m 
 
c* 
 
 -SO 
 
 REESE LIBRARY 
 
 Of THK 
 
 UNIVERSITY OF CALIFORNIA. 
 
 Received. s^^L^Zl^ rSS(/ 
 
 Accessions No.2*£~&fi*^ Shelf No 
 
 * £ 
 
KEY 
 
 TO ROBINSON'S 
 
 UNIVERSITY ALGEBRA; 
 
 CONTAINING, ALSO, 
 
 A SHORT TREATISE ON THE INDETERMINATE 
 AND DIOPHANTINE ANALYSIS. 
 
 AND 
 
 SOME MISCELLANEOUS EXAMPLES, 
 
 JBesffltie'O for 5Teacf)ers attfi SatuUents. 
 
 NEW YORK: 
 
 IVISON & PHINNEY, 321 BROADWAY 
 
 CINCINNATI: JACOB ERNSf 
 CHICAGO : 8. C. GRIGGS & CO., 39 & 41 LAKE ST. 
 
 ST. LOUIS : KEITH k WOODS. BUFFALO : PHINNEY & CO. 
 
 185 9. 
 
■ft 
 
 Entered according to act of Congress, in the year 1847, by 
 
 HORATIO N. ROBINSON, 
 
 in the Clerk's Office of District Court of the District of Ohu* 
 
\^ V OF THE ' \\ 
 
 'UNIVERSITY' 
 
 flfe 
 
 A Key, to a mathematical work, is very proper in 
 its place ; but to be constantly at hand, and consulted 
 too often, might prove injurious : we must not, how- 
 ever, confound the improper use of a thing with the 
 thing itself. Those who condemn keys, in general 
 terms, should condemn teachers also ; for a key is 
 neither more nor less than a teacher, in another shape. 
 
 The self-taught are generally sound and vigorous ; 
 but if they disregard the works and teachings of others, 
 they will be found to be wanting in that certain sym- 
 metry and polish of mind, so characteristic of educated 
 men. 
 
 So it is with an algebraist ; h'e may go through his 
 text-books, solve every problem, independent of all 
 external aid, and if he does not compare his work 
 with the works of others, he cannot know whether he 
 is skillful or otherwise ; for it is only by comparison 
 that we measure excellence. No solution of a prob- 
 lem, or of an equation, should be called good, if better 
 can be found ; hence it is important that more than 
 one standard of attainment should be before the pupil ; 
 and those who really become eminent, in any science, 
 are those whose talents and dispositions enable them 
 to gather knowledge from every possible source. 
 
PREFACE TO REVISED EDITION. 
 
 In 1857 the author and publisher thought it advisable 
 to enlarge the University Algebra by some twenty-four 
 pages, and some more than twenty additional examples. 
 
 Solutions of these are inserted subsequent to page 76. 
 Also, solutions of several other problems are inserted, 
 which were omitted in the former editions of this work. 
 
'UNIVERSITY 
 KEY 
 
 TO 
 
 ROBINSON'S ALGEBRA 
 
 SECTION II. 
 
 CHAPTER I. 
 
 EQUATIONS. 
 
 None of the questions in this chapter require the aid of 
 a key, until we come to the 15th, page 65. 
 
 (15.) (4x—4a \4 I6x—I6a 4a 
 
 /4x—4a \4_J 
 \ 3 /3~~ 
 
 = his 
 
 9 3 
 
 stock at the commencement of the third year, before his ex- 
 penses are taken out. 
 
 tj /16a? — 16a 4a \4 
 
 Hence, ( a }_ = 2a; 
 
 V 9 3/3 
 
 Reduced gives x= 14800, Am. 
 
 (16.) Put a=99, #=time past. Then a — x== time to 
 come, and per question, 
 
 2a? 4a — 4x 
 
 3 5 
 
 (17.) Let x= the whole composition. 
 Then per question, 
 
 z=54. 
 
KEY TO ALGEBRA. 
 
 2a? , 
 
 r-10=nitre. 
 
 3 
 
 x 
 
 4£ =sulphur. 
 
