucumbers, and their price. Let # = the number of cucumbers, and y = the price of one; *, 3@ = the number of eggs, — 10 and =i = the price of one egg; 3 also avy = ry — 16; or 3@y = 5xYy — 80; ‘27 "= 80; and vy = 40. 3 _ Also 5° 6e— (~y — 10) 2 tir J0), hoe or Pen — 300 = 60, and 92° — 75@ = 150; : ‘ 25\? 8625 completing the square, 92° — 75a + oy im era 1225 2 e 95 5 extracting the root, 37 — Pate +> and 3% = 30, or — 5, the latter of which is excluded by the 'nature of the problem; .. v = 10, 40 and y == — = 4. L Hence the number of eggs was 30, and of cucumbers 10; .. the price of a cucumber was 4 pence, and of an egg = = 1 penny. 32 37. A person bought a certain number of larks and sparrows ) for 6 shillings. He gave as many pence per dozen for larks as there were sparrows, and as many pence per score for 236 Examples of the Solution of Problems sparrows as there were larks. If he had bought 10 more of each, (the price of larks remaining the same,) and had given as much per dozen for sparrows as he gave per score for larks, they would have cost £1. 5s. 5d. Required the number of each. Let # =the number of larks, and .. = the number of pence per score for sparrows, y =the number of sparrows, and .*. = the number of pence per dozen for larks ; . (2 “Y _) 20Yy Gees ips oe and (az = \15 << 136)==-640. Again, if 2 + 10 = the number of larks, and y + 10 = the number of sparrows; then the price of the larks = y x aoe 5 Uwe 54+ Y a) ivi : 5 and the price per dozen of sparrows = 2 ce “2; ay? .. the price of the sparrows = ST OR 3X 12 2 an ae a BoA wera 109) = 305, 12 3 x 12 and (54 + y) 30 + 5. (y? + 10y) = 36 xX 305, or y + 16y + 324 = 36 X 61 = 2196; by transposition, y’? + 16y = 1872; completing the square, y’ + 16y + 64 = 1872 + 64 = 1936; extracting the root,y + 8=+ 44; “. ¥Y = 36, or — 52, the latter of which will not answer the conditions of the problem, 15 X 36 and #@ = ——— = 15; Y “. he bought 15 larks, and 36 sparrows. 3s. A Poulterer bought a certain number of ducks and 18 turkeys for £5. 10s.; each turkey costing within one shil- producing Adfected Quadratic Equations. 237 ling as much as three ducks. He afterwards bought as many ducks and 5 over, and 20 turkeys, giving one shil- ling a piece more for each duck and turkey than before; and found that the value of his former purchase was to the value of the latter one :: 2:3. Required the number of ducks, and the prices of the ducks and turkeys at the first purchase. Let « = the number of ducks required, and y = the price of a duck ; “. 34 — 1 = the price of a turkey, and wy + 54y —18 = 110, or vy + 54y = 128. Now at the second purchase, # + 5 =the number of ducks, y+ 1, and 3y = the price of a duck and turkey, respectively ; foul lO ce ef ot dr On Ue pep tases or 55: ay +evt+ 65y +5221: 35 ey tat 65y +5 = 165, and wy + # + 65y = 160; but vy + 54y = 128; *, by subtraction, v + lly = 32, or # = 32 — 11y, which being substituted in the first equation, 32y — lly? + 54y = 128, or 11y” — s6y = — 128, 86 128 and y? — —.y =— =; completing the square, y? — —.y + (= yo! = —s 128441 5218 43 1 extracting the root, y — Fighaa mae —, 64 and y = 7 oF % the former of which makes 7 negative, and .*. the price of a duck is 2 shillings, and the is. of a turkey = 5 shillings. Also the number of ducks =32—lly=10. 238 Examples of the Solution of Problems 39. There are three towns, A, B, and C; the road from B to A forming a right angle with that from Bto C. Now a person has to go from B to A, but after travelling a certain distance towards A, he crosses over by the nearest way to the road which leads from C to A, and when on this road he is 3 miles from 4A and 7 from C. He then proceeds to A, and when arrived there he finds that he has gone a distance, equal to one-fourth of the distance from B to C, more than he would have done, had he gone the direct road from Bto A. Required the distance of B from J and C. Let BC yy Bolj=s 4d Cee 10, yA Vis and since the shortest path from a given point to a given straight line is a perpendicular drawn from that point, draw ED perpendicular to 4C; E being the point where he leaves the road BA; .*. Dis the point where he enters the road CA; RU) ce an ia By similar A’s BA: CA:: DA: AE, en at eB and BA: BC:: DA: DE, ory: 2 3: DE= =; 32 S000: Yan ee 3=—4-, y Ya, ig or 3. (# + y) = 30 + —ay, and 27y — 24.(@ + y) = — 240; producing Adfected Quadratic Equations. 239 but 2 + y? = 100; .. by addition, v* + 27y + y? — 24.(v¥+ y) = — 140; completing the square, (# + y)’—24.(@+y)+144= 144 — 140 = 4; extracting the root, v7 + y—1l12=+2; .. @ + y = 14, or 10, the former of which only answers the conditions, .. 2? + 27y + y’ = 196, but 22° + 2y° = 2003 ., by subtraction, # — 27y + y’ =4, whence 7 —y = + 2, butz +y=14; 5. (OWE RUGIION ue as al Gk Or bes and by subtraction, 2y = 12, or 16; *°.. & = 8, or 6, and y = 6, or 8; _.. the distance of B from Cis 8, or 6 miles, and from 4 is 6, or 8 miles. 40. In a garden is a square bowling-green, a side of which is 30 yards, and near to it is a rectangular grass-plot. The number of square yards in the area of the grass-plot is 192 a mean proportional between -. , and the number of Square yards contained in the grass-plot and bowling-green together. Also the number of square yards contained in the square described on the diameter of the grass-plot is a mean proportional between 10, and the number of square : yards contained in the aforesaid square increased by the number contained in the bowling-green. Required the | area and sides of the grass-plot. | Let # and y = the number of yards in the sides ; | 2y — the area, and —— : wy :: ay : ay + 9005 192 Be 172800 | 79 °° 4 gel se 240 Examples of the Solution of Problems 192 172800 by t iti 2. ny = . y transposition, 2’ y 79 fy me completing the square, wy’? — ——ay + ( a) a —_— cused pa () mM mete ah (79)° e) 96, 3096 extracting the root, ry — er 3600 ly = 485 OF ao the former of which only answers the conditions of the problem. Again, lo: a +y i: a@+y'?: @+y' + 900; i Ci ig ATR = ae + y’) + 9000; by transposition, (# + y’)? — 10. (@ + y’) = 9000; completing the square, (7° + y’)”? — 10. (a + y’) + 25 = 9025; extracting the root, 2’ + y? —5 = 95; *, v + y’ = 100, or — 90, the latter of which will not answer the conditions ; Dlltses aye 20 *, by addition, 2 + 2ay + y? = 196, andw+y=+14; by subtraction, 2 —2ry+y>=4, ander—y=+2; *, by addition, 27 = + 16, or + 125 oe == Ce 8 0r a ae by subtraction, 2y = + 12, or + 16; ‘eu GOT ise but the positive values will only answer the conditions, and the area is 48 square yards. 41. A rectangular vat, 3 feet deep, when filled to the depth of 2 feet, holds less than when completely filled by a number of cubic feet equal to 24, together with half the number of feet in the perimeter of the base. It is also observed, that the length of a pole, which reaches from one of the corners of the top to the opposite corner of the bot- tom of the vat, is equal to one-eighth of the number of feet in the square inscribed on the diagonal of the bottom. Required the dimensions of the vat. producing Adfected Quadratic Equations. 241 Let w and y be the number of feet in the sides of the base ; then (say —22Y =) “y=Ww+r+y, and / (9 + a* + y') ==. (a + y’); Ble Sida 819 ey") =o, and (9+ a + y’)—8/(9+ 2° +") =9; completing the square, (9 + 2 + y?) —8 J/(9+2? + 7’) 16 =='95 5 extracting the root, \/(9 + a +y’) -—4=+5, and 4/ (9 + 2 +y”) =9, or —1; you ted ty) ==t8l Orel: by transposition, 2? + y? = 72, or —8, the latter of which is impossible ; but 27y = 48 +2. (@+4+y); . by addition, w + avy + y? = 120 +2. («4 + pee by transposition, (v + y)? — 2. (a + y) = 120; completing the square, (7 + y)? —2.(«@ + y) +1721; extracting the root, v + y—1=+11, and w + y = 12, or — 10, the latter of which is impossible ; + 2ay ty = 144; but 4v7y = 144; _«. by subtraction, z? — 2”y + y? =0, and #—y=0; butry+y= 12; .“. by addition, 27 = 12, and # = 6; ’. y = & = 6, and the base is a square whose side is 6 feet. 42, A person bought two cubical stacks of hay for £41, each of which cost as many shillings per solid yard as there were yards in a side of the other, and the greater stood on more ground than the less by 9 square yards. What was the price of each? Let « = the number of yards in a side of the larger, | and y = the number in a side of the less; | R 242 Examples of the Solution of Problems then 2 and #° = the number of solid yards in the stacks, and # and y? = the number of square yards in their bases ; _2—y =9, and ay + y°@ = 820; yb by ielllich ee Y xy” 4 2,2 ante and a+ 2a°y’ +y = Bare but 2 — 2a°y + y' =815 .. by subtraction, 42° y° = —2-, — 81; es 2 9,2 Pe aay SB tar F or a y* = ~—— — ——;} 7 ‘i eae +e 81 820 \* by transposition, x2*y* + mire et aaa =) = (410)? = 1681003 81\" completing the square, 2 y* + “ oe Yat (*) = 168100 6561 10764961 _ GLittapueostat ; 81 3281 extracting the root, 2’y’? + a aie LOB, the latter of which is impossible ; 4 “, vy’ = 400, or — -, LY = + 20, the positive value only answering the conditions of the problem ; 620 820 | By 20 but 27y 2540,5 and 2’? ++ y’ = .. by addition, 2? + 27y¥ +y?=81,andr7+y=+9, and by subtraction, 2 —2%7y +y°=13 .. ®@-yHHl; .. by addition, 2% = + 10, or + 8, ang @ == 5. OF coe by subtraction, 2y=+ 8, or + 10, and y= 4, or 5; . the prices were £25, and £16. ~ producing Adfected Quadratic Equations. 243 43. A and B put out different sums to interest, amounting together to £200. B’s rate of interest was £) per cent. more than 4’s. At the end of 5 years, B’s accumulated simple interest wanted but £4 to be double of 4’s. At the end of 10 years, .A’s principal and interest was to B’s as 5: 8. Required the separate sums put out by each, and the rate per cent. Let 44 = A’s money (in pounds) ; . 4. (50 — v) = B’s money; y = A’s rate of interest ; “. ¥ +1= B's rate; ae =# = A’s interest after 5 years, and oe) UO) = B’s interest after 5 years ; (50 — @) . (y +1) wy S ti Se a ae ° oO or 50y — vy + 50—x#%+20=2xy; «2 ot —= S507 +70 —@ 9. (1). Again, after 10 years, A’s capital and interest = lov + vy 5 mdb ly 21) 2xLY 5 HO — 2) eal 1d gE ERY og BO) Mo “. 802 + 8Hy = 250y — 5Hy + 2750 — 5545 by transposition, 13”y = 250y + 2750 — 1352. (2) but from (1) 15a@y = 250y + 350 — 5a; . by subtraction, 2vy = 130” — 2400 . . . (3) multiplying (1) by 13, and (2) by 3; “. 650Y + 910 — 134 = (39”y =) 750y + 8250 — 4052; by transposition, 100y = 3924 — 7340, or 50y = 196# — 3670, which being multiplied by z, 50”y = 1964° — 3670; but (3) being multiplied by 25, 50L2Y = 32502 — 60000; ES 244, Examples of the Solution of Problems “. 1962" — 36702 = 32504 — 60000; by transposition, 196”? — 69202” = — 60000; ; 1730 \* completing the square, 1964%° — 6920x# + rere 2992900 52900 = ———. — 60000 = —— ; 49 49 , 1730 230 extracting the root, 147 — = = iar i 1500 “. 14@ = 280, or —— 5 i OTS -, A’s money = 42 = 80 pounds, and B’s = 120 pounds, 196 X 20 — 3670 and y = akc A ee -, A’s rate of interest was 5 per cent. and B’s 6 per cent. 44. When the price of brandy was three times the price of British spirit, a merchant made two mixtures of brandy and British spirit, and the prices per gallon were in the ratio of 9 to 10. He afterwards mixed twice as much brandy with the same quantity of British spirit in each case, and the relative price was the same as before. Re- quired the ratio of the quantities mixed. Suppose at first 2 gallons of British spirit were mixed with one gallon of brandy in one case; and y gallons of British spirit with one gallon of brandy in the other case; then ST ried SGI NC ra 2 Se x +1| the relative prices of the first y + 3| mixtures per gallon ; yori andy +1iliiy+3: Y+3 hence, —— : a e+1i°'yti In the same manner ee) : Yad Par 9&0. L+2° y+2 and .. vy t+a4+3y+3:ay+3@2+y 43:29: 10, div’.2y +x + 3y +3:2@ —2y219:215 producing Adfected Quadratic Equations. PY +L + 3yY +3 = 18H — 18y; by transposition, ey — 17v + 21y +3=0; alsovy + 24 + Gy +12: xy + Ge +2y se 1266259 div. vy + 27 + 6y +12:4%—4y::9:1; " ny +20 + by +12 = 362 — s6y: by transposition, vy — 342” + 42y +12 =0; but ey —17@ + 21y + 3=0; .. by subtraction, 177 — 21y — 9 = 0; also 2¥y — 34” + 42y +6=0, and vy — 34” + 42y + 12=0; . by subtraction, ry — 6 = 0, Of iy 70. Now # =U 9, 21y’ + 9 oe oa “ = (ay =) 6, and 21y° + a == 102; 34 ory? + oy =; 34 9 completing the square, ¥’ ihe Fn aber gee ety ora oh 7 196 ; 3 a1 extracting the root, y + Ae eae 14? 17 3Y = 35 .. the digits are 2, 3, and 4, and the number = 234. 3. The sum of £1. 7s. was to be raised by subscription by three persons 4, B, and C; the sums to be subscribed by them respectively forming an arithmetical progression. But C dying before the money was paid, the whole fell to A and B; and C’s share was raised between them in the proportion of 3:2, when it appeared that the whole sum subscribed by 4 was to the whole sum subscribed by B:: 4:5. Required the original subscriptions of 4, B, and C. Let v — y, x, v + y, be the respective subscriptions of 4, B, and C; UHCI 3 2e== 97 ean ec. Now 5 : 2:: (C’s share =) 9 + y : the part paid by B 2 -° (9 - Y)s and5:3::9+y: the part paid by dA ==. (9 + y), -and consequently, 4 paid upon the whole 9 — y + =. (9+ y) _72—2y i — 5 3 2 63 + 2y also B paid upon the whole 9 + a 9O+ty)= romper és hence, 72 — 2y 3 63 + 2y 334: 5, and (Alg. 179, 3.) 135 : 63 + 2y 2:9: 5, and (4/g. 179, 8.) 15 > 63 + 2y 23225; *. (21) 63 -+ 24 = 75; by transposition, 2y = 12, and: 4i== 6:5 .. the sums to be subscribed originally were 3, 9, and 15 ' shillings. 248 Examples of the Solution of Problems 4. Four numbers are in arithmetical progression. The sum of their squares is equal to 276, and the sum of the num- bers themselves is equal to 32. What are the numbers? Let 27 = the common difference, and # + 3y e+y aed, x — 3y be the numbers ; then their sum = 42 = 32, and) Saw ies also the sum of their squares = 4x° + 20y? = 276, in which substituting the value of x found above, 256 + 20y? = 276; by transposition, 20y’ = 20; Sita and 7 —-h Is hence the numbers are 11, 9, 7, 5. 5. The sum of the squares of the extremes of four numbers in arithmetical progression is 200, and the sum of the squares of the means is 136. What are the numbers? Supposing as before, 7 + 3y, x + y, 2 —y, and xv — 34, to be the numbers ; then 2v° + 18y’? = 200, and’ 2a7'- 97/7 136; .. by subtraction, 16y’? = 64, and 4y = +8; ~YHt?; whence 2” = 68 — y’ = 68 — 4 = 64, WAY OR) eam. and .*. the numbers are + 14, + 10, + 6, + 2. 6. The sum of the first and second of four numbers in geo- metrical progression is 15, and the sum of the third and — fourth is 60. Required the numbers. Let x, vy, xy’, xy’, be the numbers ; in Arithmetical and Geometrical Progressions. 249 .~e@+ry=15, and vy’? + xy*® = 60, Olly. (2 -+-2y) — 60; OLiioy: — 6G; oa iat BNO te ee, and (71+ 247.—=) 37—=.153 | @ = 5, and the numbers are 5, 10, 20, 40. 7. The sum of four numbers in geometrical progression is 8. equal to the common ratio + 1; and the first term = 7 Required the numbers. Let # = the common ratio; .. the numbers are om le a a Wie ad VEER WALLY: l+e7+e42? (14+ 2").(1 + 2) 17 - 17 ‘ 1+ 2° 17 lao ee, and 1o:== 2s andi+a= and 1 = 9 ~m4= 2, Tee 4 tom Ge and the numbers are —, —, —, —. 1 ie Nes ES Hay) A regiment of militia was just sufficient to form an equi- lateral wedge. It was afterwards doubled by the supple- mentary, but was still found to want 385 men to complete a square containing 5 more men in a side, than in a side of the wedge. How many did the regiment at first contain ? Let 2 = the number of men in a side of the wedge; ”. (Alg. 192.) (@ +1). = the number of men in the wedge; mae 250 Examples of the Solution of Problems “. (@ + 1).2 + 385 = (@ + 5)’, or 2? + xv + 385 = 2& + 102 + 255 *, by transposition, 360 = 92, and 40 —\2"+ *, the number of men = 820. 9. After A, who travelled at the rate of 4 miles an hour, had been set out two hours and three-quarters, B set out to overtake him, and in order thereto went four miles and a half the first hour, four and three-quarters the second, five the third; and so on, gaining a quarter of a mile every hour. In how many hours would he over- take A? Let # = the number of hours; *, (Alg. 192.) (0 + (@ — 1) 1) x - = the whole number of miles he travelled ; but 11 + 4a = the whole number 4 travelled; 1 x , (0++@-n).%=u +42, 2 or ov += Sos ae 4 ATH 7 Tomes a and by transposition, — — + — = 225 ; Eh as bs eign 9 361 completing the square, — + — +. — = 22 + — = —_; . P 6 seen RB So Sey, +76 16 ’ 1 extracting the root, - + = = += es Wego gs 2 4 2 | . © = 8,o0r— 115 hence in 8 hours he would overtake him. — 11 not answering the conditions of the problem. 10. The base of a right- angled triangle is 6, and the sides are in arithmetical progression ; it is required to find the om two sides. in Arithmetical and Geometrical Progressions. 251 Let 6 — x, 6, and 6 + 2 be the sides; then 36 — 124 + x2 + 36 = 36 + 12@ 4+ 2” (Eucl. B. I. p. 48.) by transposition, 24a = 36, 3 and 7 = —; 2 .. the sides are - Lars. and 22, rae 2 But if 6 be the first term of the progression ; let, 6,6 + 2, 6 + 22 be the sides; then 36 + 24% + 4a” = 36 + 36 + low + 2”; by transposition, 32? + 12% = 36, or 2° + 42 = 123 completing the square, v7? + 4% + 4= 16; extracting the root, 7 +2=+4, and # = 2, or — 6, and the sides are 6, 8, and 10; or 0, 6, and — 6. The problem is not properly restricted; the algebraical ex- pression, in this instance, is more precise than the language in which the problem is stated. 11. A and B set out from London at the same time, to go round the world (23661 miles), one going Kast, the other West. A goes one mile the first day, two the second, and so on. B goes 20 miles a day. In how many days will they meet; and how many miles will be travelled by each? Let # = the number of days; L then (Alg. 192.) (w@ + 1) aa the number of miles 4 goes, and 20” = the number B goes; 2 e+e ot + com = 29661, and 2 + 41” = 47322; 252 Examples of the Solution of Problems : : 41\? 1681 completing the square, 2? + 41” + 5) 41322. ae 190969 iain . 