 6 
 
 2x 10 
 
 — -f— — 2= charcoal. 
 21 7 
 
 2x x 2x 10 
 
 By addition, — +-- f h3H =x 
 
 J 3 6 21 7 
 
 Multiply by 6, and drop 5a; from both sides, and we have 
 
 4x 60 
 
 7 7 
 
 or, 4#+21-7+60=7# . • . .#=69. 
 
 (18.) Puta=183 ; #=what the 1st received ; then 
 
 a — x=2d received. 
 
 4a? 3a — 3a? 
 
 Per question, — = #=63. 
 
 7 10 
 
 (19.) Put a=68, #= the greater part, and a — x= the 
 
 less. 84— #=3(40— a+ x) #=42. 
 
 (20.) The distance from A to B put =2#. 
 
 The distance from C to D " =3x. 
 
 Then, 3 times the distance from B to C must be 
 
 x Sx , xx 
 
 — | or the distance is, — | — 
 
 2 2 6 2 
 
 x x 
 Hence the whole distance is, 5x-\ 1 — =34. 
 
 6 2 
 (21.) Let x= the flock. 
 
 2x 
 The first party left him 6. 
 
 x 
 
 The second left 3—10=2. 
 
 3 
 
EQUATIONS. 7 
 
 (23 ) Observe that for every vessel he broke he lost 12 
 cents : 3 cents fee and 9 cents forfeiture. 
 
 300—12^=240 ... - #=5. 
 
 (24.) Had he not been idle he would have been entitled 
 to ab cents. But he was idle x days at a loss of (b-\-c) 
 cents. Hence, ab — (b-\-c)x=d. 
 
 ab — d 
 
 b+c 
 
 (25.) Put 5x= 
 Then a — 5x = 
 
 = less part 
 the greater part. 
 
 Per question, 
 
 3 
 
 a — 7x = 2 Ox (a — 5x) 
 
 or, 
 or, 
 Therefore, 
 
 7a—49x= 1 40#— 3a+ 1 Sx 
 
 204.r=10a=10-204 
 
 #=10 
 
 5^=50= the less part. 
 
 (26.) Let 8#=the price of the horse. 
 
 Then a — Sx= chaise. a=341. 
 
 5 
 Per question, 2« — 16a; — ■3#=24# (a — Sx) 
 
 5 
 or, 2a=4Sx (a— 8x) 
 
 14«=301#— 5a+40x 
 19a=341# or, #=19 
 
 8#=152 Arts. 
 
 (28.) Let 5x= his money. 
 After he first lost and won 4 $., he had 4#-f4 
 He again lost and won, and then had 3#+3-f-3. 
 £ of this must equal 20, or, 3x-\- 6=24. 
 
 x=6 
 5#=30 Ans. 
 
KEY TO ALGEBRA. 
 
 (29.) Let 3x— the income. 
 Then 2a? = the family support. 
 
 x -=-=70. Hence, .... 3a?=70-9. 
 
 3 o 
 
 30, 31, 32, and 33 require no explanation. 
 (35.) Last year the rent was x dollars. 
 
 _ . 8a; 
 
 This year it is x-\ =1890 
 
 100 
 
 (36.) is the (35) in general terms. 
 
 (37.) Let 7a?= the income. 
 
 Then x= A's annual debt. 
 
 7a? 
 
 — = what B saves. 
 5 
 
 7a? 
 
 a?=li 
 
 5 
 
 5 
 
 
 or 
 
 a?=40. 
 
 In general terms, 
 
 
 
 
 7a?=280, Am, 
 
 7a? a 
 ~5 X ~2 
 
 
 
 
 5a 
 
 x= — 
 
 4 
 
 (38.) xx 
 
 3 4 
 
 2a? 
 
 T 
 
 =a 
 
 7a?=£(35d). 
 
 (39.) Let 10a?= the income. 
 
 Then 2a?-f-100=the sum spent. 
 
 5a?+ 35= « sum left. 
 7a?-f-135=10a? the whole 
 
 or 45=a? 450 Am 
 
 (40.) Let 21a?= the income. 
 
 Then 3a?-|-«= the sum spent. 
 
 7a?-f&= " sum left. 
 