41 437 *, extracting the root, v + Pe *, #@ = 198, or — 239; “. they travel 198 days; A goes 19701, and B 3960 miles. 12. A traveller sets out for a certain place, and travels one mile the first day, two the second, and so on. In 5 days afterwards another sets out, and travels 12 miles a day. How long and how far must he travel to overtake the first ? Let # = the number of days ; then # + 5 = the number the first travels, and .*, (Alg. 192.) (vw + 6). “te = the distance he travels, and 124 = the distance the second travels ; 2+ 5 er (x + 6) a = 122, and 2 + 112% + 30 = 242; .. by transposition, v*? — 13% = — 30; : 16 16 4 completing the square, 7 — 13%” + a = “ —30= =; : 13 extracting the root, # — mr = + Z Un Pipe) 2a ae .. they are together at the end of 3, and 10 days after the second sets out; and 36 and 120 miles is the distance travelled. _ 13. A and B, 165 miles distant from each other, set out with a design to meet; A travels one mile the first day, two the second, three the third, and so on; B travels 20 miles the first day, 18 the second, 16 the third, and so on. How soon | will they meet ? in Arithmetical and Geometrical Progressions. 253 Let x = the number of days required ; Menli+2+3+... . + @ = (1 + #).< = the number of miles A travelled, x and 20+ 18+ . - . . + 20-2” +2= (42 — 24).5 = the number B travelled ; L 2 L *, (42 —2av).—-+ (1+ 2).—= (43 — 2).-— = 165, 2 2 2 Or @ — 432 = — 3303 a 184 184 52 completing the square, v? — 43v + a = at — 330 = =; 3 extracting the root, 7 — “ = = ; .. & = 10, Or 33. | Hence it appears that they meet in 10 days. On the loth day B travels 2 miles, and the next day he rests; the following | day he returns 2 miles; the succeeding day 4, and so on, in- creasing two miles every day; and on the 33d day he again comes up with 4, who has been travelling forward, every day’s journey being one mile longer than that of the preceding day. 14. There are four numbers in arithmetical progression whose continual product is 1680, and common difference is 4. Required the numbers. Let # + 6,# + 2, # — 2, and w — 6, be the numbers; then (#@ — 36). (xv? — 4) = 1680, or 2° — 402” + 144 = 1680; | .. by transposition, v* — 402° = 1536; completing the square, 2* — 402° + 400 = 1936, extracting the root, 7? — 20 = + 44; ao) Ge 1045 OF —_ 245 and # = + 8, or + 24/ (— 6), jand .. the numbers are + 14, + 10, + 6, + 2; the two other values of 2 being impossible. 254: Examples of the Solution of Problems 15. The product of five numbers in arithmetical progression is — 945, and their sum is 25. Required the numbers. Let 7 + 2y, 2+ Y, 2,0 — y, @ — 2y, be the numbers ; then 64 = 25, and .. 7 = 53 . (a — y’) . (2 — 4y’) = 945, or dividing by # = 5, (u? — y"). (aw — 4y") = 1889, or x — 5a°y’ + 4y* = 189, and 4y* — 125y” + 625 = 1893 by transposition, 4y* — 125y* = — eg 125 15625 8649 completing the square, ay —125y? + (—) = 67 0 =e 125 3 extracting the root, 2y’ — mare ee + 2 also, #@ and the numbers are 9, 7, 5, 3, 1. 16. A Gentleman divided £210 among three servants, in geometrical progression; the first had £90 more than ‘hey last. How much had each? Let vy’, vy, 2 = the number of pounds each had; then vy’ = &% + 90, and # + wy + vy’ = 210; or 2@ + vy + 90 = 210; by transposition, 27 + wy = 120. Now from the first equation, 7 = 7 we 7 and from the last, 7 = ites 53 E20 tae 100 Se ~ y + 2 = y’ 25 1’ 4 3 in Arithmetical and Geometrical Progressions. 255 4y°—4=3y4+ 6; by transposition, 4y° — 3y = 10; . 16 completing the square, 4y’? — 3y + ws E19) poe pa 16 16 16 3 13 extracting the root, 2y — lea Pike AIS 5 and 2y = 4, OFS» 5 and ich YY = 2, On 2 120 whence 7 = == 30; or 160; Yr 2 and the sums are 120, 60, and 36 pounds. 17. The sum of three numbers in geometrical progression is 35; and the mean term is to the difference of the extremes as 2to 3. Required the numbers. Let x, vy, and wy’, be the three numbers ; woe Pb vy + 47y” = 36; and vy: vy’ — #3: 23 3, OF Ys) YAS BAe 35 a Gr ei Ue Tie waive oe 3 by transposition, y’? — Y= 13 ats erp Sl Sg be pee completing the square, y Abin IK Pea ATOR Sc mehENe : 3 5 extracting the FOO Yeo £73 ] } ae? “. ¥ = 2% or — 5, which last does not answer the conditions ; .(@ + 27444 =) 72 = 35; oe V= 55 and the numbers are 5, 10, and 20. 256 Examples of the Solution of Problems 1s. There are three numbers in geometrical progression, the greatest of which exceeds the least by 15. Also the differ- ence of the squares of the greatest and least is to the sum of the squares of all the three numbers as 5: 7. Required the numbers. Let x, vy, xy’, be the numbers; then vy’ — # = 15, and a? y* — a? ; x y* + ay? + 2° 35557, ory —liyi ty tlii5i73 SY tl eee el, by” 2 and yw#—1= +5; by transposition, y* — -. y= '5: 25 25 121 i MP Uke See 25 é completing the square, sid Aen hammer ke 2 5 11 extracting the root, y? — i= + Fee “. y’ = 4, or — *, which last is impossible, 11th a) et .. from the first equation, (47 — v7 =) 34 = 15, and # = 5; .. the numbers are 5, 10, and 20. 19. The sum of three numbers in geometrical progression is 33, and the product of the mean and the sum of the ex- tremes is 30. Required the numbers. Let the numbers be ? a, and ry; aL SD aaes hae tl x and ( +2y) sah 308 .. by transposition, 13 — x =F +ay= = in Arithmetical and Geometrical Progressions. 257 and 1342 — x” = 30, or & — 134% = — 30, “4h6 16 4 completing the square, #? — 134 + rae = —30=— : 13 extracting the root, # — Tone = c, ande,”. @ == 10, OY 3: 3 tea ss GO eas Salsa 10> or 3 + 3y° = 1043 by transposition, 3y’ — loy = — 3, Or ye ay els 2 ail a 3 completing the square, y’? — -- Y + = = = —-1= ~; : 5 4 extracting the root, y — Hie Gare 1 Aas Yy = 35 or 3? and the numbers are 1, 3, 9. If the other value of x be taken, the corresponding values of y are impossible. 20. There are three numbers in arithmetical progression, and the square of the first added to the product of the other two is 16; the square of the second added to the product of the other two is 14. What are the numbers? Let « — y, x, x + y, be the numbers ; then 227— avy +y’ = 16, and 2”? — y? = 143 .. by subtraction, 2y? — vy = 2, and by addition, 42° — vy = 30, or 2y°= 2+ 2y, and 427 = 30+ @y;3 .. by multiplication, sz’y? = 60 + 327y + 2°*y’; S 258 Examples of the Solution of Problems by transposition, 77°y’ — 32v”y = 60, 32 60 or ay’? — — .ry=—;3 je completing the square, 777? ste L Ba sae at » 2 ae p £ q 9 VY 7 y 49s 40 16 26 extracting the root, vy — ia = a3 ' 10 ee DT ie ore {oy = 24+ ry = 8, and y? = 4; yor), and 42” = 30 + #y = 36; aie POD te Be PLOY ia mai .. the numbers are 1, 3, 53 or —5, —3, —1. The other value of wy was introduced in the operation, and does not answer the conditions of the question. 21. The sum of four whole numbers in arithmetical progres- : : : . eae sion is 20, and the sum of their reciprocals is ay Re- quired the numbers. Let 7 — 3y,2—y, 2% + y, 2 + 3y, be the numbers; then 4% = 20, ORs : 1 1 1 1 25 ey NEN 5 V—y + v+y TE Sy oe 4g" — 200y" OB oY. {7 = a —=102°y? + 9y* 24’ *, 25 x (9y* — 2504” + 625) = 24 x (500 — 100y’), or 9y* — 250y° + 625 = 24 x (20 —4y’); by transposition, 9y* — 154y? = — 145, completing the square, 9y* — 154y? + —— ata == ih aoe in Arithmetical and Geometrical Progressions. 259 extracting the root, 3y? — 2 = =, 145 and 3y? = 3, or rang! .y =1, = sSCAdE and y = + 1, or —Y—— and .*. the numbers are 2, 4, 6, 8. 22. There is a number consisting of 3 digits, the first of which is to the second as the second to the third; the number itself is to the sum of its digits as 124 to 7; and if 594 be added to it, the digits will be inverted. Required the number. Let the digits be represented by 2, vy, xy’; then 1007 + lovy + ay :x+ay4+ uy’ 22 124:7, orl00+ loy+yr>:ltyty 2124273 div’.99 + 9y:1+y+y i: 117: 7; or Ik -pyei at y ty’? ts 13273 *. 13y? + 13y +13 =7y +773 by transposition, 134? ; 6y = 64, a or y’ Le pre Soars — 84] completing the square, y’ AC eet (2 ve es + aye 169 extracting the root, Rs =+— cae = 2, or rae oe Y=2; 13¢ also 100% + lowy + wy’? + 594 = 100%y’ + lowy +2; by transposition, 99” + 594 = 99ay’, OF @4-16 = wy = 423 .. by transposition, 6 = 32, and oi as .. the digits are 2, 4, 8, and the number is 248. s2 260 23. Examples of the Solution of Problems There are five whole numbers, the three first of which are in geometric progression; the three last in arithmetic progression, the second number being the second dif ference. The sum of the four last = 40, and the product of the second and last = 64. Required the numbers. Let x = the first, and y = the common ratio of the three first ; .. the numbers are a, vy, vy’, vy’ + vy, vy’ + 2@Yy;3 spy? + 42y =140, and x*y* + 22’y" = 64. Multiplying the first equation by xy, and the second by 3,. 3a°y P42 y? = 4074, and 3a°y° + 6a’y? = 192; by subtraction, 2a’°y’? = 192 — 40ry; by transposition, 22°y? + 4ory = 192, or vy’ + 20"%y = 96; completing the square, xy’ + 20a”y + 100 = 1963 extracting the root, vy + 10=+14; *, vy = 4, or — 24. Now from the first equation, vy. (3y + 4) = 40, or 4.(3y + 4) = 403 °. 3yY +4= 105 by transposition, 3y = 6, and y == 2; {. also ime, and the numbers are 2, 4, 8, 12, 16. There are two casks A and B; of which, A the greater holds 312 gallons. Into A a certain quantity of wine is put, and B is filled with water; then water is conveyed out of B into 4 in the following manner. First, a number of gallons is taken, which is less by two than the square root of the number of gallons in A, then a quantity less than the former by two gallons, and so on. Now when B is in this manner exactly emptied, A is exactly full: and it in Arithmetical and Geometrical Progressions. 26] is known that 8 gallons were taken out of B at one time, after which the quantity left in B was 12 gallons. Re- quired the number of gallons of wine in A. Since the quantities taken out of B are in a decreasing progression, whose common difference is 2, and one term of this progression is 8, therefore the next terms are 6, 4, 2, the sum of which is = 12; and therefore the quantity last drawn out is 2 gallons. Let 2? =the number of gallons of wine in A, then 2 — 2, x —4, &c. are the numbers of gallons drawn each successive time; and the number of terms is evidently = — ip and therefore the whole quantity drawn from 8B is pG-)-2-5 [~] x x e°@ a + pF a oe SIA 4 2 : 5 L j or ins, es ch Leg 4 2 i ~ 22 1248 Lv — —r% = —; 5 5 : 2 1 1248 1 6241 completing the square, 2? — —# + — = — + — = —_ ; P 8 d 5 BL 25 5 25 25 ” : 1 extracting the root, 2 — gic a and # = 16, or — * which last will not answer the conditions; therefore #? = 256 =the number re- quired. 25. The diagonals of 4 squares are in an increasing geome- trical progression, and the product of the squares of the diagonals of the extremes is to the product of the dia- arals of the means as a side of the third is to the square root of the common ratio divided by 44/2. Required the diagonal of the third square, and the common ratio, supposing their difference equal to 45. 262 Examples of the Solution of Problems Let ? x, xy, and xy’ = the diagonals ; then since the diagonal : a side :: 2213 the side of the third = pan 2 and ex. wf ive x HADES Y V2 4/2” oo By tit ae /y 213 . ay = 42 SY; and wy? = 4. Now y — vy = 45, or y — 4y2 = 45; completing the square, y — 4y? + 4= 49; extracting the root, y2 —-2=+7; “, y? = 9, or — 5, which last does not agree with the con- ditions; and .. y = 81; 4 4 whence # = — =-; wo Seay and .*. the diagonal of the third square = 36. 26. Two persons, 4 and B, traded together. 4 gained every year £3 more than the preceding year, and the last year he gained £17. His whole gain was £57. 8B in the first four years gained £52, and if what 4 put into the common stock be added to what B gained the second year, the sum will be £13. How many years did they remain in trade, and what were their original stocks ? Since 4’s annual gains are in an increasing arithmetical — progression, whose common difference is 3, the last term 17, | and sum 57, if n = the number of terms, then (Alg. 192), fai—(n—1).3$.- = 57, or 372 — 3n” = 114; in Arithmetical and Geometrical Progressions. 263 completing the square, n’ — = n + 1S 20 ACU ee ee. extracting the root, n — = == ob 19 *. nm = 6, or ia hence they remained 6 years in trade, and consequently A’s gain the first year was £2, and his gain in four years was £26. (Let ..°. # = A’s stock, and 26 ; 52: 2: B’s stock = 22, and A’s gain the second year being £5, Gael te ater ROA el ° Pb oe woe Cad BE anda 3 *, A’s stock was £3, and B’s £6. 27. A pyramidical pile of cannon-balls, the base of which was an equilateral triangle, was all used in an engagement, except the three lowest layers, and 4 balls of the next layer; these were afterwards formed into a pile with a rectangular base, having as many balls in one side of the lowest layer, as there were in the side of the lowest layer of the pyramidical pile, and 4 in the adjacent side. What was the number of balls; and what the number of layers in each pile when complete? Let x = the number of balls in a side of the lowest layer ; Pi De aot = the number of balls in that layer, and (7 — 1) = = the number in the next, L—1 : and (7 — 2). Pee the number in the third ; , 3a°— 34742 264: Solution of Problems. Now since there were only 4 balls in one side of the second pile, there can only be four layers, which will contain 4a, 3.(%— 1), 2.(# — 2), and w& — 3 balls respectively ; 32° — 34 +% *-———___—__ + 4= 10% — 10, 2 or 327 — 34 +2+8= 20% — 205 by transposition, 32° — 234 = — 30, ee or 2? —- —xr =— 10; 3 : 23 23\? 529 169 completing the square, 77 — —.w# (22) = — —10= — ; ater a 3 ae 36 36” 23 13 extracting the root, v — eens ra re 5 . . e Ande — oer = which last cannot answer the conditions of the problem. Hence there were 6 layers in the first pile, and they contained 1, 3, 6, 10, 15, 21 balls, respectively; .. the whole number of balls in the first pile was 56, and in the second 50. (32.) In the preceding solutions it may be observed, that, in many instances, values of the unknown quantities are de- duced, which do not agree with the conditions of the pro- blems. This is always the case when the roots of the equations are negative; and the circumstance arises from that peculiar quality of an algebraic expression, by which it is denominated either positive or negative. The product of two or any even number of such quantities, whether all of them are positive or all negative, will only be affected with a positive sign: thus the quantity vy will represent the product of + #@ x + y, or of —«#x —y; and a’, of +ax +4, or of —a x —4a; con- sequently, in the reduction of such quantities to their con- stituent factors by the rules of division or evolution, these factors may be considered either as all positive or all negative. But in common language, in which the conditions of a problem are expressed, quantity or number is from its very nature what in Algebra is meant by the term positive, 7. e. it increases any Solution of Problems. 265 homogeneous quantity to which it is added, and diminishes any one from which it is subtracted. Hence it may be understood, why, when quadratic equations are formed to express the con- ditions of a problem, the resulting roots may exceed in number what appear to be required as answers to the problem, and why such as are negative cannot be applied to its conditions. These roots or values, however, though inapplicable in their present shape, will, if assumed as positive, become correct answers to the problem under a different modification of the conditions. In the equations thence deduced, these former negative values will appear as positive roots, and the former positive values as negative roots. Thus, if Prob. 12, page 200, be transformed into the following, “A detachment from an “army was marching in regular column with 5 fewer in depth “than in front; but upon the enemy coming in sight the front was increased till it became = 845 — the original front; and “by this movement the detachment was drawn up in 5 lines. © Required the number of men;” from the solution of this problem the number is found to be 3900, answering to the num-. ber which would be found from using the negative value of 2 in the original problem; and the equation for determining this (7 — 542 = 4225 — 52) differs from the other only in the sign of x. | ( In Prob. 19, page 204, v is found to be equal to + 4, where the negative value shows, that if the trading vessel had turned out of its first course in a direction contrary to CE, or on the opposite side of the line AC, it would have been taken after ' sailing 4 miles in that direction. In Prob. 1, page 212, the negative value of 2 is found to be — 130. But if the problem be modified so as to become “A merchant sold a quantity of brandy, by which he lost “£29 more than the prime cost, and found that his loss “was as much per cent. as the brandy cost him. What was “that price?” the equation for determining the price, is 2 q00 = 29 + 2: which is deduced from the equation to the 266 Solution of Problems. original problem by changing the sign of x, the positive value of which is in this case 130. Also, in Prob. 4, page 213, the negative value of 2 is — =. Now if the problem were, “ Bought two sorts of linen, “for the finer of which I gave 6 crowns more than for the “other. An ell of the finer cost as many shillings as there “were ells of the finer. Also 28 ells of the coarser (which ‘was the whole quantity) sold at such a price, that 8 ells “cost as many shillings as one ell of the finer. How many — “ells were there of the finer; and what was the value of “each piece?’ an equation arises differing from the equa- tion to the original problem only in the sign of x, and whose o,° » ad positive root is =; whence there were 73 ells of the finer at 7s. 6d. per ell, the whole price of which was therefore £2. 16s. 3d., and the whole price of the coarser was £1. 6s. 3d. — And in the very same manner, all the other problems may be transformed. (33.) The same reasoning will apply to the case in which all — the roots of the resulting equation are negative. None of its | values can in this case be applied to satisfy the conditions of the | problem; but if the conditions are properly modified, equations may be deduced, of which these values rendered positive will become roots, and will satisfy such conditions. (34.) The same observation holds, if the resulting values be the square roots of negative quantities, with this exception, that such roots can never be applied to satisfy the conditions of the problem under any modification whatever. SECTION XI. Examples of the Solution of Equations, where the number of unknown Quantities exceeds the number of Equations. (35.) Ir has before been observed, that when the number of unknown quantities exceeds the number of equations, some of these quantities cannot be found except in terms of the others; the values of which, being undetermined, may be assumed at pleasure; thus admitting a number of answers that will be in- definite. A problem thus not properly limited is called an in- _ determinate problem. In such, however, it is not unusual to annex the condition that the values of the numbers sought ‘should be positive integers, or at least rational; or by other limitations to lessen the number of answers. In the different _kinds of these indeterminate problems, different expedients will be made use of; and different artifices be found appropriate to questions differently circumstanced. These, however, are best | learned by practice. (36.) In the case of a simple equation expressing the re- ‘lation of two unknown quantities, whose corresponding integral ) values are required, the common rule is, to divide the whole | equation by the less coefficient, and to assume that part of the ‘quotient, which is in a fractional form, equal to a whole num- ‘ber. A new simple equation is thus obtained, in which a repe- tition of the process takes place. And this is similarly continued with each new equation, till the coefficient of one of the quan- tities becomes unity, and that of the other a whole number. An integral value of the former may then be obtained by sub- stituting zero or any whole number for the other; and then from the preceding equations, integral values of the quantities proposed may be ascertained. 268 Examples of the Solution of Indeterminate Equations. EXAMPLES. 1. Having given 2z + 3y = 35, to find the corresponding positive integral values of x and y. Dividing by 2 the least coefficient of the unknown quantities, 35 — 3 —1 ge Pt yy — 2 ; y—1 2 Assume = whole number = m, Y= 2M+4+ 15 whence # = 16 — 3m; and here 3m must be less than 16, or m not greater than 5; the question .*, will admit of 6 answers. Letun = 0; ‘then vr 16%and 44, m=1,; v= 13 Y = 3; m= 2, v= 10 y= 5, nis; ign #6 y =7, m= 4, v=4 Y¥ = 9, m= 5, P= y= 11. 2. Given 72 + 11y= 100, to find the corresponding positive integral values of # and y. : Dividing by 7 the least coefficient of the unknown quantities, 4y —2 L=14—-Y— a ; Assume -t— * = whole number = m; OT Ae oe ah dgaee tele shai ; m+1 Again, assume = whole number = 7; | . m=2n— 1, and y =7n— 3, £=19 —11n; | where must be less than 2, and can .*, only = 1; : in which case v = 8 andy = 4, Examples of the Solution of Indeterminate Equations. 269 3. Given 9% + 13y = 200, to find the corresponding positive integral values of # and y. Dividing by 9, y ne = 22—-y— ot. Assume are = whole number = m; “. 2y¥=9m +1, and y = 4m + an Again, assume an = whole number = n; .m=2n— 1, and y = 9n — 4; oe & = 28 — 132. _And since 9 must be greater than 4, and 13” less than 28; .. 2 must not exceed 2, nor be less than 1. elm le then dg — loan 3.5, n= 2, v= 2. y = 14. 4, Given 1142 + 13y = 190, to find corresponding positive integral values of # and y. Dividing by 11, 190 — 13y 2y — 3 Ce oe ; ll ee 11 2y — 3 Assume < = whole number = m; m+ 1 yom + 1+ ——. m+ Again, assume = whole number = 7; then m= 2n—1, and y = 11n — 4, Uy 290 — 1371s 270 Examples of the Solution of Indeterminate Equations. 22) . . where 7 cannot exceed 737 & it cannot exceed 1, and must be greater than 0. Tpeten ta hee ae 9,0 ee Ifin =0; 7 — 37; andy ==: if” = 2, 7 = — 4, and y — 18, &e. 5. Given 13” + 16y = 97, find corresponding positive in- tegral values of # and y. Dividing by 13, SADArs LY yenayee 3y — 6 13 er I mie’: 13 v Assume = whole number = m; ve YIM + 2; and 2 = 5 — 16m, where it is evident that m cannot be = 1, to have a positive value of x. Let — 0, then — Sanday ee It =, tien ¢ — — LF, dud 7—15,- and so integral negative values may be obtained. 6. Given 17@ + 23y = 183, to find corresponding positive integral values of # and y. | Dividing by 17, pet Sermo el (0 eee ees 17 Ly, 647 —13 = whole number = m; “. 6Y = 17m + 13, and y = 3m 4+2—"—. arnt! m Assume = whole number = n; “ m=6n+1, and y =17n + 5, L=4— 23025 Examples of the Solution of Indeterminate Equations. 271 where » = 0 is the only value which will give positive integer values of # and y; viz.2=4andy=5. 7. Given 19% +5y= 119, to find corresponding integral values of # and y. Dividing by 5, 1g 1 92 aad = + = 24 47 — ——., y 5 5 L—1 Assume = whole number = m; =5M +1, and y = 20 — 19m; where 19m must be less than 20, and .*. m cannot exceed 1. Letm= 0, .*. a= 1, and ¥ = 20, m=1, .. &©=6, y= 1, which are the only integral values. 8. Given 46% + 3y = 3668, to find corresponding positive integral values of 2 and y. Dividing by 3, L—2 Yaar hee a iat | L—2 Assume = whole number = m; °. 7=3m + 2, and y = 1192 — 46m; 1192 where m must be less than —, and cannot .*, exceed 25. Let m= 0,:then 2 = 2, and: y= 1192; m =, Bs 5, = 1146, Mm = 2, L = 8, y = 1100, and so on, by assuming m equal to every number up to 25, we ‘obtain pairs of integral positive values of x and y, the former ‘increasing by 3, and the latter decreasing by 46. ) 9. Given 3% =8y— 16; find the least corresponding positive integral values of x and y. 272 Examples of the Solution of Indeterminate Equations. Here » = SY" = 3y—5 —Y Assume = whole number = m, y+] 3 .y =3m—1, and v = sm — 8, where m must be greater than 1. Assume m = 2, then # = 8, and y = 53 the least numbers which will satisfy the conditions. If m = 3, x = 16, y = 83 and other values of m will give corresponding values of # and y. 10. Given 77 — 9y = 29; find the least corresponding positive integral values of # and y. Here w= TU aa ty + ais. Assume aE. = whole number = m; ° 2y=7m—1, m—1 and y = 3m + : m—1 Let ane ae whole number = m; ._m=an+ i, and y = 7n + 3, L= 9M + 8. Let m=0, then x =8, and y=3; which are the least whole numbers which satisfy the equation. By assuming other values of m, corresponding values om x and y may be obtained. » | 11. Given 92—7y=6; find the least corresponding positive integral values of # and y. 97 — 6 27 +1 =wf—1+ . Here y = Examples of the Solution of Indeterminate Equations. 273 27+ 1 Assume = whole number = m; 7m—1 m— 1 = = 3m + = ee 1 = whole number = n; .m=2n+15 whence # = 7n + 3, and y = 10n + 3. If n=0, x =3, and y=3, which are the lowest integral values. Others may be obtained by assuming different values of n. 12, Given 11a — 17y =5; find the least corresponding in- tegral values of 2 and y. 5 Pin pga Here x = “¥** — Ysa; 1] ei —1 Assume % f= whole number = m; ete al Layee and #7 =17m + 2. And if we assume m = 0, # = 2, and y = 1, which are the Teast numbers. Other values may be obtained by assuming different values of n. 13. Given 14v4 =4y +73; determine whether positive in- tegral values of # and y can be found. -_ 2 1 Here y = oo = Leg ye Sgt, Cane, 4 4 1 | Assume at = whole number = m; ve 22 — 4m eae 15 which is impossible; .*. no integers can be found to answer the sonditions. 14, Given 19% = 14y — 11; find the least corresponding dositive integral values of x and y. dd 274 Examples of the Solution of Indeterminate Equations. 197 + 11 5@ +11 Leese ee pale a pg Assume he +11 TH whole number. = ™; 14m — 11 m+ 1 ee es TR me . 5 5 Ter ae 1 = whole number = 7; .m= 5N—13 whence # = 14n — 5, and y = 19n — 6; where to obtain positive integers, m cannot be = 0. Let n = 1, then # = 9, and y = 13, the least values. 15. Given 23” — 9y = 929; find the least corresponding positive integral values of # and y. 23% — 929 5” — 2 Here y = Wargo Poet cee oa : 5k — 2 Assume = whole number = m; m—2 7. £= 2m — —. 5 m—2 Let zp whole number = 7; °. M=5n + 23 whence # = 9n + 4, and y = 23n — 93; where any value of 7 less than 5 will give y negative. Let .*. m = 5, then 2 = 49, and y = 22, the least values. 16. Given 54 + 7y + 112 = 224; find all the positive in- tegral values of x, y, and z, which satisfy the equation. 24 — 7y — awe a 1a 2; ok aha sell revere : 5 i) Examples of the Solution of Indeterminate Equations. 275 = whole number = m; then z == 5m — 2y — 1, and 7=47 + 3y— 11m. If then we assume different values for m and y, correspond- ing values of # and z may be obtained. meretnen 7% — 1, and y—1; .. # = 39,2=—23 any other values of y will give z = 0, or negative. et m ==, and 7 == 1, then 2 = 28, and z= 7, y = 2, e&= 3), z=5, y = 3, & = 34, = 3, y = 4, v2 = 37; z<=l. Other values of y will give z negative. It is evident that m must be greater than te AS Y ld iete 7 and with this limitation other values may be found. and less than 17. Given 17x + 19y + 212 = 400; find all the positive in- tegral values of #, y, and z which satisfy the equation. ey ead 2 4z2— eres bili py, Spas el ei a 17 17 2y+42—9 Assume = whole number = m. l ae Yy — 8m rea | 22 + 4 + ant Let mo = whole number = 7; “ M=S=N—15 whence y = 17n — 4 — 22, and # = 28 — 19n + 2. Substituting for z and n different values, we obtain integral values of # and y. r 2 276 Examples of the Solution of Indeterminate Equations. Ifn=1, andz=1, r= 10, y= 11, — 9, 711, fa Os 2=3, f= 12,y7= 7, &e. z2=6, ©=15, y= 15 8 + 2 y) and m cannot exceed , nor be less than 1. find the corresponding po- sitive integral values of 2, y, and 2. 18, Given 2 + 2y + peered da + sy + 62 = 47) From the first equation, 27 + 4y + 62 = 40, but 4@ + 5y + 62 = 47; *, by subtraction, 27 + ¥ = and y = 7 — 223 whence 32 = 20 — 14+ 42 —®@ =6 + 32, and 2z=2-+ @#. Let .. @ =1, then y = 5, and z= 3, Tei; ij 12; fa v= 3; y=1; 2=535 but if 2 be assumed greater than 3, y becomes negative. 19. Required the positive integral solutions of the equation vy +20 + 3y = 42. Here (vw + 3).y = 42 — 2a, pices al ee, SA e+3 L+3 : - v + 31s a divisor of 48. Now the divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Let thenv +3=™m™; c= Mm -— 3, and y = 48 and y = eins We m .. must be greater than 3, else # will be negative; and if less than 24, y will be negative; if = 24, y will =o. : Examples of the Solution of Indeterminate Equations. 277 Let m= 4, thenz = 1, and y = 10, m= 6, L= 3, y= 6, m= 8, v= 5, Y= 4, Dial 29. J We 216; Ti emetASP e— wal If all the integral solutions be required, Peri — ol, then’ 2 — —"2, and y ="46, eee. L=—1, = 23. m= 3, v= 0, y= 14, Mm = 48, Te AD, y=. 20. Required the positive integral solutions of the equation lary = 52 + 7y + 15. Here (12@ —7).y = 5@ + 15, ie BY +15 Y 12%@—7’ aa aha a 215m gece 19g Fete for 7 and .°. 122 — 7 must be a divisor of 215. But the divisors of 215 are 1, 5, 43. And since if m=127—7 : a aT, such values of # only will answer as when increased by 7 will be divisible by 12, it is evident that 5 is the only one ‘lof the divisors which can be used ; .t=1, and y =4. 21. Required the integral solutions of the equation ey + xv =2H + 3y + 29. Here (v@ + 3).y = 27 — 2 + 29, 20 — 2 +29 26 SO Y=) ae tte gaa WEF 1) eas and .*. 2 — 3 must be a divisor of 26. het il aij a te 1 tes 278 Examples of the Solution of Indeterminate Equations. Now the divisors of 26 are 1, 2, 13, 263 and om ela 4, aa} m= 2,= 5,Y= 7, m=13, ©= 16, y=— 15, m = 26, # = 29, y= — 29. 22. Find a number which, divided by 2 and 3, leaves re- mainders respectively 1 and 2. Let 2 = the number; aC l e then is a whole number; let it =m; ee mh Py pote vw Sara 2 . But is a whole number ; . 2m—1 Zc. = whole number = 7; 37 + 1 n+ m= =n + ° 2 n-+1 Let amas! i ._nm=2p—1, andm=n+p=3p—135 whence #7 = 2m+1=6p—1. Assume p= 1, thena# = 5, p= 2; L=115 and by assuming different values of p, other numbers are ob- - tained which answer the conditions. 23. Find the least whole number which, when divided by 3. and 5, has its respective remainders 1 and 3. | Let 2 = the number; &e— 1 = whole number = m; then aa L= 3m + 1. Examples of the Solution of Indeterminate Equations. 279 Rm 3 But ey whole number ; 3m — 2 ap eee oe whole number = n. 5n + 2 on +2 and m= "7 =n + : Let nt ’ — whole number = OS ~ n= 3p—1, and m=n+2p=5p—1; C= 3M +1 = 1p — 2. Let p = 1, then = 13; the least whole number which will answer the conditions. 24. Required a number which, divided by 11, leaves a re- mainder 3, but being divided by 19 leaves a remainder 5. Let « = the number ; LX—3 then = whole number = m; Jit = 1m + 3. U—5 l1l1m—2 But —— = ——— = whole number = n; 19 19 19n +2 8n + 2 2 PEE SM Eants, 11 li 4n-+ 1 Let : ll “42 = llp—1, 1lpy—1 a+1 and n = —f—— = 3p egaant 4 4 is et Assume / a fee To I and 2 = 117 — 3, m= 19" — 5; whence vw = 2097 — 52. 280 Hxamples of the Solution of Indeterminate Equations. Let 7 = 1, then # = 157, the least number which can answer the conditions. Other values may be obtained by assuming Peat ROPER fed eh 25. Find the least whole number which, divided by 19, leaves a remainder 7, and divided by 28 will leave a remainder 13. Let # = the number; pal | then = whole number = m, and w= 19m + 7. Saar ees 19m—6 Also = whole number = n; *, 19m = 28n + 6, and m=n + valaat es 19 3 2 Let “ = whole number = p; 19p — —2 then as ee by ies Sasa 3 3 p—2 Assume = whole number = 7; © then p = 3r + 2, n= 197 + 12, m = 287 + 18, and # = 5327 + 349. Assume 7 = 0; then x = 349, the least number that will answer the conditions. Other numbers, however, may be de- termined by assuming different values of 7. 26. A certain number, when divided by 5 and 4, leaves a | remainder 1; but when divided by 3, leaves no remainder. Determine the number. Let v = the number; oO) then = whole number = m; . = 56m, +1. Examples of the Solution of Indeterminate Equations. 281 ee | 5m m Also apy aicassial aa m+ Vs whole number. m Let 7A whole number = n; eee tt and 7 = 20n +1. 20n +1 i—1 —- = 7rn— ne But - = whole number = n— Assume thenvn= 37 --- 1, m= 12p + 4, and # = 60p + 21. If then p be assumed = 0, # = 21, the least whole number. [EE i ele and so on for every number which may be assumed for p. 27. Find a number which, divided by 3, 4, 5, respectively, ‘shall leave 2, 3, 4 for remainders. Let # = the number; “2 then ae whole number = m; ° fom 3m + 2. 2-3 3m—1 Also or. aa ae whole number = 7; 4m + ] ie ae m= -———— =n 3 ST; n+1 and ~ = whole number = p; | “. N= 3p — 1; and m= 4p — 1, whence # = 12p — 1. L—4 12p—5 2p But —— = ——— = whole number = 2p —1 + =, : ot = whole number, and £ = whole number = 7; 282 Examples of the Solution of Indeterminate Equations. Ie or, and 7 = 12p —1=60r—1. Assume 7 = 1, then 7 = 59; r= 2, & = 1195 a= 2p Fy WAN RL u ii 28. Find a number less than 400 which is a multiple of 7, and upon being divided by 2, 3, 4, 5, 6, always leaves 1 for a remainder. Let # = the number; aL— 1 then = whole number = m; a ee oe L—) 2m , and es whole number = mig let it =n; m= ee ‘he & ee — 9 — 2° Assume — = 7p; noma 207, mM == 37, and v7 = 6p + 1. Rap Me at Pa Fe whole number ; 4 4 2 <7 . = whole number = g, and 7) ==: eae) 7 ae a Again << = ae = whole number = 2q + “f. Let2 = ee 5 BET Bny and # = 607 + 1. G—1 607 Also Pre ee 107 = whole number. Examples of the Solution of Indeterminate Equations. 283 By assuming .*, 7 = 1, 2, 3, 4, 5, the values of x will be found to be 61, 121, 181, 241, 301, 361, which are all less than 400. But 301 is the only multiple of 7; the only number .*, which answers the conditions. 29. Divide 25 into two parts, one of which may be divisible _ by 2, and the other by 3. Let 24 = one part, and 3y = the other; then 22 + 3y = 25, 25 —= 32 uw—l and @ = ——"# = 12 —y— 4, y—1 Assume % ak whole number = m ; oe y == 277 + l, and #@ = 11 — 3m. And since 11 — 3m is a positive number, 3m must be less than 11, and .*. m less than 4. mem = 0, then y = 1, and = 11, .*, the parts are 22 and 3; pa I, Buia, v= 8, .. the parts are 16 and 9; m = 2, jo", %=5, .. the parts are 10 and 15; m = 3, jem e e%=2, .°, the parts are 4 and 21. And these are the only divisions which can be made. 30. How many ways are there of paying £7 with crowns and seven-shilling pieces ? | Let 2 = the number of crowns, ‘i required to pay and y = the number of seven-shilling pieces, the sum. | then 52 + 7y = 140, 2 and « = 28 — y — ~4. ca) Let e ce i i 2 Y= 5m, ) and wv = 28 — 7m. 28 And as m cannot be = Ree 4, there can be only 3 answers, or 3 different ways of payment. 284 Examples of the Solution of Indeterminate Equations. Suppose m= 1, then #= 21, and y =5; Mm = 2, 24a oie m = 3, L=7,; y = 15. 31. In how many ways may £80 be paid in sovereigns and guineas ? Let # = the number of aire required to pay y = the number of guineas, the sum. then 20% + 21y = 80 x 20 = 1600, and #@ = 80 —y—<. But i. = whole number = m; fai = 120 ans and # = 80 — 21m. And since 80 — 21m must be a positive whole number, .. m 80 | must be less than ore and cannot .*. exceed 3. There are .* only three ways of payment. Suppose m.= 1, then # = 59, and y = 20; M.= 2, L = 38, ii 40% m = 3, &= 17, y = 60. 32. Can £100 be paid in guineas and moidores? If it can, let x and y be the numbers of each. then 214 + 27y = 2000, 2000 ee UE akin 6y — 5 21 paca? ye and #7 = But “y—* = whole number = m; (es oy =3m + 3m 5 Assume t: which is impossible; and .*. the payment cannot be made. Examples of the Solution of Indeterminate Equations. 285 33. In how many different ways is it possible to exchange 11 bullocks which are worth £12 each, for sheep which are worth £2. 5s. each, and pigs which are worth 12s. each? Let x and y represent the numbers of sheep and pigs which must be given in exchange ; then 452 + 12y = 11 xX 240, or l5v@+ 4y =11 X 80 = 880; v age Dee iar titer: 3 L Assume ae . &®=4mM, and y = 220 — 15m; _ where it is evident that m cannot exceed 14, nor be less than 1; there are .