 10a?+a-K>=21a?= the whole. 
 
 («-H) 
 
 x= . 
 
 11 
 
EQUATIONS. 
 (41.) 2#-h4 : 3x-f-4 : 
 
 (42.) Let x* — 7= the number. 
 
 Then, per conditions, x — l=Jx* — 7 
 
 a*— 2a?+ 1=^—7 
 
 or, x=4 and a? 2 — 7=9. 
 
 (43.) A's rate of travel is J miles per hour. 
 
 B's rate of travel is f miles per hour. 
 
 A is in advance when B sets out, 5 / miles. 
 
 Let a;= the hours after B starts. 
 
 5x 7x 56 
 
 Then, — = — 1 Reduced gives . • . • . a?=42. 
 
 3 5 5 
 
 CHAPTER II. 
 EQUATIONS IN TWO UNKNOWN QUANTITIES. 
 
 (6.) Add the two equations together, representing (x-\-y) 
 
 by s, and we have 5$-f-3s=50 or 3=5*3. 
 
 But aH-9y=21-3 
 
 Subtract x-\- y= 5*3 
 
 8y=16-3 or, y=6. 
 
 (7.) By adding the two equations we have 
 55=50 
 or, x-\-y= 10 
 
 but 4#-{-2/=34 
 
 Hence, 3a? =24 or, «a;=8. 
 
 (9) and (10) are resolved same as (6) and (7). 
 
 (11.) From the first equation we have 
 
 y=2x~- 80. 
 
 Transpose —8 in the second equation and we have 
 
 x-\-v x 2y — x , 
 
 9 1 -=— +35. 
 
 5 ' 3 4 
 
10 KEY TO ALGEBRA. 
 
 Multiply by 60 and we have 
 
 12aH-122/+20;r=30i/— 15z+35-60 
 or 47a?= 183/+35-60 
 Substituting the value of 18y, we have 
 47#=36;r— 1 8-80+35 -60 
 or ll;r=— 240-6+350-6=110-6 
 Hence, x=60. 
 
 (14.) Bringing unknown terms to the first members of the 
 equation and we have 
 
 4 4_ 4 2__3 
 
 x y y x~~2 
 
 2 1 
 
 By addition, -=- or x=4. 
 
 x 2 
 
 (15.) Put a=50. 
 
 Then, #+3a : y — a : : 3 : 2 
 
 And x — a : y-\-2a : : 5 : 9 
 
 2£+6a=3y— 3a (1) 
 
 9x— 9a=5y+10a (2) 
 
 Multiply (1) by 5, and (2) by 3 ; then, 
 
 10a?+30a=15?/— 15a (3) 
 
 27#— 27a= 1 5y + 30a (4) 
 
 Subtract (3) from (4) and 
 
 17a? — 57a=45a 
 
 17#=102a 
 
 ar=6a=6-50=300. 
 
 (16.) Divide the numerator of the second member of 
 the first equation by its denominator, and we have 
 
 — lla>— 14y+127y 
 
 3*+6 y +l=3*+6 y +l+ to-J+8 
 Hence, llar + 142/=127 (1) 
 
 Multiply the second equation by (Sy — 4) and we shall 
 have 
 
EQUATIONS. 11 
 
 (151 — 16a?) (3y— 4) 
 9xy—l2x=± '-XZ --f-9ay— 110 
 
 (151— 16#)(3y—4) 
 
 or, 110— 12z=- ili L 
 
 4y— 1 
 
 440t/— 48#y— 1 10+12#=453y— 48#t/— 604-f-64a? 
 0=13i/+52;r— 494 
 or, 4#+y=38 (2) 
 
 Add (1) and (2), and we have 
 
 l5(x-\-y)=165 
 or, a?+y=ll (3) 
 
 (3) from (2) gives 3#=27 x=9. 
 
 (17.) Multiply the 1st equation by 14 and we have 
 42;r— 7y=49 
 Add — a;+7y=33 
 
 41x =82 or, «=2. 
 
 (18.) 2 
 
 aH-g2/=16 
 
 5 
 
 2 3 
 
 Subtract -yA — =19 
 
 3* 5 
 
 Kty+9y=19-15 y=15. 
 