*. 14 different ways of exchange. Let m= 1, then 2 4, anda = 905; M2, aks. Te and so on, the values of x increasing by 4, and those of y decreasing by 15. 34. A company of men and women club for the payment of a reckoning; each man pays 25s., and each woman 16s.; and it is found that all the women together pay one shilling more than the men. How many men and women were there? Let x = the number of men, and y = the number of women ; then 16y = 25% + 1, ov + L and y = @ y TE Ek 16 Gb Let ae = whole number = m; <3 9v = 16™ rma 1, 21 + 1 9 . and # = 2m — = whole number = 7; 9m-+i1 | Assume = eoroe 286 Hxamples of the Solution of Indeterminate Equations. then 2m = 9n — 1, n—1 and m =4n + n—1 2 then n = 2p + 1, m=9p + 45 whence @ = 16p + 7, and y = 25p + 11. Let p = 0, then # = 7, and y = 11; the least numbers which answer the conditions. Other values of x and y may be ob- tained by assuming different values for p; the number of men increasing by 16, and the women by 25. Let ie 35. A person distributes 4s. 2d. among some beggars, giving 7d. to some, and a shilling each to the rest. How many were there ? . Let 2 = the number of those to whom 7d. was given, y = the number to whom 1s. was given; then 7% + 12y = 50, 50 — 12y sy — 1 ape eee re and # = 7—Yyr- But = = whole number = m; . by=>7M+4+1, 2m + 1 and y = m + ———_. 2m + 1 And to ae hole number = 7; n—1 mM =2n + But “—" = whole number aie 9 . m=2p4+1, and m = 5p + 23 whence y= m+ 2n=7p + 3, and #7 =7—y—m=2—12p. Examples of the Solution of Indeterminate Equations. 287 And as 12y must be less than 2, no whole number substi- tuted for it will answer the conditions: but if p = 0, 2 = 2, and y = 2, the numbers required. 36. A wishes to pay a debt of £1. 12s., but has only half- crowns in his pocket, while B has only fourpenny pieces. How may they settle the matter most simply between them ? Suppose 4 to pay # half-crowns, and receive y fourpenny pieces ; | then 30% — 4y = 32 X 12, or 152 — 2y = 192; v Bi Ick rm OO ree e Let >= m; “Tr + =a, and y = 15m — 96; whence m must be greater than 6. Let m=7; .. v= 14, and y = 9; the smallest number of coins which will answer the conditions. 37. It is required to divide 24 into three such parts that if the first be multiplied by 36, the second by 24, and the third by 8, the sum of these products may be 516. Let x, y, and z be the three parts; thenv +y + 2= 24, and 364% + 24y + 8z = 516, or 9% + 6y + 22 = 129; but 22 + 2y +22 = 48; .. by subtraction 7” + 4y = 81, and y = 20 — 2% + ar Let +1 ++ 14: 5, to find the value of x. UN ieee Given ee! : aa —2”7::5:4, to find the value of x. ANS. s2.8. Given 16% + 5: Tr °$ 362 + 10: 1, to find the value of x. ANS. 2 = 5. 42 +3. 6x2 — 43 © of 2. ANS. & = 8. Given 1:32” +19: 3x2 —19, to find the value Ee en 10z” — 18 Given 52 = eae ch Fale) 27 +3 , to find the value of x. ANS. &@ = 3. Given (/ (107 + 35) — 1 = 4, to find the value of 2. ANS. &@ = 9. Given \/ (97 — 4) + 6 = 8, to find the value of 2. IANS eat Ae Given \/ (v + 16) = 2+ 4/2, to find the value of 2. ANS "9. Given (/ (2 — 32) = 16 — 4/2, to find the value of z. ANS. # = 81. Given \/ (4a + 21) =2°/@ + 1, to find the value of 2. ANS, & = 255 64. 65. 66. 67. 69. (70. one unknown Quantity. 303 Given ay/ (bu —c) =d./ (ex + fx —g), to find the value of 2. ac—d'yg @b—d’.(e+f) ANS. &2 = Given «/ (a + ¢ ee ak: ) to find the value of x. (2 + b) ANS. dealt WAC ian Given / ( (a + 2) = A/a? + 5ax@ + 6’, to find the value of zx. Se ee ail) 3a Given a + 6.\/ (x + d) =c, to find the value of z. C—a m ANS. @ = ( “aed 18+ to find the value of z. /5t + 3 2 ANS. & = 3. 304: Simple Equations involving y A, 72. Given ing wi nian ad 2, to find the value of 2. 3 64—3% 3 #-—2 14 ANS. #2 = —. 13 812° — 9 a 3 2n7'°—) 57 —aee - 0 ares 2 @ 73. Given Tk WW abe EMS 2 ; 2 (3@—1).(@ +3) to find the value of 2. ANS. & = 10. 74. Given \/{1+2/(2’ + 12)? =1+ 4, to find the value of a, NN Gees 2 ) 75. Given So (ca +d’) + a = ex, to find the value of a. ON ier ey. Pee cli be a’ d 2abce — 36 » Gi — 9) = ———_.,, to find th faa 76. Given /2+4/(4 — 9) 7 ey) o find the value of #. ANS. & = 25. 77. Given = ff (v? + 39% + 374) — f/f (wv + 20% + 51)§ ae & + 22 a7 ic % *) to find the value of 2. ANS. & = 78. Il. Semple Equations involving two unknown Quantities. 1 GIVEN@ + 1y= 53, | andy + 3x = 27,{ fe = 8, ANS. NS ‘te S to find the values of # and y. we ot ge + 10, | o two unknown Quantities. 305 Given 47 + oy = ey) to find the values of x and y. and sv — l3y = 9, [xv = 6, ANS. ly = 3. Given = + 2 = 6, | to find the values of x and y. and — + a= 52; | ie fe wee ly = 16 Given = + 8Y = 194, to find the values of x and y. and 7 ew —= 131, ANs. Ley ly = 24. Taped eae dbs to find the values of wv 3Y — 5 and y. and +22 — 8 =72, ANs. ee y= 5. | to find the values of w and y. ANs. le ly = 10. 306 Simple Equations involving | 7+o 27 y _ ° Gi eo ke as eS a d nesne t: 4 34 — °> | to find the values of # 5 —: | and -4 fy “ta 18 — 52, one ) | ANS. t, we : WS syt47 _ | _ oy +33 | ale oe 14 ’ | to find the values Bom AY a Deel Y Sale of x and y. er i a eee eo s. Giveney+1— and y — 3 — ANS. : 15 — &@ 7H+11 | 9. Given 4a 4+ ——— = caer ne | - 4 etmoene Neos eae the values Bey ee tice | of x and y. | ANS. Eons | ly = 4. | 10. Given 4 — al +17 =5y + ae to find the © 22—6y 5x 21 sy +5 values = =i | d Pees SR GAGn ee ice Ts xz and y, | ANS. ie ‘ ly 3 20 PY | 90 = BY AD), AU SY 2 esr eee 16 11. Given wee Bde aT i A es | to find the 5 values of | wand y. ANS. se He yea 15. ' 12. 114. two unknown Quantities. 307 ek b+y 344 2° | to find the values of 2 and y. and av + 2by =¢, | _ 26°— 6a’ +c 5 3a : _ 37 —h +e oer 36 Given & ANS. Given 12) +--6 BUS UE emai 3Y —2& fat sess = 39 — ——~ — t "| and 32 + 4: 2y —3:: 5:3, | to find the values of 2 and y, Ogee ly =9. ° 1 —— — i a, SEE A eo a Rd Adee 8 2 6 3 Y+7 . 3y—8 and fam ee ees to find the values of x and y. ANS. Ne ome ly =4. Given’? +y:4¢+Yy:24:7, to find the x en) 2 + — values of and. ————— = i xz and y. Given Sood tS ee Yi FSS 10 15 5 and Yt 5%@—8 _ @ty _ 7e@ +6 12 4 11 to find the values of x and y. ANs. ie fa Yor oe x2 308 Simple Equations involving 4y—17 +2 15 —3H7_ 12y +11 i7. (Given 13% + = : : lo@ + 7y + 28 “SEN pray? ue ow eS 12 oh rOy iia AY So te oe 4 5 8 15 to find the values of x and y. ANS. | ee q Beye sR: | | — 2b). | 18, Given 32 + A ek Urea | a? — 6’, | and ee — 2 +(atb+c).by=e + (a+ 2b) ab, to find the values of x and y. ) 20 + Yy 74 + 6a 411 yi 3 WY) —_——— + URS Se ee I 18 6 . sat 3y+2, 9y +6 | ; Ranger to find the values of # and y. ANS. ae ly =4. 19. Given and. lise 32 —5y 247—8yY—9 1 90. Given — mae 12 prs and = + 24 1}: 4 —F— 24 1238 2 3h to find the values of # and y. Auteur ly = 4. 21. 22. 23. 24. 25. two unknown Quantities. 809 Given (v7 + 5). (y +7) = (w+ 1). (y—9) + 112, } and 27 + 10 = 3y + 1, i to find the values of 2 and y. fe =3, ANS. ys Given 6x + 9 3U-T5Y = 34 + hpi | to find the 4 42—6 2 Hp it toesy oes values of of Mehvsae eae ; an -e Ley : 4 +4 5 {| x and y ANS. oe ly=9 Peenidy — sap — <9 4 137 _ to E8 | to find the 27 — 6Y values of 21—4y 184+ 13 ad3 et ee es OF T dy. an “ana : 21, wand y = 7 ANS. : ie att ae 2 : Given 167 + 6y—1= ebilae EER ay to find the 8H —3y+2 ee, values of and eS ee ee wand y. 20+ 2 +3 3@ + 24 —] ANS. [ee ly = 5. oe ely ly 2 G; 2 __ l6u*+12V7y—8x+5y +28 | Mel an a ee igo ee Bertie t oe es 3y pos ey te 108 | 40+ 6y +3 to find the values of x and y. ANS. (as, ly =2. 310 Simple Equations involving A — 247" 26. Given 34+ 6y+1= SR SOS \ toe mae 2N—4yY +3 ata | Bye values of and by i heir asic hss DL ak, xv and y. 4y—1 3y —4 ANS. [eases ae 40° —y.(@ + 3y) 2—Yy+a4 and (2 + 7).(y—2) +3=2xey —(y—1).(@ + )); to find the values of x and y. 27, Given 3y + 11= + 31 — 4a, | Ans; doin ©? oa 3. y 4 7} +5 -y—2 [ af 3 | Aone tn 28 ey he mAs orale tented ai (x y) | 10 mit ANS. Peps I= 2 y lle oa — +6 ’ &—6 4 20. Cie SEE a Oe SN a 7y 24 6 42 56Y 13 86 14 and 12” — 15y + —-: 1oy — 8@ + — 3:93 —9@ : 6a@——, to find the values of # and y. ANS. . te malt eS ” two unknown Quantities. 3ll 72 3y +6 347 —2 Me Wp evaHIE NI 9 — By +, 30. Given — : a ts ae | 5 8 16” 32 2y as and —+—~+2):-—#+4-::10i:11 ee ait “2 SG a cls to find the values of x and y. ANs. [ares Jee ; 40 —2 Se Nis a an 4s '5 x 1 1, py SEERA Sl all, Badass Sue IAD PE 3 7 4 5 rf | gy US ee 1 a :y—2 55 oS ae ere ee ee and 27—y+15:y—2"+4 MEL pe oe eo on Gee to find the values of w and y. ao jv = 18, ly = 24. ay — EF 2 “eu + low +13 32. ities acm Gk a RM ey ah gyi tae to find the values of # and y. ANS. - ma y = 2 3. Given «/y — \/ (y — 2) =4/ (20 — z),] to find the values and 4/ (y — #):4/(20 — 2) ::3:2,f ofwandy. ren fe =e Ly = 25. 312 Simple Equations involving 34. Given? + y= a4, | v + 2= 0,7 to find the values of x, y, and z. yYytez=e, | [eat @+ 0-0) Ans.jy =4.(@—6+4 0), aero aie tina 35. Given” —y— z= 6, | | 3Y —e@—-Zz= a to find the values of x, y, and z. | 72 —yY— v= 4, ease ANS. 4 ¥ = 21, zg 12. 36. Giveny+ y— 2=8, 2% —- y+ 3z= a to find the values of x, y, and 2, 42+ 3Y — 2% = 17, see ANS. (Y¥ = 5, eee S7. Given = + —= 2, yey aS 1 1 3 ia to find the values of w, y, and z. 1 1 7 yt ea 38. 39. 40. 41, 42. three unknown Quantities. 313 Given @ + 2y + 3z= 17, y + 2z + 3x = 13, ; to find the values of 2, y, and z. 1 ul (eaee 1, ANS. Iwo “= 4, Givenv— y+ z= a 8“ —4y + 2z = 50, ; to find the values of x, y, and_z. We%—oy+3z2= All ANS. i = % ie = 364. Given 30 — y+ 7z=15, 5x2 + 3y — 22 = 16,; to find the values of a, y, and z. 7@+4y —52= 11, aa 4, ANS. yY = 2, <= 5. Given 2v + 4y — 32 = 22, 4@ — 2y + 52 = 18, ‘ to find the values of @, y, and z. On -- 7Y — Gee fe 35 IXNAS 4 = 7; Nevah Given 327 + 2y — 2 = 20,} 2v+3y+t6z= yy find the values of x, y, and z. e— yt 6z= 41, (aes . -_— ANS. ° 314 Simple Equations involving three unknown Quantities. 43. Given 7@ + l2y + 42 = 128, 32+ 3Y + 72 = 60, 6x2 + y+ 52 = 68, to find the values of x, y, and z. | Peet ANS, Yy —= 5, | ie = 3. : 44, Given 62” + 3y — 42 = 22, : 4x — y + 6z = 20, ; to find the values of 2, y, and z. 52 + af 6Zz2= wel ya bara: ANS. heise \2 = 2. [ 45. Given 11a — 1oy = —2 =, : to find the values of 2, V+Z—2yY z2—-y-!) : ye pe eae y, and 2. 3V =Y+Z2+7, ae ] ANS. 4y = 1), | lee | sou aera ip ee, ; | 46 CNL oe pd ig nh ots ABE x Zz . —+%—== 23, to find the values of #, y, and 2. je EN ct | Fas ahs a [Sey 420s ANS. 5 ¥ = 60, panies 2 3. or e 6. iy Pure Quadratics, &c. 315 III. Pure Quadratics and others which may be solved without completing the Square. GIVEN 32” — 4 = 28 + 2’, to find the values of z. ANS, & = 43 Gj jess wen’? +ysy: ey to find the values of # and y. cha ee g, ANS jem & “ly =+3. Given @—yiyii4:5,| and a? + 47’ Cala: [est ANS. i ie ah, to find the values of # and y. Given2 +yi:ev—yiia:b,) and ay = c’,| v=e./ (4+), ANS. to find the values of w and y. Given a’ + yi 2? —yiti7: A to find the values of x and x#y’? = 45, and y. ANS Feary ys Gi — 2Y = 6 Od “td +] to find the values of x and y. and ry — y? = 18,{ hep el to aaen Sani kE Givneg+y:v2—yil: 4,| to find the values of x and ay = 21,f andy. fx = 7, or — 3, ANS. ly = 3, or — 7. 316 Pure Quadratics and others which may be solved gs. Given az + bay =c',| And @.— YoU Lt Mes Bl Ge es ghee n— m Le BO Crees, 9. Given #* + y*°: 2 — y° 2: 559: 127,] to find the values of . and a’y = 294, # andy. to find the values of # and y. ANS. ANS. Pea ly = 6. 10, Given a — wy: xy —y’ ::3:7,) to find the values of 2 and ry’ = eh and y. { ANS. [ae eee | 1. Given /@ + /y: Se — VY 3: 4: 1,) to find the values and 2 — y = 16,f of v and y. [uv = 25, ly =9. 12. Given x — Sy = — and «/@ 4- ES ah to find the values of w and y. [vi 625, | ly = 16. ANS. ANS. 13, Given — i Ay 8 1] to find the values of and «/w7y =15,/ andy. {v = 25, or 9, ANS. ly = 9, Or 25. 14. Given #* — x’; xy — xy’ +: 7: 2,) to find the values of # andw+y=6,{ andy. Neca fe =4, or 2, ly = 2, or 4. without completing the Square. ; Weeks ws) 15. Given - +-= 7 y to find the values of x and y. 2 and — = 2 x 9 eee [@/==16, or 3; 16. Given z‘ — y‘ = 369, ave : y ’ 95] to find the values of wv and y. and 2” — 7’ = 9, f tae Mi e= = ly=a4. 17, Given a’ — 7° = 56,) | 16 > to find the values of # and y. andw—y=-, | LY oo [v= 4, or — 2, ly-=2,0r— 4. ; eat hE 18. Ses Sara mesa) aie to the values of 2. ANS. @ = i 2 19. Gi . OS aby meee ey a9 Die al to find the values of x and y. and vy? + Y= 14, | 2 le = 5 or2/2 ANS. - { > ee 20. Given Viz+V7y = 6] and # + y = 72, == O40 bes ANS. J eke ly = 8, or 64. 317 find 318 21. 22. 23. A 26. Pure Quadratics and others which may be solved Given 42? + a Ag ah 16Y, | 2 ey to find the values of # and y. | iL ober ene | ANS. frat st | ial | Given pada aN Seg + HOE tied ie, ax, to find th | 1 2 + /(2—2") e— fee): 3 | values of 2. Ans. o = ./ (41), | | | | | | Given VAC ar = 4, to find the values of 2. ab ANS. 2 = + Vb +1)" Given \/ (}2 + 2) —V/ ($e —2) =o (w@ +3) —VS/(@-3), to find the values of z. ' Ans. ®=2+5. | Given 3o/f (49-2) +3 SA Y—2) =A (Ry — 2), 4 and § / (2° — 6y) + of (y* — 92) 3 of (#* — by) 2219 1) to find the values of # and y. | : ANS. [a= 7, y = 8. | i ¥ 4 lues of . 8 va | ST Ng) ALN a 40 y 4° x 9 Ys | pipette | ANS : 28. 29. 32. Tiv iar, without completing the Square. 319 Gi 3+ y3 = 20 aia 7s i! to find the values of # and y. and 23 + ys = 6, Jukes (vous, or + \/3, ' Ly = 32, or 1024. Given a* + 2a’y’ + y§ = 1296 —4ry. (@ + ry 4+ ¥’),| and #7 — y= 4, J to find the values of # and y. eee {v= 5, or — 1, = 1,0r— 5. Given (a? + 1) . (vi — 1)? =2. (# + 1), to find the values of 2. 2b 2 ANS. @ = (CS) : Given Va—/ (4-2) = a, to find the value of z. Jat /(a—2) 2 ANS. & = a ; (a +1) Given Va aaa at =4,] to find the values of x and fa: Syii Vy i 4 | f 625 Ans. ! 16” y = 25 ieee Given —— —= - 2 \ to find the values of # and y.. and wy — xy’ = 16, jen se=4, or — 2, ly = 2, or — 4. 3820 Pure Quadratics and others which may be solved V (4a +1) + V4e I to find the value of a. ; J/ (42 +1) — i Peas th 33. Given 4 ANS. % =-. 9 fick Bat vou! Rae ae at+ae—/(2ax + 2’) (6 = 1) | TGA | : 34. Given to find the values of a. ANS. @=+t 1 1 35. Given (S a 2) +- i 2) ates 4a a 2 , to find the 27 — 3 94 +3 13 42 values of 2. aii ANS. & = —. 14 es » ind 36. Given V(@—y) +3 @+N=FE— | of and 2° + y?: wy 3: 34:: 15, wand y. ANS. te ae : | y= 3. 37. Given 2‘y’? — 2°y' = 216 , : an >| to find the values of # and y. and zy — ary’ =6, J 8 {v= 3, or — 2, ANS. : ly = 2, or — 3 38. Given 2 + @q/ (vy’) = 208,) and y* + y ¢/ (ay) = 1033, ANS ee = 8 ly = 27. - to find the values of # and y. 39. Given #3 + viyi+ y= 1009, | to find the values of @ and 2° + ay3 + y° = 582193, | and y. ee) fz == §1,forelG; ly = 16, or-8t. 40. 41. a 42. lisa without completing the Square. 3821 Given 2 + y’ + vy. (v7 + y) = 68,] to find the values of and 2° + y*°—3a°=12+3y,f wandy. pen {v= 4, or 2, ly = 2, or 4. Given vy . (v7 + y) = 84, pnd 23? . (2* +4") 2600, | to find the values of # and y. we {w= 4, or 3, "ly = 3, or 4. ete YY, etven —— r a Y to find the values of w and y. and = = ba pious a ANS. {z sii? ly = 3. 43. Given / e + 3) —/ <— 3) — (2): to find the values of x. Ans. =+9//2. 4. Given f/x —SYy=V2. (fat Vy)» | to find thevalues and (v7 + y)?=2.(@—y)’,{ of # andy. ANS. a ly = (3-202). Y alm — n)? + g-imn Given fees iano *, to find the value of 2. ze 1 A Sat Use a ] \(m +2)? ANS. & =[— a : at —=1 $22 Adfected Quadratics involving only 46. Given y +30/y. \./(a+ b2) —V7y). ve + ta) aa 30/y. weal a— bx ) 3 Fata a— bx Cera rua / ( )s to find the values of w and y. _a—1 ANS. Fs hou y = §0/(2a +1) + 12°. \ f and AT IV. alerted Quadratics involving only one unknown Quantity. a. Given v? + 4v = 140, to find the values of 2. ANS. C= 10, or — 14. 9, Given 2 — 6x + 8 = 80, to find the values of 2. ANS. x = 12, or ae 0. 3. Given a — 10% + 17 = 1, to find the values of 2. ANS) ot 55,0102; 4. Given 2 — x — 40 = 170, to find the values of 2. ANS. # = 15, or — 14. 5. Given 32? — 92 — 4 = 80, to find the values of wz. ANS. & = 7, or — 4. 6. Given 72? — 21” + 13 = 293, to find the values of z. ANS. & = 8, or — 5. . ee eae 7. Given Py ee Te 154, to find the values of x. 5 ANS. 2'= 9. or — one unknown Quantity. 323 | , 22° L gs. Given ie +3i= s + 8, to find the values of z. 9 ANS. @ = 3, or — a io, Given? + 4+ ae = 13, to find the values of z. ANS. @ = 4;0r' — 2. . 36 — @ 10. Given 47 — ae tee find the values of z. 3 ANS. av = 12, or — re me Given 16 — 5— — Bo enh + 3a, to find the values of x. Lv 6 ANS, = 35 or 5 “+3 16— 22 12. Given + ——— = 51, to find the values of z. 2 20 — 5 6 ANS, # = 5, or so 10 ; : 40 13. Given 14 + 4% — —— = 3@ + oe to find the values of x. ANS. & = 9, or 28. +4 —v 42 14, Given —— = Meh wists — 1, to find the values of z. ee Sl i a net 21 OF oe ; 15 — 2 12—3@ 23x@ + 60 15. Given —— — ~~ = 7x — Ef to find the 4 40 — 5 fi values of 2. 229 ANS. # = 3, or —. 148 324: 16. 17. 18. 19. 20. 21. 22. Adfected Quadratics involving only : e+ill 9o+42 Given Te eae ss ae 7, to find the values of z. 1 ANS. # = 3, or — 3 ; Lk 42 — 3 34” — 16 Given DEEN NO Nino erica es + —, to find the values 9 42 + 3 Ole 1 ANS. i = 6, or ms WH “. Given ———— = saath, to find the values of 2. +60 3H%—5 ANS. # = 14, or — 10. , 32 — 42 — 10 Given EAS Sagres ame 31, to find the values of 2. YU+5 10 ANS. & = 7, or — ae Given — ——~ = 21, to find the values of 2. vL—1 22 4 : 8a 20 Given — 6 = —, to find the values of 2. L+2 30 2 ANS. #& = 10, or — i : 40 27 Given z a aa 13, to find the values of #. 15 13 23. 24. 25. | 26. 27- one unknown Quantity. 325 5X2— 12 3H — 24 (a 34 Given ———— + ——— = 9 — ——_,, to find the values 42 —12 15 Uline. 4 184 ; 2” — 5 Given + — = 81, to find the values of 2. —4 L— 3 0 ANS. # = 6, or —. : x LX Given eel tied 1, to find the values of 2. 10— 2% 25—34 : WANS? @ = 8, 0r 1322. Pe: 42 —5 32 — v + 23 «yd SR al kite, to find the values of x. YH 302 +7 132 154 ANS. C= ee 25 or Pe) ae 45 : 8x" + 16 122 — 11 Given 2x7 + 18 — Peas eerie 27 — ——_——., to find the 42 +7 22 —3 values of 2. ANS, 2== 8) Ors, rye ax’ “+ 20 20 Given — ae ac naa pears to find the values C+9 20%+18 2 of x. ANS. & = 4, or — 2, Given —— eas + cabal Sabeae to find the values of x. L+6 27 + 4 3H +4 2 ANS. & = 8, or — A 326 30. 31. 32. 33. 34. 35. 36. 37. Adfected Quadratics involving only Given ———— eee eges Lara to find the values of 2. ov +3 5H@+18 5x 4 ANS. & = 6, or — 2. : 82 —1 4 Coven ee ie a ee etme to find the values of 2, 9+5”0 2+4” +412 | 137 ; 8 32 Given “ & , to find the values of 2. 5—-B@ 4-2 #42 ANS.@ == 12, DF sh 13 , 2" — 1 — x Given ee kash al + z to find the values of z. 3-2, Wael 2 : it Z 3 6 11 ee he ee oe CO TIN the Values Ones ee én — at ee+oe 5a" i ANS, # = 3, or 7 40° +7H 5e—2' 4x? Given —_—_ = , to find the values of 2. 19 3+2 9 ANS, & = 3, or — mi 10 “ton +8 Given A Ree nr = 2 + x + 8, to find the values of 2. ve +e—6 14 ANS. @ = 4, or — ry V+ 12 & L+-12 ANS. # = 3, or — 15. Given 5 =, to find the values of 2. 38. 39. 40. 41. 42. | 43. 44, 45, one unknown Quantity. O27 Given \/ (42 + 5) x «/ (7@ + 1) = 30, to find the values of x. ANS. Pe oe 28 VE +9 _ V9 — 37 Given = , to find the values of z. re 9— Cie ANS, # = 25, or Sida 400 Given ~ aoe <4 to find the values of 2. L— Vi i 4 —3+f/—7 ; PONS. e == 4, OF 1, OF 2—/ (ex +1) B+ / (e+ 1) 8 ANS.” = 8, or rer Given = =, to find the values of #. ; 32 —1 2 - Given 5 . ——__~ + —— = 3/2, to find the yalues +5fu Va ’ of 2. 1 ANS. 2==¢1,-0% 3 Given (/2* — a = 32, to find the values of wz. ANS. # = 4, or (— 5)3. Given 23 + 7v3 = 44, to find the values of 2. ANS. # = + 8, or + (— 11)8. Given 4x3 + # = 39, to find the values of #. is \* ANS. & = 729, or (o>) : 328 46. 47. 48. 49. 50. 51. 52. 53. Adfected Quadratics involving only Given 32° + 424° = 3321, to find the values of 2. sos ANS. @ = 3, or — 41. Given -- +2= ae to find the values of x. 2 ANS. @ = 4, or ee 4 Given x3 + ane se 3 + x8, to find the values of z. Lv ANS. # = 4, or (— 7)3. 6/are ane many a Given yf = ee =. = “v=, to find the values of 2. 27 ANS, e= 1, or — nS Given 32" or — a. Given adx — acxz’ = bcx — bd, to find the values of 2, 87. | 54. 55. 56. 58, 159, 60. one unknown Quantity. 329 2 rn2 Given < — ° +> as = 0, to find the values of 2. ANS. 2 = — a) bsy@—0) a Given 9a‘ b*x’? — 6a°b’x = 0’, to find the values of z. a’? 4 ANS. “= NaS VAN a 3a 6 Given (a + 0) .2° = cx a S55 to find the values of x. eee ck + se) .(@ + 4) Given 3 // (112 — 87) = 19 + / (34% + 7), to find the values of 2. ANS. & = 6, or ie) 625 Given (/ (22 +7) + (34 — 18) = / (7% + 1), to find the values of 2. ANS. & = 9, or — =. Given 7. / 5) — Je + 45 = Gv (ioe +56), to find the values of 2. 14568980 ANS. & = 20, or sarah Given 16—46/r 88+ 33/@ 2? — 5x” +11 SEE ee * @—3V2). (+ Vay to find the values of z. PENA tae 935,01 (7. 330 61. 62. 63. 64. 65. 66. 67. 68. 69. Adfected Quadratics involving only Given 54-0 a _ 230 — Vv, "2 —3u +4 L+24/z 6+ f/x (v7 +2/x) x (6+ Sa) to find the values of 2. 32 ANS. & = 5, or — —. 15 Given a2 + /e7i:7—-Srir3Vet+e: 24/2, to find the values of 2. ANS. # = 9, or 4. Given 2 + 11 + 4/ (x + 11) = 42, to find the values of 2. Ans. v = +5, or + V/ 38. Given (7 — 5)° — 3. (v — 5)3 = 40, to find the values of a. ANS. & = 9, or (— 5)3 + 5. | Given 2 + 4/(# + 6) = 2+ 34/(wx + 6), to find the values of x. ANS. # = 10, or — 2. Given (x + 5)? — 42” = 160, to find the values of a. ANS. @ = + 3, or + «/ (— 15). Given x? — 72 + 4/ (x —7x + 18) = 24, to find the values of x. , ak ANS. £2 = 9, or — 2, oh tse Given 9” — 427 + / (42° — 97 + 11) = 5, to find the values of 2. oe te ANS. £& = 2, or be or a Given a + 4/ (5% + x’) = 42 — 52, to find the values of a. —54 221 one unknown Quantity. 331 2 J (#@ +2) _ 17 came ‘2 regan Choe + CACeeyy. to find the values of x. . Given 3 ANS. # = 6, or wary a L 4 21 Given ee a. Se a ae to find the values of xv. a ANS. # = 12, or — 3, or ae eet to Se eG Given oe on — = aa to find the values of 2. ANS. # = 7, or — 3. - Given 22 pimp ® erases ——., to find the values of x. 3H—5 3H +5 a pores Bey FSi hg NT OF ee “v+@r—4 Va , to find the values of 2. Givne + /rt+e= ANS. @ = 4, OF. 1. 2 _ ; i eee , to find the values of 2. w—4 258 39 ; 2 Given (« + =) +e=42— to find the values of 2. —7+AY 17. 2 WAM Mer 45 OF 2, OF 332 ree 78. 79. 80. 8l. 83. Adfected Quadratics involving only Given v7 + 4—2 a) 52. = ——, to find the values of x. ANS. Qpeeet tot Olat Aa Lys Given iia + J («" ~ 2) =, to find the values of x. Given \/ {(v — 1). (# — 2)} +/{(@ — 3). (#-—a)} = V2, to find the values of 2. TNS xe ay OLE. Given z*. ( + =) — (3° + x) =70, to find the values — of 2. aca, sa ANS. #@ = 3, or — an or alls Vas | 3 6 Given 2? — = +] es a ae to find the values of 2. naan Ge a ANS. # = 4, or — 8, or ate (on), Given Wa ra 92") a P ee sal find the values of 2. ANS, = 55, or a Y (15661) Given : ; e@+ir—s 2+27—8s 27—13%—8 find the values of 2. DNS) imeet fOr als 84. 85. t 86. b 87. 88. 39. } 0. ide } one unknown Quantity. 333 Given 3. {(v — 1)’ — v7}? + 24 = 341 + 2. (x — 1)’, to find the values of 2. 34/3 +4/(— 109) 24/3 ; Given 2 — 243 + ox — \/x = 6, to find the values of z. —5+4/(—11) mayer ett ANS. & = 5, or — 23 or PONS ata 45,0Fc1, OF 132° Given 2* + — 39x = 81, to find the values of x. —13+ — 155 PUN Ssrae oct. S108 en alee Ure Given 2? — 24% + 4 = 24/ (x — 1), to find the values of x. Ans, = 4+ 4/6, or + \/—2. 2 f= aay aie : e+s 4046 Fi Given - _ _ are as to find the values of x. ANS. & = 4, or — 23 or —1 + 4/ (— 3). Given /x — iets ae to find the values of z. v / L—2 + a PUN Sy 1G, Ofe1,. 08 tee Given 7 — anil at Sy + LM kes find the values of x. AT oat ast Vie —3FV—7 ANS ae 9, OF 4, 0F = REY Given 42* + = 44° + 33, to find the values of @. 3 Bitar — 43) . ANS. 2 = 2, or — 53 Fass ten 304 Adfected Quadratics involving only g2. Given @.(/# + 1)? = 102. (@ + eres) [ 27 ANS. < 640 4. two unknown Quantities. 307 oy Eta ig fae crag ry 5v 5 3 and “24 ¥ 22 PY + 2, to find the values of wv and y. 3 xv = 6, or — —, 266 ANS. J | Bee gee oTBE Mee ick hear Given 2’? — 7’? =’, | to find the values (w+y+ 6)? + (w@—y +t by =20',f of v and y. —b+// (2a — BB + 20’) Oe Naa re eae an eae CR Sear weirs es a ahs c i: = / (207 — 8? + 20’) rs : fy uF oy eae) (geen a ee) Given 2’ + y:27—yi::9:7, Ie find the values of z andl +a? ;y+4::5y +7: 3y, and y. jpatoorti/(-2), 14 y = 2, or — —. 19 ANS. Given x’ + 22°y = 441 — xy’, | to find the values of x and vy = 3+ 2, Wai hop v= 3, or — 7, or —2+4/(— 17), ANS. 5 $4/ (— 17) ren he |\y sr ore, or Given x’ + 4y’ = 256 —4xy, ) to find the values of « and 3y? — 2? = 39, J and y. fv =+6, or + 102, ANS. ; ay == 5, or 59; 338 Adfected Quadratics involving 9. Given (@ + y)? — 3y = 28 + 3a, | to find the values of # and 27y + 3% = 35, j andy. — 5 +4/ (— 255) 4 ] vi eae ie or | 7 Ue (= 255) 4 y == 2, oF =, OF ANS. 10. Given (27 — 4y)’ + @ — 2y = 5,| to find the values of # and 2? — y’ = 8, Jo gandiy. ly. ANS. ee 7 Y — 1, or 3° : to find the values | a | We Given 3 @—) =1 4+ GQryy | of # and and J@+ytVe-n=s | ut te 13 eee ANS. 5 ly Tig" ; 2 Ay — ., \ to find the 12. Given a+ 10%7+y=119—2 Sy x (x + 2), | a: of and # + 2y = 13, 2 andi —6o9+ , es = 5, ree or ie Ree VAN 2 4 ANaH 121 = 4/ (241) ‘ ly = 4, or - or Se ee ; 2a" 39 eh Given Ld gy al | y" = y j 49° \ to find the values of x and y. | and a + y’ = 65, ANSWER, = 4, 0r + 44/(— 65). opp (= 450. 30 3410) G 5 as 14. 16. two unknown Quantities. 309 Givenew +y+/(e + y) =6,) to find the values of wx and xz? + y* = 10, f and y. 9+ /—é61 9 b - v= 3, o0ril13; or ANS. v + — 61 etl Or as or NT Given 2 + 44/ (@’ + 3y +5) = 55 — 3y,| to find the values and 64 — 7y = 16, J of wand y. — = ce a 5 a phe (389 y —7o+ 5 ly = 2, or — ca or bad ACE ) 7 49 he = 5, or ANS. . Given # + 34 + y = 73 — 2xy,| to find the values of # and y’ + 3y + @= 44, f andy. Ae (@ = 4, or 16; or — 12 + 1/58, ‘ly = 8, or—73 or—1 4 58. Pee Vee M7 ) to find the wy) Y — 4o/(@ + y)’ values of ae J wand y. lie 6, Obi; An elie IOP ANS. oe LY = 2, or 1, or re SnkVAMoePB ETE 8 Given y — y? = 16 — cea to find the values of # and y. and 28 —-y=2 + 422, 58|" ANS. 16,.0F res yp ee: 289° Z2 340 19. 20. 21. 22. 23. Adfected Quadratics involving . 4 4 —_ br Given 2 + y' = 975] to find the values of # and y. andw+y= 5+ — 151 | fee aor as or == a se (= 151) 2 y = 2, 0r 35 OF a gl ee to find the Su JC )+ Mga —2y % values of | and 2? BA rea ony | x and y. = 6, or 3, ANS. 3 (ee or —. Z | Given v +4\/e+4y=21 $3Vy ta ay "Plo find the and (/a + /y =6, J values of x and y. = 25 ANS. | 169 Gi 2 ; "y) = (% — 4). iven 37 + 2/ (wy’ + 92°y) =(@— 4) -¥%| to find ti and 62 +yiyii@+-5:3 J ! values of # and y. [v= A NS. fy Cee | to find the values of Givenv+y=5, 1] = 455, x and y. aa and (a + y’) x (#@ + y’) | ) two unknown Quantities. 341 24. Givenz + y — J()-,; and #’ + y’? = 41, ae or 3.4/8, piano 4, rot of 2, Y : Salle ay | 95, eke a of Sten 136 = — Hg Te | to find the values of # ay to find the values ey of #@ and y. ANS, andvy+4=14—y, | ANS Pee caae ors £5 / (— 8), : ANS. - ly =4, or6; ors¢5./(—© Ms. Given @t¥_ 2-9 _ 4 \?. eet y ree to find the values of ery (252) 4 wand y. OS (@ YY 3 5 & = 3, or —3 or, or ANS 35 1 135 HT EE 05) A emer ecenceer al DT ok ea ibe | J = Kage ha leet ae. Given | to find the values ee and y. | andy’ =12+2xy, |{ Ro fewer y = 6. Adfected Quadratics involving 2 +> 29. Given an ae pe Deere & + ey | to find the values of y y z ee. | wand y. and 4y’ — vy = 2, ~ 50 faahahe my er ANS. | = ] or = a Y > 5 Given /iQit a? tye} tia ty t=4) and (4 — 2”)? = 18 — 4y’, | to find the values of # and y. Oe Ol / 105 30. — 2 AL ce RV ey ee to find the 31. Given = = o 2 40 © A a he tay values of and yy — fay’ = G? x and y. 196 289 puke ks OY =, OF eS ok & Y = 4; 9 ) 9g’ 3° D ABN/AM Era a— Jf (ev —y) | e Gi , — 2 9 ee J (2 — y’) Caiaves VAD e a) and oat) ene ee to find the values of wv and y. ANS. ; two unknown Quantities. 343 xv —15y—14 : 33. Given 5y + Vv ; J ) = “ — 36, to find the 5 eae e EN cea y values of erage. (F+5)-% w and y. ANSWER, + 5 5+ ie or — aL ore vases: or 4 WD 5 20 4 i = are es br i oan / iy maces 3 4/ (3849) 12 120 8 ; e+ #) eee a yx ( 42 ) 34, Given /( as togicg- 5 Gea and Acai Cir Bees 2 <= 7 1; Va —/(w@—y—1) to find the values of w and y. (amt or #3 or es Ge a ANS, . os a — ly = 2, or — 1: ee 35. Given aa" + yb" = 2 (ax)? : (by)2, to find the values . and wy = ab, of w and y. 2n S m—n genta, “= bmtn : La 2 a AeA ae a4 OE aa n ANS A eee 0 mH Pant 2s \a 20 +b of (an ay puny bmn weet Yt Ve y') 9 . 30. = —, 36 eee fy fe 4") ay (7+ y), and (@’ + yy +ev—y=2x.(xv? + y) + 506, to find the values of # and y. me as [a= sor Hat anes) 20g) 6¢ gins ——' 1209 ly =, or— 25 or va ). J ANS. 344, Adfected Quadratics involving 37+ Given 4 = JE+ ev. Ji “i”? | to find the values ie e of w# and y. and ——- — ——= = s 16 625 625 @ = 4, or —, or ——-, or 144” ANS. : | ("4 (= Ne y = 64, or — or ; L 61 38. Given J2 + /f= —= +1, | to find the values of # y Vay yin ees and y. and /a'y + / yx = 78, dl 371 spi ae or 163 or — ——_ +347; 12 ANS. 371 re ake SR or eae. 39. Given @ + 4/(3y? — 11 + 2v@) =7 + 2y — to find the e+y values of | and 3y—“e+7)= : 4 J (3y ae 2 anda ANS. fe ai = 40. Given # + y*=1+ 2%y + 3a°y*,| to find the values of | anda®+y=oaye+oy+a+1,f wvandy. | v= 2,0r—1 | Ans. | ; ° | y=1. : 2 ss a Pee 41. Given v’y—4=4ahy a | to find the values of # andy. Li— 3 = why}. (43 — y3), {v=1], ANS. ly = gt t L a4, 145. two unknown Quantities | 42. Given 5—2\/(y +2) = 2 4 Bde ata t/ha y Y to find the values of 2 and y —{Va-—3VyP B) {e~=4, ANS ly =}. EN Bae ACh a VALE ed 43. Given ae Pes Ga) cae and —.y‘ =y’4 — 1, | to find the values of z and y 5 85 ‘ [erg Oleg? NS 5929 lyase, rt /(-3 + ame ) Given (7 — 2 y—Vfxy.(y —1) =27? —2, to find the Pee MALY — 13 values of ae Dae ry —138 ° wand y. eae eae ANS. ee ees ie Flee )=a2-y, ronindatie ] f ma Ve +9) 3 _ 28-3 Hay, values o 22 J (@ ty) ae xv and y. 345 346 Adfected Quadratics involving 46. Given 2’? —y’?=3, | and (af +y'+a7y.@—y)y+e—-y= 328, { to find the values of x and y. — 2+2 — 13 ; yg —— aw Pp (8) Bm /—1,0r# SAAR 4 ANS. - $2 / Woe por 82 ona, on ey eee ). 47. Given y= 2’. (ay — Oz), and 2’? = ax — by, fe =a-+ br, sort ly =ar-+ br’; } to find the values of # and Yo | where r =p V/p’?—1, a—b+fae+20b+58' and p = A; =y.(l— ay"), | to find the values 3 + 22” —- 4x" 48, Given ——_,.——_ x —1 and (2”’ — 1). (2y’ — 1) =3, | crt NS. ane he or + /—1; or & 44/ (5+ V 33). : ) of x and y. ; U4 49. Given —# +4—40y° =140—y’. J (a - and at — 2. (2 4 ise Mad Sid, LS) y y to find the values of # and y. two unknown Quantities. 34:7 50. ee, y + 9) Ve. to find the x y LY a Jj A values of Y es PAN igeTS: re and vemiares AE | xv and y. ANS aoe: La == 95. Bee Ay typi Aleit oy es even y < Vu ’ | to find the values and a (y +1) =36.(y° +), jendev ep roy ANS. — lyati(re/®), or 2. (1 Given (#° + 1).y =(y? + 1).2", and (y° + 1).27=9.(# +1).y°, J ANS. 53. e=at// (a? +1), wherea=+ Given J and /at/is.(y—VW2)—4t= to find the values of # and y. ANSWER, = 4 or = or — or yah sty ’ 9° Q5° 3 — 3° 5 ys 13 y = 3, or 55 or =, or — 1; Nene yay See a? Bl aR A ital Eatin en vx of # and y. | 1. to find the values of # and y. VA (3 TY 1), y= b+ (b° —1), where b = 34/3.4/ {3+ V3}. 3 —- J y +1; : 788 + 24 / 644 2 25 37/644, or ea 5 348 Problems producing Simple Equations, 1 54. Given —.(#?—1) + 4. (223—1) = (yi + £3) 3 3 3 oe, 3 Tee oat 133 1 2 Yi GN 3 y Y3 36 | ¥f3 V3 U3 1S to find the values of # and y. ANS. ) 2 ie ly =8. VI. Problems producing Simple Equations, involving only one unknown Quantity. 1. Waar number is that, from the treble of which if 18 be subtracted, the remainder is 6? Ans. 8. 2. What number is that, the double of which exceeds four- fifths of its half by 40? Ans. 25. 3. In fencing the side of a field, whose length was 450 yards, two workmen were employed; one of whom fenced 9 yards, and the other 6 per day. How many days did they work? Ans. 30. 4. A Mercer bought 4 pieces of silk, which together measured | 50 yards; the second was twice, the third three times, and the. fourth four times as long as the first. What were the respective | lengths of the pieces ? Ans. 5, 10, 15, 20 yards. and afterwards 17 bushels at the same rate; and at the second time received 36 shillings more than at the first. What was the price of a bushel? | 5. A Farmer sold 13 bushels of barley at a certain price | Ans. 9 shillings. involving only one unknown Quantity. 349 6. A person bought 198 gallons of beer, which exactly filled 4 casks; the first held twice as much as the second, the second twice as much as the third, and the third three times as much asthe fourth. How many gallons did each hold? Ans. 108, 54, 27, and 9 gallons. 7. A Silversmith has 3 pieces of metal, which together weigh -48 ounces. The second weighs 12 ounces more than the first, and the third 9 ounces more than the second. What are their respective weights? Ans. 5, 17, and 26 ounces. gs. A Vintner fills a cask, containing 96 gallons, with a mix- ture of brandy, wine, and water. There are 20 gallons of water more than of brandy, and 17 more of wine than of water. How many are there of each? Ans. 13 gallons of brandy, 33 of water, and 50 of wine. 9. A Gentleman buys 4 horses; for the second of which he gives £12 more than for the first; for the third £6 more than for the second; and for the fourth £2 more than for the third. The sum paid for all was £230. How much did each cost ? Ans. 45, 57, 63, and 65 pounds. 10. A poor man had 6 children, the eldest of which could earn 7d. a week more than the second; the second sd. more ‘than the third; the third 6d. more than the fourth; the fourth 4d. more than the fifth; and the fifth sd. more than the ‘youngest. They altogether earned los. 10d. a week. How ‘much could each earn a week? ANS. 38, 31, 23, 17, 13, and 8 pence per week. 11. An express set out to travel 240 miles in 4 days, but in consequence of the badness of the roads, he found \that he must go 5 miles the second day, 9 the third, and 14 the fourth day, less than the first. How many miles must he travel each day? Ans. 67, 62, 58, and 53 miles. 550 Problems producing Simple Equations, 12. There are 5 towns, in the order of the letters, 4, B, C, D, E. From Ato Eis so miles. The distance between B and C is 10 miles more, between C and D is 15 miles less, and between D and E 17 miles more than the distance between 4 and B. What are the respective distances ? Ans. From A to B 17; from B to C27; from Cto D2; and from D to E 34 miles. 13. A gentleman gave 27 shillings to two poor persons; but he gave 5 shillings more to one than to the other. What did he give to each? Ans. 11, and 16 shillings. 14. What number is that, the treble of which is as much above 40, as its half is below 51? Ans. 26. 15. Two workmen received the same sum for their labour; but if one had received 15 shillings more, and the other 9 shil- lings less, then one would have had just three times as much as the other. What did they receive? Ans. 21 shillings each. 16. ‘Two merchants entered into a speculation, by which one gained £54 more than the other. The whole gain was £49 less than three times the gain of the less. What were the gains ? Ans. £103, and £157. 17. The perimeter of a triangle is 75 feet, and the base 1s 11 feet longer than one of the sides, and 16 feet longer than the other. Required their respective lengths. Ans. 34, 23, and 18 feet. is. A company settling their reckoning at a tavern, pay 8 shillmgs each; but observe, that if there had been 4 more, they should only have paid 7 shillings each. How many were there ? . Ans. 28. involving only one unknown Quantity. 351 19. Divide the number 46 into two such parts, that one of them being divided by 7 and the other by 3, the quotients may together be equal to 1o. Ans. 28 and 18. 20. A certain sum is to be raised upon two estates, one of which pays 19 shillings less than the other; and if 5 shillings be added to treble the less payment, it will be equal to twice the ‘greater. What are the sums paid? Ans. 33, and 52 shillings. 21. Having bought a certain quantity of brandy at 19 shil- lings a gallon, and a quantity of rum exceeding that of the brandy by 9 gallons at 15 shillings a gallon, I find that I paid one shilling more for the brandy than for the rum. How many gallons were there of each? Ans. 34 of brandy, and 43 of rum. / 22. ‘Two persons, 4 and 3B, have each an annual income of £400. A spends every year £40 more than B, and at the end of 4 years, the amount of their savings is equal to one year’s ‘income of either. What does each spend annually? Ans. £370, and £330, respectively. 23. 75 Ye) s ii eae 2. Given 5” + 8y = 153, find corresponding positive in- | tegral values of w and y. eee fv=2, y= 1, fv ==, 135 7 == 11, la = 21, y = 6, [aoe 5, y = 16. 3. Given 52 + 21y = 2000, find corresponding positive integral values of x and y. [az = 3795 Y= 5, lw = 358, y = 10; Frith 17 other corresponding pairs, the values of x decreasing by 21, and those of y increasing by 5. ANS. | 4, Given 82 + 3y = 17, find corresponding positive integral y values of x and y. i ANS 0194, yin 3: 5. Given 10# + 17y = 71, find corresponding positive in- ‘tegral values of @ and y. =o ANS. ee ; ' ly =3. 6. Given 112+ 7y = 108, find corresponding positive in- ‘tegral values of # and y. ANS. # = 6, y= 6. pd2 404 Indeterminate Equations and Problems. 7. Given 11% + 15y = 1031, find corresponding positive integral values of # and y. AS se = /OlgHnG 7/12, lw =76, and y/== 13> with five other corresponding pairs, the values of # decreasing by 15, and those of y increasing by 11. gs. Given 132 + 7y = 141, find corresponding positive in- tegral values of # and y. ANS. i rae Y = 9. Given 13% + 14y = 200, find corresponding positive in- tegral values of # and y. 5R = 0) ANs.| ; MS ky 10. Given 172 + 7y = 310, find corresponding positive in-— tegral values of # and y. ; fe= 3, [v= 10, [v= 17, ANS. ay ==137,0 |) 120, sy ee 11. Given 27” + 16y = 1600, find corresponding positive integral values of w and y. ROT ee. 16, ea 32, {@ = 48, Y = 73; y = 46, | = 19. 12. Given 71” + 17y = 1005, find corresponding positive integral values of # and y. ANS age? “ly= 9 13. Given 99% + 19y = 1900; find corresponding positive integral values of # and y. | [v= 19, ANS. < ly = 1. | Indeterminate Equations and Problems. 405 14. Given 52—7y =3; find the least corresponding po- sitive integral values of w and y. ANS on NS. yaar 15. Given 7e—12y = 19; find the least corresponding positive integral values of x and y. 16. Given 11v —18y = 63; find the least corresponding positive integral values of # and y. fa=o, ANS. ly = 2. 17. Given 132 — 17y = 54; find the least corresponding _ positive integral values of @ and y. 1s. Given 19% — 117y = 113; find the least corresponding positive integral values of x and y. milf ANS. é 4 Ue De 19. Given 14% —5y = 7; find the least corresponding positive integral values of # and y. (v= 3, ANS. ly ae _ 20. Given 177 —7y = 13 find the least corresponding positive integral values of x and y. {v= 55 ly = 12. 21. Given 242 = 13y + 16; find the least corresponding ANS. positive integral values of w and y. | fv=5, ANS. iN ale 406 Indeterminate Equations and Problems. 22. Given 3% + 7y + 172 = 100; find all the positive in- tegral values of x, y, and z, which eitisty the equation. | {Gama Vi 16, PWS EAL he ono pas Ieee <= ies pl eee ae 5 ' 23. Given lov + lly + 122 = 300; find all the positive — integral values of x, y, and z, which athe the equation. ht fate = 20, V=21, v= 22, : TANS2 t/t, 0) = nD ae tome 4 js Tey pire i OE Sabet 24. Given 1747 + 23y + 3z = 200; find all the positive in- is tegral values of wv, y, and z, which ade the equation. Cor 3, @©= 2, %7= 1 &%= 6, ANS Sa aoe Te att 9 eh 3, 0 7/— e zou, 2tbah See ioe 25, 25. Given 20% + 15y + 6z = 171; find all the positive in- tegral values of 2, y, and z, which cbt the equation. op SaP 3a ae G5 ANS. {y= 1. soya Fee: z2=16 2=>6, 26. Given 35” + 43y + 55y = 4000; find all the positive: integral values of #, y, and z, which satisfy the equation. % Li 105, pee ob ree A \"= 5.8/0 yak SU See z= 2 <= 3. <= 1, 27. 6a + 7y + 4z = 122,)to find the corresponding positiv e lle +sy—6z= | integral values of a, y, and z. ; ANS see = 0506 8s, eae ‘ (i Vy the corresponding 28. Given 37+ 5y + 72= 560, ositive integral values ox + 25y + 492 = 2920, P g § of x, y, and z. fv=15, y=82, 2=15, =50, y=40, &£= 30. ANS. Indeterminate Equations and Problems. 407 29. Required the positive integral solutions of the equation 2ey+L+_y=195. pice fve= 8 #=11, ; Ly = li. y= 8. 20. Required the positive integral solutions of the equation sry —4y + 37> 14. ren a 3 and 2, mk Bnd 4¢ 31. Required the positive integral solutions of the equation bey = 2H 4+ 3y 4+ 18. (v= 5, [ve=3, [x=7, ANS. me ly =10. ly =2. ly=1. 32. Required the positive integral solutions of the equation 72y — 5x2 = 3y + 39. | ee fe = 1, 3, 5, 21. (y== Tl, 3, 2, Ie 33. Required the integral solutions of the equation 2Vy —3a°+y=1. ANS. deipris? ly = 4. 34. Find the least whole number which divided by 3 and 7 leaves remainders 1 and 2. ANS. 16. 35. Find a number which divided by 6 leaves a remainder 2, and divided by 13 leaves a remainder 3. Ans. The least number is 68. 36. Find the least whole number which divided by 17 shall Jeave a remainder 7, but being divided by 26 the remainder shall be 13. ANS. 143. 408 Indeterminate Equations and Problems. 37. Find the least whole number which being divided by 28 will leave a remainder 17, and being divided by 19 will leave a remainder 13. ANS. 241. 38. What number is that which divided by 23 gives a re- mainder 22, and being divided by 37 gives a remainder 36? ANS. 850. 39. Find a number which divided by 39 leaves a remainder 16, and divided by 56 leaves a remainder 27. ANS. 1147. 40. Find two fractions whose denominators shall be 7 and ‘ l 9, and their sum ~. 4 3 ANS. ae} and oe if 9 41. Find a number which divided by 2, 3, 5, respectively, leaves as remainders 1, 2, 3. ANS. 23, 53, 113, &c. 42, Find a number which divided by 4, 5, 6, respectively, shall leave 3, 3, and 5 for remainders. ANS. 23, 83, 143, &c. 43. Find a number such that when divided by 11 there shall be a remainder 3; when divided by 19 there shall be a remainder 5; and when divided by 29, a remainder 10. Ans. The least number is 4128. 44, Find the least number which being divided by 28, 19, and 15, leaves remainders 13, 2, and 7. ANS. 97. 45. Find the least number which divided by 6, 5, 4, 3, and 2, respectively, shall leave 5, 4, 3, 2, and 1, respectively, remaining. ANS. 59. Indeterminate Equations and Problems. 409 46. Find the least number which divided by 3, 5, 7, and 2, shall leave remainders 2, 4, 6, and 0, respectively. ANS. 104. 47. Find the least whole number which divided by 16, 17, 18, 19, and 20, shall leave 6, 7, 8, 9, and 10 remainders re- spectively. ANS. 232550. 48. Find the least whole number which being divided by the nine digits respectively, shall leave no remainder. ANS. 2520. 49. Divide 100 into two such parts that the one may be divisible by 7 and the other by 11. ANS. 56 and 44. 50. Divide 100 into two such parts, that dividing the first by 5 there may remain 2, and dividing the second by 7 the remainder may be 4. 82 and 18, ANS. i" and 53, 12 and 8s. 51. In how many different ways may £1000 be paid in crowns and guineas? — Ans. 190 different ways. 52, In how many different ways can £43. los. 6d. be paid with half-guineas and sovereigns ? Ans. In 3 different ways; paying 81 half-guineas and 1 sovereign 3) 41 29 93 22 39 29 1 39 3) 43 9? 53. In how many different ways is it possible to pay a bill of £351 with guineas and pieces of 27s. each ? Ans. In 36 different ways ; _ paying 333 guineas and one piece of 27s., and diminishing the _ former number by 9, and increasing the latter by 7. 4.10 Indeterminate Equations and Problems. 54. In how many different ways is it possible to pay £20 in half-guineas and half-crowns ? Ans. 7 different ways. 55. In how many ways can £100 be paid in guineas and pistoles (17s. each) ? Ans. There are 6 different ways of payment ; paying 7 guineas and 109 pistoles, 99 «24 9 §8 9 increasing each time the number of guineas by 17, and diminish- ing the number of pistoles by 21. 56. A Gentleman has to pay £1000, and has only two sorts of coins, guineas valued at 21s. 6d. and Louis d’ors at 17s. each. How many different ways may the payment be made with these coins ? Ans. 27 different ways ; paying 32 of the former and 1136 of the latter, and increasing the former by 24 and diminishing the latter by 43, in each case. 57. If I have 9 half-guineas and 6 half-crowns in my purse, how may I pay a debt of £4. 11s. 6d.? Ans. By s half-guineas and 3 half-crowns. 58. In how many ways can an equivalent for 13 dollars, at 3s. each, be given in English crowns and seven-shilling pieces ? Ans. Only one, viz. 5 crowns and 2 seven-shilling pieces. 59. A company of men and women spend £50. The men pay each 19s., and each woman 13s. How many men and | women are there? Ans. 2 and 74; or 15 and 55; or 28 and 36; or 41 and 17. 60. A Farmer laid out the sum of £1770 in purchasing — horses and oxen. He paid £31 for each horse, and £21 for — each ox. How many horses and oxen did he buy? Ans. 9 horses and 71 oxen, OF'S0 a eee OF 81 3" eA eee ea Indeterminate Equations and Problems. 411 61. ‘Two women have together 100 eggs. One says to the other, When I count mine by eights, there is an overplus of 7. The second replies, If I count mine by tens, I find the same overplus of 7, How many eggs had each? Ans. The first had 63, and the second 37; or the first had 23, and the second 77. 62, What quantities of tobacco at 16d. and lod. per pound, may be mixed with 50 pounds of tobacco at sd. per pound, so that the whole may be worth a shilling per pound? Ans. 51 lbs. of the former, and 2 lbs. of the latter; 52 lbs. 4 lbs. “ and so on, increasing the former by 1, and the latter by 2. 63. A has moidores only; B only crowns. How can B pay A £1. 12s. most easily ? Ans. By paying 28 crowns, and receiving 4 moidores. 64. A having only one-pound notes, owes 12s. to B, who has only seven-shilling pieces. What is the least number which 4 and B must respectively interchange, so that the debt may be discharged ? Ans. He must give 2 notes, and receive 4 seven-shilling pieces. 65. A person buys some horses and oxen. He pays £31 for a horse, and £20 for each ox; and he finds that the oxen cost him £7 more than the horses. How many horses and oxen did he buy? Ans. The least numbers are 3 and 5: and other answers may be obtained by adding 20 to the former, and 31 to the latter continually. 66. Find two numbers, such that their product added to their sum may be 79. ANS. 1 and 39; 3 and 19; 4 and 15; and 9 and 7. 67. Find three integers, such that if the first be multiplied by 3, the second by 5, and the third by 7, the sum of the pro- ducts may be 560. But if the first be multiplied by 9, the second by 25, and the third by 49, the sum of the products may be 2920. ANS. 15, 82, and 15; or 50, 40, and 30. 412 Indeterminate Equations and Problems. 6s. It is required to buy 20 fowls for 20 shillings; viz. geese at 4s., quails at 6d., and pigeons at 3d. each. ANS. 3 geese, 15 quails, and 2 pigeons. 69. Thirty persons, men, women, and children, spend 50s. in a tavern; the share of each man is 3s., that of a woman 2s., and that of a child 1s. How many persons were there of each class ? ANS. 1 man, 18 women, and 11 children; and there are s other answers, the number of men and children increasing by 1, and the women decreasing by 2. 70. and 87. Determine three numbers, such that their sum shall be a square, and the sum of their squares a perfect fourth power. ANS. (119)?, 2. (119). (180), and 2. (180), whose sum is (349)*, and the sum of their squares (281)‘. ss. Determine three square numbers which shall be in arithmetical progression, but the sum of their square roots shall be a cube. Aws. (169), (845)’, and (1183). APPENDIX I. I. Problems in Arithmetical Progression. 1. DeTermineE the esth term of the series 13, 122, 123, &c. 2. Having given the first and last terms of an arithmetic progression, and their common difference; determine their number of terms. 3. Having given the first and last terms of an arithmetic progression, and the number of terms; determine the progres- sion, in general; and in the particular case where the first term is 100, the last 1, and the number of terms 19. 4. Find the series in arithmetic progression having 29 terms, of which the first 1s 3, and the last 17. 5. Having given the first term of an arithmetic series = 2, the number of terms = 7, and the common difference = 1; find the last term. 6. Show that +, +, and — 4, are in arithmetic progression ; and find the 11th term of the series. 7. Show that +5, 1, 4, are quantities in arithmetic pro- gression; and find the sum of s terms of the series of which they are the first three terms. 8. In an arithmetic progression, it is observed that the ‘fifth and ninth terms are 13 and 25: what is the 7th term? 9. The sum of the two first terms of an arithmetic pro- gression is 4, and the 5th term is 9; find the series. 10. If three quantities are in an increasing arithmetic pro- 416 APPENDIX. gression; show that the second will have to the first a greater ratio than the third to the second. ine 12, Find the sums of the following series: 1+34+5+7 + &c. to n terms. 1+5+9+ 13+ &c. to n terms. 1+4+7+410+ &c. to 12 terms. 5+7+9+ 11+ &c. to 50 terms. 2+ 24+ 22 +3 + &c. to 13 terms. ban vc < + &c. to 16 terms. 4. 8 eee to 12 terms. > + 1 + 12 + &c. to 8 terms. —9—7—5 — &c. to 20 terms. —5—3-—1, &c. to 8 terms. 11+8s+5+ &c. to 8 terms. l 5 —~—]1——— &c. to 29 terms. 2 2 44 43 15 + oc + o + &c. to 16 terms. = 4+ <4 ot &c. to 19 terms. a Pate = 4+ = -— + &c. to 8 terms. 16." 15 Pim Mey Min oe aaa Resear n n na—b+(n—1).a+ (n—2).a+6+4 &c.tonterms. (a+ 27)? + (a+ 2’)+ (a—2z)’ roe to n terms. Ce) He aa Gis Oni Sat a+b a+b lad 2 Show that + &c. to n terms, is een at ih ees mn APPENDIX. AlL7 13. Show that the sum of the (m— n)™ and (m + n)™ terms of any arithmetical progression will be equal to twice the m™ term. 14, It is required to divide (a) into n parts proportional to the numbers 1}, 2, 3, &c. 15. Having given the first and last terms, and the sum of an arithmetic series; determine the common difference, and apply it to the case where the first term is = 1, the last = 50, and the sum = 204. 16. Having given the first term = 1, the number of terms =n, and the sum = s, of an arithmetic series; determine the common difference. 17. Having given the n™ term of an arithmetic series, and also the sum of m terms; determine the series. 18. If the first term of an arithmetic series be = 1, and the common difference = m, the sum of n terms of the series is}. {mn? — (m —2).n. 19. How many terms of the series — 7 — 5 — 3 —, &c. ‘amount to 9200? 20. If the first term of an arithmetic series be = 1, the ‘common difference = 4, and the sum = 120; determine the ‘number of terms. _ 21. Ifthe first term of an arithmetic series be = 31, the ‘common difference = 14, and the sum = 22; determine the number of terms. 22. The first term of an arithmetic series is 3, the common idifference is 4, and the sum of 7 terms is 1081; find », and ex- plain the double answer. 23. If the first term of an arithmetic series be = 11, the common difference = — 5, and the sum = 5; determine the number of terms. 24, There are five numbers, of which the first two are 21, Ee A418 APPENDIX. 3-2;, and each number exceeds the preceding one by the same fraction ; find the numbers and the sum of them. 25. The sum of an arithmetic series is = 1455, the first term = 5, and the number of terms = 30; determine the com- mon difference. 26. The sum of an arithmetic series 1s 91, the common difference 2, and the last term 19; find the number of terms. 27. The sum of a certain number of terms of the series 21 +19 +17 + &c. is 120; find the last term and the number of terms. 28. If a be the first term, 4 the second, 7 the last term of an arithmetic series, the sum = ; = ; decoded 29. In the expression s = {2a + (n—1).d}.4n; if n be negative, point out the form of the series which satisfies that condition; and in finding », show when the values, one, both, or neither are congruent values supplying interpretable so- lutions. Determine the series in the case of a =7, d= 2, 5 40: 30. If abe the first and 7 the last term of an arithmetic series, in which the common difference is 8, and the number of n. (1? — a’) 2.(n—1)8 31. The sum of » terms of any increasing arithmetic series, whose common difference is equal to the least term, will be terms 2; then will the sum of the series = equal to the sum of (m + 1) magnitudes, each of which is half | the greatest term of the progression. 32. Ifthe number of terms of an arithmetic series be odd; show that the sum of the series is equal to the middle term multiplied by the number of terms. i n— ? od. The n™ term of an arithmetic progression is : and the sum of n terms is — . (32 + 1); find the series. APPENDIX. 419 34, The first term is n? — n+ 1, and the common difference is 2; prove that the sum of » terms is n°; thence show that P=1, 2=>3+4+5, 2=7+9+11, &c. If the first term be n”-!—n-+ 1, the sum of n terms is n”; and thence show that =i, 2=7-+9, 3° = 25 + 27+ 29 + &e. 35. Thesum of n terms of an arithmetic series is pn + gn’, whatever be the value of 2; find the m™ term. 36. In an arithmetic series the first term and the common difference are the same, and the sum of the series is always equal to the number of the terms + the square of that number; determine the series. 37. If from any square number (n’) there be subtracted the sum of an arithmetic progression beginning from unity, having a common difference unity, and continued to as many terms as there are units in the root of the number (n); the remainder will be the sum of the progression continued to (2 — 1) terms. 388. The sum of an even number of terms of any arith- metic series whose common difference is equal to the least term, will be four times the sum of half that number of terms di- minished by half the last term; the first term bemg the same in each case. 39. The latter half of 2” terms of an arithmetical series is equal to trd of the sum of 3” terms of the same series. 40. Prove that 1, 3, 5, 7, &c. is the only arithmetic pro- gression beginning from 1, in which the sum of the first half of any even number of terms bears to the sum of the second half the same constant ratio; and determine that ratio. 41. The sum of » terms of the series 1, 3, 5,7, &c. is to the sum of (nm — 1) terms of the series 2, 4, 6, &c. :: mi n—13 required a proof. 42, The difference between the sums of m and n terms of an arithmetic progression: the sum of (m +) terms :: m—n :m+n. Ee2 4.20 APPENDIX. 43. ‘The two first terms of an arithmetic progression being together = 18, and the three next = 12; how many terms, beginning with the first, must be taken to make 28? and explain the reason of the double solution. 44, The first two terms of an arithmetic series being to- gether = 18, and the third term = 12; how many must be taken to make 78? 45. The first and third terms of an arithmetic series being 21 and 14 respectively; find the second term and the sum of 19 terms. 46. The (n+ 1)" term of an arithmetic progression is ma—nb a—b 47. How many terms of the series 1, 3, 5, 7, &c. must be added together to produce the (2m) power of a given quantity 7? ; required the sum of the series to (27 + 1) terms. 48. Find » terms of the indefinite series 3, 5, 7, &c. whose sum may be the (m)' power of n. 49. Inthe two series 2, 5, 8, &c. and 3, 7, 11, &c., each continued to 100 terms; find how many terms are identical. 50. Having given (a) and (0), the (m)™ and (n)™ terms of an arithmetic series; determine the value of the (#)™ term. 51. Having given as before; determine the sum of (p) terms of the series. 52. Having given (a) and (6), the (m)™ and (n)™ terms of an arithmetic series, and its last term (a + 0); determine the first term, common difference, and number of terms. 53. The sum of m terms of an arithmetical series is m, and the sum of n terms is m; show that the sum of (m+) terms is — (m+n), and the sum of (m—n) terms is (m—n) . ( + va ‘ 54, If the m™ term be n, and the n term m; of how many ] | APPENDIX. 421 terms will the sum be 3 (m+) .(m+n— 1); and what will be the last of them. 55. In an arithmetic series, if the (m + n)™ term = p, and the (m—n)™ term = g; show that the (m)™ term =3.(p+ q), and the (n)" term = p — (p—q). —. 56. If 2, y, and z be respectively the p, g, and 7** terms of an arithmetic progression; show that (p — g).z+ (r—p).y +(qg—7r).v%=0. 57. There are two series in arithmetic progression, the sums of which to n terms are :: 13 —7n:3n+ 13; prove that their first terms are as 3 : 2, and their second terms as — 4: 5. 58. If S, S’, S” be the sums of three arithmetic series, = the first term of each, and the respective differences be 1, 2,3; prove that S + 8” = 2S’. 59. There is a series of m quantities in arithmetic pro- gression whose first term is a, and common difference d; show that if there be taken v terms of the first, n — 1 of the second, &c. to the nv inclusive, the sum will be n.(n + 1) ites - {3a + (n— 1). ad}. 60. If S,, S,, S,.... are the sums of m terms of different arithmetic series having the same first term, and common dif- ferences 1, 2, 3.... respectively; show that S,, S,, &c. are in arithmetic progression ; and when that first term is 1, prove that mn’. (n + 3) Seer. +S = ; 61. If there be (p) arithmetical progressions, each begin- ‘ning from unity, whose common differences are 1, 2, 3... show that the sum of their (n)™ terms is = 3-t(v—1)-p' + (n+ 1) - ph. 62. Ifaandd are respectively the first term and common difference of an arithmetic series, S, the sum of terms, S_, , , 4.22 APPENDIX. the sum of (n+ 1) terms,&c. prove that S,+ 8,,,+85,,, + &e. a = —1).2.—— — 2).(7— 1).n.———. to m terms = (37 — 1). erste 2).(~—1).n ae ONDA S54.) S53 Uy emac rs S, be the sums of (p) arithmetic progressions continued to » terms, and their first. terms be 1, 2, 3, 4, &c. and their common differences 1, 3, 5, 7, &c.; show that S,, + S, + S, + ..0.. + 8 = 3. (np + 1) np. Ode LEO os iy eet S, are (p) arithmetic series each con- tinued to m terms, whose first terms are the first (py) even numbers, and common differences the first (p) odd numbers respectively ; show that m.(n + 1 (n+ Dy 65, 1858), 85, 94. +--: S,,, be the sums of n terms of 2” arith- . metical progressions, whose first terms are the same, and com- — mon differences d, 2d, 3d....2nd; then will (S, + S,.....+ 8,,) —(S, + Sot... + 8,,_,) = 3. (n—1).d. 66. If S_ denote generally the sum of m terms of any arithmetic progression; prove that / (Src | S 9= 8,2. n@ —S.n.(n— 2). 67. Find three arithmetic means between 1 and 11; and — seven between 1 and — 4; and fifteen between 3 and 47. | 68. If between all the terms of an arithmetic progression f the same number of arithmetic means be inserted; show that — ‘ ae the new series will still form an arithmetic progression. d ie 69. Insert 6 arithmetic means between 4 and 2; and find — their sum. 4 70. Insert ” arithmetic means between a and 8, and apply i the general expression to insert 3 terms between 5 and — a5 find the sum of the series for 20 terms. % APPENDIX. 423 71. There are ” arithmetic means between 3 and 17, and the last is 3 times as great as the first; find the number of means. 72. There are n arithmetic means between 1 and 31, and the 7th mean is to the (n—1)™:: 5:9; prove that the number is 14, 73. Between aand 4, a being less than 4, insert m means such that a, — a, a, — G,,...... 6 — a, form an arithmetic progression whose common difference is d; and find the limits between which the value of d must lie. 74. The sum of n arithmetic means between 1 and 19 is to the sum of the first (n — 2) of them:: 5:3; determine the means. 75. If S be the sum of an arithmetic progression, a, J, ¢, d, &c. to n terms; determine the sum of the series Sta, St(a+bd),St(a+b+0), &. 76. In any arithmetic progression of which a is the first term and 2a the common difference; prove that the number of terms which must be taken to make a sum S, 1s ih 8, S being a assumed such that is any square number, but no other. 77. Determine the sum of » terms of the triangular num- bers 1, 3, 6, 10, 15, &c. the terms of which series are 1, 1 + 2, 1+2+ 3, &c. the successive sums of 1, 2, 3, &c. 78. Determine the sum of terms of the pyramidal num- bers, 1, 4, 10, 20, 35, &c. the successive sums of 1, 3, 6, 10, 15, &e. 79. Find the sum of n terms of the series 1’, 2’, 3”, &c. 80. Solve the equation (v7 — 1) + 2.(¥— 2) + 3 (w — 3) to 6 terms = 14. 81. Having given the first term and common difference 424, APPENDIX. of an arithmetic progression; find the sum of m terms and the sum of their squares. 82. Having given the sum of (2”) quantities in arithmetic progression, and the sum of their squares; determine the quan- tities themselves. 83. In the series 1, 2, 3, 4..... 100, determine the sum of the numbers which are not squares. 84. Prove that the sum of the series 1’ + 3’ + 5’, &c. to nm terms = “ . (42? — 1). And determine the sum of the series a’ + (a+ 6b)? + (a + 26)’ + &c. to n terms. 85. The sum of the series 0, 1, 2, 3, &c. continued to an unknown number of terms, being = 1225; determine the sum of their squares. 86. The sum of 9 terms of the series n’ + (n + 1)’ + (n + 2)? + &c. = 501; determine the value of n. 87. Determine the sum of 10 square numbers, whose roots are in an arithmetic progression, the least term of which is = 3, and the common difference = 2. 88. The square of any number of digits less than ten, each of which is unity, will, when reckoned from either end, form the same arithmetic series whose common difference is — unity, and greatest term the number of digits in the root. Required a proof. 89. A man is employed ina certain manufacture, where the quantity of work which he produces in the 1st, 2nd, srd,.... days are 1, 3, 5, 7,..... For the first day’s work he receives a shilling, and afterwards in proportion to his work. If a new workman is added every day under the same conditions, how much money will have been paid to the men after n days? 90. A gentleman owed to each of two persons, A and Bb, an equal sum of money, which he discharged as follows: to 4 he paid £8 the first payment, £12 the second, £16 the third, ~ Poa oe APPENDIX. 425 and so continued increasing £4 each payment. Now B at his first payment received but £1, the second £4, the third £9, increasing according to the square of the number of payments. Determine what he owed each person, and the number of pay- ments required to discharge the debt. 91. Compare the sum of the numbers 1, 2, 3, 4, &c. with the sum of their cubes. 92. Find the sum of » cube numbers, whose roots are, in arithmetic progression, the least term of which is a, and com- mon difference d. 93. Determine the arithmetic progression, the number of whose terms is 11, their sum 220, and the sum of their cubes 147400. 94, Prove that when 1 is indefinitely great, a’ + (a+ 6)’ + (a+ 26)" + &.tonterms JB mn’ +1 wa r+ if 95. Find the sum of (m) terms of a series of polygonal numbers, which numbers are formed by assuming any arith- metic series that has its first term 1, and difference a whole number, and by making generally the (m)' polygonal number equal to the sum of (n) terms of the arithmetical series. 96. If the first term of an arithmetic series be (a), the last term (J), the common difference (d); and S,, S,, S,...... S_, be the sums of the 1st, end, ard,....... (m — 1) powers 10min of the terms; prove that (7+ d)”—a"™=mdS,,_,+m. ; m—-1m— 2 Ag? Same ec 3 oS, 5 Mt: OF GeLeb. d,- @ AT, @:—s27 cand. 0,0. 7.0 + 7+ ot 75 t ke. an inf. 63. Given =, : the first two terms of an arithmetic pro- gression; find the sum of 15 terms. And if the same quantities be the first two terms of a geometrical progression; find the sum of 15 terms. 64. If the terms of an arithmetic progression, a, a+7, a+a2r, &c. be multiplied by the corresponding terms of a geometric progression, 0, bd, bd’, &c. of the same number of terms; determine the sum of the resulting series. 65. Determine the sum of 7 + 5 + &c. an arithmetic series of 12 terms; also of — : ab Z — &c. a geometric series of 5 terms: and also of the infinite series, whose terms are the products of the corresponding terms of these series con- tinued in inf. 66. There are two infinite geometric progressions, each beginning from 1, whose sums are o and o'; prove that the APPENDIX. 437 ” sum of the series formed by multiplying their corresponding 1 Ooo terms 1s ———, : oto-—1 6%. If o, represent the sum of a geometric progression continued in inf., «, the sum of the squares of the terms, o, the sum of their cubes, &c.; then will l a. , day lh SS A Es — Hi om oO» 4 asa 1 a+r 68. IFS =1 + 24+ 2424 &e. ining 99 RE 8 3 5 i A As) —— 7 y. and S, = 1 tee, 5 + Ke. tn inf: Prove: (hats S665 0505509751, 69. If 8 =1 4+ + &e. ining. 1 1 1 S,=1—--+ —— KC eceeeee y jie 1 ] 1 S,=l1+— 57+ 7+—...... s=lt+stats prove that S, x S, = S,. 70. Determine the sum of the series ar + 3ar? + 6ar + loar* + &e. which arises from multiplying the terms of a geometric pro- gression by the corresponding terms of a series of triangular numbers. 71. Find the sums of the following series: 1+ 227 + 327+ &c. to n terms. 2 3 4 ex: 1+-—-+—+-—>+ &c. in inf. Lie edad ee 4 2a" 80 i Ag joke. a Sat ig home: + &c. in inf. a a a’ 438 APPENDIX. 1.2.04+2.3.a°+3.4.a° + &c. in inf. and to n terms. 1.2.3.0+2.3.4.2°+3.4.5.a° + &c. in inf. and to n terms. 1.2.3.4.” + 2.3.4.5.a" + 3.4.5.6.a"% + &c. ininf. & ton terms. 1.2+ 32? + 52° + 72" + &e. in inf. 1.27 +52? 4+ 94° + &c. in inf. 1.2.2 +3.4.47 + 5.6.2 + &c. in inf. 1 1 o 22.3.0 +4.5.6.2° +7.8.9.2° + &c. in inf. .3.04+2.4.27°+3.5.2° + &e. in anf. 5.27 +3.6.2° + &c. in in. 6.07 +5.8.x° + &c. in inf. 1.3.0+4.6.a°+7.9.x2° + &c. in inf. 5.0? + 5.7.0 + &e. in inf. 72. Prove that the sum of m terms of the series 1° + 3° + 5° + 7° + &c. is a hexagonal number whose root is 7’. 73. Prove that the sum of the reciprocals of the n powers of the odd numbers is to the sum of the reciprocals of the same powers of the even numbers :: 1 ; 2”—1. - 3242424 ...4.% 74. Reduce to its lowest terms ———————_._, 1—2+4+4— &c. Lie COD se mitts P) ES) 1 — oe, tee Soe O27 Jarek ‘ A t—444-—&c. III. Problems in Harmonic Progression. 1. Kxpxuarn the nature of harmonic progression; and continue in both directions the series 2, 3, 6. 2. Continue the harmonic progression....3, 4, 6....up- wards and downwards. How far can it be continued either way? 3. Continue to 3 terms each way the series 1, 14, 12. APPENDIX. 439 4. Ifaand 0 be the first two terms of a series in harmonic progression, continue the series to three more terms, and find the n‘ term. ~~ 5. Prove that the reciprocals of quantities in harmonic progression are in arithmetic progression. 6. In any harmonic progression, the product of the first two terms is to the product of any two adjacent terms as the difference between the two first is to the difference between the - two others. 7. In any harmonic progression, the difference between the first two terms is to the difference between any two others as the second term diminished by (m) times the difference between the first and second is to the last; where » =the number of terms between the first and last. 8. In any harmonic progression, the second term dimi- nished by (mn) times the difference between the first and second is to the last as the product of the two first is to the product of the two last; n as before. 9. Any term of an harmonic progression is equal to the product of the first two terms divided by the difference between [the second] and [z times the difference between the first and second |. 10. Thesum of any two terms of an harmonic progression is greater than twice the intermediate mean term; and this excess is greater, as they are the more remote. 11. Find a fourth harmonic proportional to 6, 8, 12. : tb bon Mm 12. Find the sum of n terms of the series —, —, —, nm mn 3n — 2m a Se a7 oo * &e. mn 13. If the two extremes and the number of terms in an harmonic progression be known; the intervening series may be found. 44.0 APPENDIX. 14. Insert two harmonic means between 2 and 4; two between 6 and 24; four between 2 and 12; six between 1 and 20; six between 3 and -°,; and seven between 10 and 12. 15. Find two numbers whose difference is 8, and the har- monic mean between them 1+. 16. Insert 2 harmonic means between a and 6; and if a, be the first harmonic mean, prove that Te Oee Ce ey es e ee e ] e 17. The difference between two numbers ts 18, and 4 times the geometric mean is equal to 5 times the harmonic mean; find the numbers. 18. Prove that a geometric mean between two quantities is a mean proportional between an arithmetic and harmonic mean between the same two quantities. 19. If (a) be an arithmetic, (6) a geometric, and (c) an har- monic mean between two quantities; show that @ is greater than 6, and J greater than ec. 20. The difference of the arithmetic and harmonic means between two numbers is 1+; find the numbers, one being 4 times the other. 21. The arithmetic mean between two .numbers exceeds the geometric by 13, and the geometric exceeds the harmonic by 12; what are the numbers ? 22. If the arithmetic mean between two quantities # and y be m times the harmonic mean; then will Tey Wy a he ey he ee 23. If the geometric mean between two quantities x and y, be to the harmonic as 1 : 2; show that eiyiitY/i—-w):1-Sf—n). 24. Ifm harmonic means be inserted between a@ and pa; prove that the ratio of the first : the last is = to the ratio of m+pr:mp+i. ei ete eae APPENDIX. 441 25. Ify be an harmonic mean between # and z, and # and z be respectively the arithmetic and geometric means between a and 4, show that 2.(a + b) Y= er {OOF Ua) Aa) 26. Having given (a) the sum of three numbers in har- monic progression, and (d) their continual product; determine the numbers. 27. There are three numbers in harmonic progression; if 1 be subtracted from the first, the progression becomes geo- metric; and if 4 be subtracted from the third, it becomes arithmetic. What are the numbers? 28. From each of three quantities in harmonic progression, what quantity must be subtracted that the three results may be in geometric progression ? 29. There are four numbers, the first three of which are in arithmetic and the last three in harmonic progression; prove that the first has to the second the same ratio which the third has to the fourth. 30. The sum of three consecutive terms of an harmonic progression, whose first term is 3, is = 14,;; determine the progression, and continue it both ways. 31. If S and s be the sums of two infinite series, the common ratios of whose terms are R and r respectively; then S, s, R, r are in harmonical progression, the form of each series being (r, 7’, r*, &c.) and r and R fractional. 82. Given M and N, the m and nv terms of an harmonic progression; find the first and second terms. 83. Having given the two first terms of an harmonic pro- gression; determine the (m)'" term. 34. Having given the (m)™ and (z)™ terms of an harmonic progression ; determine the (m + nm)" term. 442 APPENDIX. 35. Ifa, b, ¢ be the p*, g™, 7 terms of an harmonic pro- gression; then will (p — g).ab+(r—p).ac+ (q—r).bc=0. 36. Ifa™ = bY = c* = &c., and a, b, c, &c. be in geometric progression ; then will 2, y, z, &c. be in harmonic progression. 37. Compare the lengths of the sides of a right-angled tri- angle, when the squares described upon them are in harmonic progression. APPENDIX ILI. 1. Form the equation, whose roots are Pi ABE oe WY Air (eo bones 2. Also whose roots are + «/— 2, 3, and 4. 3. Also whose roots are 1 + 4/—2, and 2+ (/—3. 4. Form the biquadratic equation, two of whose roots are 1+ fa’, and — /—28, Also if two of the roots be 4/3, and — «/—5. Cr 6. Determine the equation whose roots are is al apa aad ( io 54 rs , and — a. 7. Form the equation, of which the roots are the different values of a@ + 4 b. 8. Given that an equation has one root; show that it will have as many roots as it has dimensions. 9. If any coefficient in an equation be changed; prove that all the roots will be changed. an APPENDIX. 443 10. Ifa be a root of the equation oP arth ise. + Px’ + Q2?+ Rx + S=0, hey R, andif> + R=R,— + Qa Q,, &e.3 h S r show that a R,, Q,, &c. are integers. 1l. Ifa, 3, y, &c. be the roots of the equation ES) EN 1 me ea Qe +8 R= 03 show that pQ—nP rit OA ae Lalkeehl egatie ceatieat erage ae Pa day teal yp 12. The roots of the equation id oat 1 da ec — Qv + R=0, being a, B, y, &c.; show that 2 2 2 2 2 2 Se ULE tingid eet Als gt yarn ag abet Boy BN oy a B = (p29). —P. 13. Ifa — pa"-! + qv"? — &. + W= A, and abe any root of the equation 2” — px"~! + gz”—~?— &c. + W=0, prove that x — a is a divisor of the expression x” — px-1+ ga"? — Kt W. 14. Take away the second term of the following equa- tions: a’ — 92° + 2067 — 34 = 0. v—3e@ +47 —5=0. a+ 242° — 124° +47 —30=0. wt+sa4t+a2’?’—r—10=0. mw no = a So. CHa. 15. Take away the third term of the following equations : 1 v2 — 62" + 97 — 20=0. 9. #—427+57—-2=0. 44,4, APPENDIX. 16. Prove that the third term of the equation v— pe +qx—7r=0, cannot be taken away by the common method, if p’ be less than 3g. Show how it may be taken away in this case. 17. In an equation of m dimensions, show that the second and third terms may be taken away by the same transformation, when the square of the sum of the roots is to the sum of their Squares (7: 1. 18. Exterminate the last term but one of an equation of five dimensions by the solution of a simple equation. ip das Mt De m i oD into one whose coefficients shall be integral. 19. Transform the equation 2 — 20. Transform the equation y* — 2py’ — 33p’y + 4p” = 0, into one whose coefficients shall be numerical. 21. Transform the equation a” — at pan} + gx? — a?ra"-> 4+ &c. = 0, into one whose coefficients are rational. 22. Transform the following equations into others whose terms shall be alternately positive and negative : : SR 1 eee Tes MM 2 2 + a — 192° + lla + 30=0. 23. Transform the equation a’ + x2 — 10% + 4 = 0, Into one whose roots shall be greater by 4 than the roots of the given equation. 24. Transform the equation 2* — 42° + 6a? — 12 = 0, into one whose roots shall be greater by 5 than the roots of the given equation. 25. Transform the equation 2 — 6x? + 9% — 12 = 0, into one whose roots shall be less by 6 than the roots of the given equation. APPENDIX. 445 26. Transform the equation 32° — 1247 + 15” — 21 =0, into one whose roots shall be treble the roots of the given equation. 27. Transform the equation v* — 2”? — 3% + 4=0, into one whose roots shall be one-eighth of the roots of the given equation. 28. Ifa, b,c, &c. be the roots of the equation wen — per} + Caen mw &e. = 0; transform it into one whose roots are ma, mb, mc, &c. 29. If the roots of the equation 2 — px’ + ga —r=o0, be a, 0, c; transform it into another whose roots shall be atbat+cbo+e. 30. Transform the equation #* — 40x + 39 = 0, into one whose roots shall be the sum of every two roots of the original equation. 3l. Transform the equation 2° — px’ + gx — r = 0, whose roots are a, 6, c, into another whose roots are 1 1 1 a+b ate b+e° 32. Ifthe roots of the equation v—px+gqxu—r=o0, bea, b,c; determine the equation whose roots are ab, ac, bc. 30. Transform the equation 2 — pa’ + qx —r=0, into one whose roots shall be mean proportionals between the roots of the equation, and a given quantity (m). 34, Ifthe roots of the equation 2 — px + ¢ =0,beaandd; determine the equation, of which — /a, and — /0 are roots. 35. Ifthe roots of the equation xz’ —px+gq=o0, bea and b; determine the equation whose roots are an arithmetic, a geo- metric, and an harmonic mean between a and 0. 44.6 APPENDIX. 86. Transform the equation 2° — 22° +27 —4=0, into one whose roots are the squares of the roots of the original equation. 87. Transform the equation 2° — px’? + qx — r = 0, whose roots are a, J, c, into one whose roots are a’, 0’, c’. 38. Transform the equation 2 — px’ + gv —r = 0, whose ; oe) roots are a, 5, c, into one whose roots are GP BP? oF 39. Transform the equation x* — px’ + gx — r = 0, whose roots are a, 6, c, into one whose roots are @7+0,e70+0¢,0 +c’. 40. ‘Transform the equation 2° — pa’ + gx — r= 0, whose roots are a, 6, c, into one whose roots are 1 4 1 1 + 1 1 it. 1 a b?? a ce? & C 41, ‘Transform the equation 2 — 6x’ + 114 — 6 =0, into one whose roots are ageless : G0” Ft ¢ b? + ¢c 42. Transform the equation a* + a + #’?+4#+1=0, into one whose roots shall be the squares of the roots of the given equation; and show from the roots themselves that the trans- formation is correct. 43. ‘Transform the equation 2° — px’ + qx — r = 0, whose roots are a, 0, c, into one whose roots are a b a Cc b Cc (5 + Z)> (G+ 3)» ana (3+ 5): 44, Transform the equation 2° — pa’ + gx — r = 0, whose roots are a, b, c, into one whose roots are c b a at+b—cCat+e—Vb+e—a APPENDIX, 44,7 45. Transform the equation v— px +qr%—r=od, whose roots are a, b, c, into another whose roots are atb+tabat+ec+ac,o6+c+4 be. 46. Transform an equation into one whose roots shall be the squares of the differences of the roots of the original equation; and show by means of this transformation how the number of impossible roots in an equation of five dimensions may be detected. 47. Transform the equation #*—pa"—1+ qa"~?— &c.=0, into one whose roots are the reciprocals of every (x — 1) roots of the original equation. 48. If the roots of the equation # — px’ + qv —1r = 0, be a, b,c; transform it into one whose roots are a’, b°, c’. 49. If 2 — 22° + 1=0; deduce the equation of which the roots are the cubes of the roots of the original equation. 50. Transform the equation x" — pa”-1+ gxu"—-*— &c.=0, into one whose mm" term shall be a given quantity. 51. Determine the roots of the equation at — 40/2 + 62° 4— 44/8 +2=0. 52. Solve the equation x#° — 4@° — 3% + 12 =0, one root of which is of the form / a. 53. One root of the equation z* — 6a + 6% + 8 = 0, being 1+ / 3; find all the roots. 54. One root of the equation 2 — 11” + 37” — 35 =0, being 3 + 4/2; find all the roots. 55. One root of the equation z* + #2 — sx2° — 167 —8 = 0, being 1 — 53 find all the roots. AA8 APPENDIX. 56. Solve the following equations, two of whose roots are equal : ‘ lL. #@—72? +164 —12=0. 2 v+8a2?> + 207+ 16=0. 3. @—5a°+8r7—4=0. 4, @—52°—8H +48=0. 5. PP — 2 — 82 + 12 = 0: 6 pO pa neste © F 2 16 ri Pe pee eae 7 9261 57. The equation 32° — 102° + 15% + 8 = 0, has three equal roots; determine them. 58. Solve the equation a — 144° + 614° — 84¥ + 36 =0, whose roots are of the form a, a, 0, 0. 59. Solve the equation av — 132° + 672° — 1712” + 2162 — 108 = 0, whose roots are of the form a, a, a, 0, 0. 60. The equation a°—2x°+6z‘— sx’ + 124°—87+8=0, has equal roots; determine them. 61. Solve the equation a + pa’ + qa’? + r# + s =0, which has two pairs of equal roots. 62. Solve the following equations, which have two roots of the form + a, — a, 40° — 3227 —xv+8=0. wv — 50° — 54" + 454% — 36=0. v + 32° — 72 — 27H —18=0. v+a—11H? +07 4+ 18 =0. m0 te 63. Solve the equation a — 102#* + 292° — 10x? — 6227 + 60 = 0, two of its roots being 3 and 4/2. 64, The equation 2 — 15z* + 66a — 80 = 0, has two roots whose sum is 133 find all the roots. APPENDIX. 44,9 65. The equation 2 — 45a’ — 40x + 84 = 0, has two roots whose difference is 3; determine all the roots. 66. The roots of the equation 2° — 152° + 66% — 80 = 0, have a common difference; determine them. 67. In the equation #° — 6x? + 11a —6=0, one root is’ double another; determine all the roots. 68. The product of two roots of the equation av + a — 62”? — 80% + 1200 = 0, is 30; determine all the roots. 69. Determine the roots of the equation xv — 17x"? + 94” — 168 = 0, two of them being in the proportion of 2 : 3. 70. In the equation x — 102 + 27” — 18 = 0, the greatest root is double of the second, and the second treble of the third; determine all the roots. 71. One root of the equation # —52*°—#+5=0 1s 5; determine all the roots. 72. The equation x° — zy aa ly — 1 =0, has two roots of the form 4a, =; determine all the roots. 73. The roots of the equation 6a* — 432° + 1072” — 1084 + 36 = 0, are of the form a, 5, : and ; ; determine them. 74, Determine the roots of the equation a — 102° + 352” — 50a + 24 = 0, they being of the forma +i1,a—1,6+1,0—1. Gg 450 APPENDIX. 75. Solve the following equations, whose roots are in arithmetical progression : lL &@—62? —447 + 24=0. 2. #@ — 9x”? +234 —16=0. 3. @—62°> + 11% —6=0. 4. #2 —327+67+8=0. xv’ — 102° + 35a? — 50” + 24=0. Or ° 6. vt —sa’°+ 142° 4+ 8a@— 15=0. Pol et ae — 1 her” 1 921+ 18& 0. 76. The roots of the equation # — px"—! + ga”~? — &e. = 0 being in arithmetical progression ; prove that the least root is pes lara fa am tT) usa) NAt ni— 1 iF ” 7 and the common difference 2 {(m_— 1) .3p' — 6g]. a4 n> —1 li 77. Solve the following equations, whose roots are in geometrical progression : lL #—72’ + 14%7—8=0. 2 @ — 1327 + 39% —27 = 3. &@ — 142? + 562 — 64 =0. 4, x — 26a" + 156% — 216 = 0. 5. & — px? + gx—r=o. 6 @+pert+qe +re+s=0. 78. Ifthe roots of the equation a” — par} + Ob ae oe &e.= —_— be in geometrical progression; having ae p = 15, q = 703m find n, 7, &c. APPENDIX. za Ae | 79. Solve the following equations, whose roots are in har- monical progression : l wv — 112? + 367 — 36=0. 2. «#— 132° + 547 —72=0. 3 Do Es Shoat ie ae eS er 0; 2 6 4. 82 —62?—3H7+1=0. WL a ee aes Sena a eet — a 4 6 ax’ — ba’? —c#+1=0. 80. -In the common cubic 2° — px? + gv — r =0, if the roots are in harmonic progression, and p, q, 7, integer numbers, then 7 is the square of the greatest root. Apply this to solve the equation 2° — 23a° + 135” — 225 = 0. | 81. If the roots of the equation 2° — pa’? + qv~—r=0, be in harmonic progression; show that » py—3r p— spar + or tre ms 2+ ¢ 0, Lv contains the greatest and least. 82. Ifthe roots of the equation Deep og oe ea + Qa’? — Pe +L=o0, be in harmonic progression, then will the greatest and least be nf (n+1).L J (n+ 1).P = 9/ $3. (w= 1)?. P= Gn. (w= 1). QLY J/(m+1).P+46/33.(n— 1)’. P? —6n.(n—1).QL2 83. Solve the equation x* — 31a’ + 300% — 900 = 0, whose roots are successive triangular numbers. 84. Explain the method of finding the equal roots of equations, and apply it to the equations lL #@—92?+4%4 12=0. 2. &# — 132" + 672° — 1712" + 216% — 108 = 0. Gge2 452 - APPENDIX. 85. Having given the equation 22° — 12”° + 19”? — 6H +9=0, determine whether it has equal roots. 86. Ifthe equation v + ga — rx2’® — t =0 has two roots equal to each other; prove that one of them will be a root of ; Sha. 47 . 2q drat ye eS, ee Se 4 the quadratic x#* + = 2+ oF FE: 87. Show that if an equation have two equal roots, and the terms be multiplied by the terms of an arithmetic progression, the result will = o. 88. If an equation have (nm) equal roots, the equation formed by multiplying the terms by the terms of an arithmetic progression has (z — 1) of them. 89. Having given xv — px + gx — r = 0, whose roots are a, 8, c3) and w — p'v’ + g'v — r' = 0, whose roots are a, b,c’; J find c, and c'. 90. One root being common to the two equations xv — 9x + 267 — 24 = 0, and 2 — 7a? +7x# + 15=0, find the remaining roots of each. 91. Determine all the roots of the two equations which have one common root v— 32° + 1le~—9=0, e— 52? +1lw7—7=0. 92. Solve the equation 2° — 1 = 0; and show that its roots — are of the form a, 6, 67, | 93. The roots of the equation x2 + px’? + 1=0, must be of the form a, 8, - 33 exhibit them in that form. APPENDIX. 453 94. Solve the following recurring equations: 1 2t—3a° + 24° — 347 +1=0. 2. 2 + 6ax* — 200°2" — 6x4 + a =0. 4 53 2 5 vol us ee a ae aS eae 4. wtil =o. 5. av? — 21at + 372° — 372° + 2127 —1=0. 4 15 37 ay 15 Goa ae a age a: T 2 2 2 Pe Ane 100 e190 e 4a eo a Se 0. 8 a ae ns — 0 95. Reduce 9 8 7 6 5 4 3 2 —* e—pxe+qer—ra4+sae—sae+re’—qe+pr—1=0, to an equation of four dimensions. 96. Exhibit the quadratic factors of the equation wm +t1i=o. 97. Show that when m is a prime number, the roots both real and imaginary of the equation 2” + 1 = 0, are different powers of any one of its roots. 98. In any recurring equation # — pa"! + &c. = 0, whose roots are a, 6, c, &c.; prove that 2 6? 2 2 ae — ata ta tat ke. = (p'— 29+ Vn). (p'—29—- V2). 99. The roots of the cubic equation 2° — gv” + 7 = 0, are g° y? real when = exceeds is 100. The solution of the cubic equation 2 + gv +r=0 is dependent on the solution of the equation #* — 1 =0. 454: APPENDIX. 101. Having given, 1, a, (3, the cube roots of 1, i ce Yea a and A ; Ng, of) f? and B = is Seth: & = aN L 277 | + prove synthetically that 4 + B, ad + PB, and B4 + aB are the three roots of the equation. 102. If two roots of the cubic equation a —qe+r=0 beat bY —3, and a—bY/—3; By 3 then will — = + Wea = (6 — a)’. 103. Explain in what case, and for what reason, Cardan’s formula for the solution of a cubic equation does not enable us to determine the roots. 104. Determine whether Cardan’s rule is applicable to the solution of the equation 2° — 23747 — 884 = 0. 105. Show that Cardan’s rule for the solution of a cubic equation is applicable when all the roots are possible, and two of them equal; and by means of it, find the roots of the equation x + 6a? — 32=0. 106. Solve the following equations by Cardan’s rule: e—o9v—l4=0. xv — 62% —40=0. xv — 9x + 28 =0. v+32° +9" —13=0. xv—6xr° + 347— 18 =0. a’ — i2n”? + 57% — 94 =0. e+ 6x2? +20” + 15=0. av — 122° + 367 —7=0. a" — Hem + ge" —r=0. 107. Find by the doctrine of permutations, the roots of the equation v — ga+r=o. APPENDIX. 455 108. Assuming the quadratic factors in Des Cartes’s solu- tion of a biquadratic to be 2? + aw + b=0,and 2 —axr+c=0; find the reducing equation in (0) or (ce): and show that it may be depressed to a cubic. 109. If a@ be a root of Des Cartes’s reducing cubic, the four roots of the equation z + ga’ +rxv+s=0, are vB A (iss Ut to ye 87 2), 2 2 qY 110. Prove that Des Cartes’s solution of a biquadratic succeeds when all the roots are possible, and two of them equal; and apply it to solve the equation e—6e+ 82°? +6%—9=0. lll. Find the roots of the following equations by Des Cartes’s method: lL #@—427° — 8x + 32=0. vt — 327 — 4% —3=0. aot — 62° + 527 +27 —10=0. e+ o7°—~ 72" — 8x + 12 = 0. me OG wv a en, Oe 112. Give Euler’s solution of a biquadratic; and show that the cubic involved in that method has its roots four times less than the roots of the cubic in Des Cartes’s. 113. Solve the equation 2 = 12¥ + 5, by the method attributed to Waring. One root of the reducing cubic is 2. 114. Ifa+ BW/—1 bea root of the equation e+pet+ge+re+s=0, two roots are i{- (p+ 2a) 4 pay -2(¢ B+ em) |. 456 APPENDIX. 115. Prove that if (uw) the last term of any equation be resolved into prime factors a, 9, y, so that w =a” 2” y’, then the number of divisors of uw will = (m + 1). (mn +1).(p +t 1). 116. Find by the method of divisors the roots of the following equations: lL #—62? +54 +12=0. v— on? + 227 — 24=0. av — 62° — 16x + 21=0. wv’ — 42° — 84 + 32 =0. av’ + a — 2927 — 97 + 180 = 0. 32° — 262? + 34% — 12 =0. 82° — 2627 +1la7+10=0. sx — 452” + 734 — 30 = 0. CON Oo KF W Wb Perr ae eee en ee ee 117. In the method of divisors, show how the number of substitutions may be lessened :—and in the equation av — xv — 162? + 55” — 75 =0, determine whether 3, 5, and — 5 are roots. 118. Apply the rule for quadratic divisors to the equation xt — 172° + 88x — 172% + 112 =0. 119. Solve by the method of divisors, the equation 62° + 532° — 952" — 2h7 +42 =0. 120. It is always possible to find those roots of numeral equations which are whole numbers or rational fractions with- out the aid of formule of approximation. 121. Ifa be an approximate root of the equation e+ px’ + qx =r, so that a + pa’+qa=r, a.(r—7,) TOV Ty 20° + pa r or r, being used according as a is greater or less than 1. Approximate by this formula to the value of # in the equation xv — 20 = 5, prove that 7 =a@+ 5 very nearly ; APPENDIX. 457 122. Three given quantities (¢ + 2), (a@+2)+h,(a+z2) +h, approximations to the root @ of an equation, being substituted for the unknown quantity, give results n, n + 6, n + 0’; show that z will be very nearly found from the equation 2. (hd’ —h’d) + 2. (WV s’ —h*8) + nhh’. (h’ —h) =0. 123. Approximate to a root of the following equations : lL #@—x7—50=0. 2. v—2727—5=0. 3. 2+ 27 —30=0. 4. e@+a’+r2=90. 5 av —6H +1=—0. 6 #@—20°+37—4=0. ret ae = 3. 8. wv —12% + i= 0. 9. 22° — 162° + 402° — 307 ++1=0. o. {ot eras 2Vy—y =2. a: , Nik ae YE y —x“v=6. 12. ar Meee |e +y=s. 124. In the equation 2 + 92° + 4% = 80, approximate to the value of x by means of a series of converging fractions. 125. Express the roots of 2° —7x# + 7=0, by continued fractions ; and determine the accuracy of the approximation of any converging fraction deduced from these. 126. If a be an approximate value of w in any equation, and J, c be the results when a is substituted for z in the ori- ginal and in the limiting equation; then will =a-— a near! = y- 458 APPENDIX. 127. Determine the number of positive and negative roots in the equation #* + 42* — 192° — 34v” + 60x + 36 =0, of which all the roots are real. : 128. Determine the same in the equation x — 5a — 152° + 852" — 26H — 120 = 0. 129. Prove without resolving the equation into factors, that if two numbers (a) and (4) when substituted for the un- known quantity in the equation v — px"! + ga"—?— &c.=0, give results affected with contrary signs, there is at least one real root between (a) and (0). 130. If P represent the sum of the positive terms in any equation, and a series of quantities be successively substituted for the unknown quantity; prove that the successive increments of P may be made less than any assignable quantity. 131. If P and WN be the greatest positive and negative coefficients in the equation eee 12. sd eee Pa ites: ee TA ata Re 0; then a superior limit to the roots is N + 1; and an inferior limit is ea or oe according as u 18 positive or negative. 132. If M2"-™ be the first negative term of the equation B+ 0087 cn — Ma” — &e.= 0, and if P be the greatest negative coefficient, then1 +

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