 (19.) Divide 2d by the 1st, and x—y~Z 
 But x+y=8. 
 
 (20.) Multiply the first equation by (x+y), and the sec- 
 ond by 9, and we have 
 
 4(z-j-y) 2 =9(z a — y*) 9(z*— y*)=9-36 
 4(a?-{-y) 2 ==9-36 
 Hence, x+y=9 
 Divide the 1st equation by this last, and we have 
 • x — #=4. 
 
12 
 
 KEY TO ALGEBRA. 
 
 (21.) 
 
 x= Jy 
 
 3 
 
 a*= 
 
 64i/ 3 
 
 64y» 
 ~27~ 
 
 ■^=37 
 
 27 
 372/ 8 =37-27 y=3. 
 
 (22) and (23) require no remark. 
 
 (24.) The first equation gives 
 
 #4-242/= 91 (i) 
 
 Add 40tf-|- y=763 (2) 
 
 Multiply (1) by 40, and subtract equation (2), and 
 
 959y=2877 or, y=3. 
 
 (25.) From 1st equation take the 2d, and we have 
 
 2hx+5y=60. 
 Divide by 2\ and we have x-\-2y=24 
 But ix+2y=\9 
 
 \x = 5 or, a:=10. 
 (26.) Add the two equations, and 
 
 2(*+y)+£(*+y)+ 20 =*+y 
 
 or, -+-+20=$ 
 2 3 
 
 By 2d equation, 
 
 5=120. 
 
 £(120)— 5=y=35, JSm. 
 
 CHAPTER III. 
 
 EQUATIONS OF THREE OR MORE UNKNOWN QUANTITIES. 
 
 (7.) Given 
 
 '2x=u-\-y-\-z 
 
 3y=M-f-a?-fz 
 
 4z=u-)rx-\-y 
 
 w=a?— 14 
 
 Subtract 2d from the 1st, and 
 
 2x — Sy=y — x or 
 
 Subtract 2d from the 3d, and 
 
 4z — 3y=y — z or 
 
 > to find u, x, y and z. 
 
 3x=4y (1) 
 5z=4y (2) 
 
EQUATIONS. 
 
 Add 3d and 4th and 4z=2x-\-y — 14 
 
 Multiply (3) by 5 and (2) by 4 and 
 
 202=10aH- 5y— 70 
 202= 16i/ 
 
 13 
 
 (3) 
 
 
 0=10a?— lly— 70 
 
 or 30#— 33i/=210 
 
 But, equation (1) 30#— 40iy= 
 
 7i/=210 
 
 y= 30 
 
 
 ~x y z * 
 2 3 4 
 
 =a+b 
 
 (8.) 
 
 3 4 5 
 
 b 
 
 J>=a — 
 
 4 
 
 
 x t y z 
 .4 5 6 
 
 =a — b 
 
 To avoid numeral multiplication, and really, to under- 
 stand algebra as applied here, observe that 62+38=100. 
 Put a=50; then 62=a-f-12=a+6. 
 Clearing of fractions, we have 
 
 6aH- 4y+ 3z=12a-f-12& (1) 
 
 20#-f- 1 5i/-f- 12^=60a— 156 (2) 
 
 15a?+12y-r-102r=60a— 606 (3) 
 
 Multiply (1) by 4 , and subtract (2). 
 
 Then, 4#+y=636— 12a (4) 
 
 Subtract (3) from (2,) and 
 
 5x+3y+2z= 456 
 
 3 
 
 Subtract 
 
 15#-f9i/4-6z=1356 
 12a?-j-8y-f-6z= 246+24a 
 
 (5) 
 
 3x+ y =1116— 24a 
 Subtract (5) from (4), and we have 
 
 #=(12a— 486)=12(a— 46)=12-2, Ms. 
 
14 KEY TO ALGEBRA. 
 
 That is, a?=24 or 26. 
 Now, equation (4) gives us 
 
 8b-\-y=G3b—ba 
 
 y=(55— a)6=5-12=60. 
 (9.) By adding the three equations and reducing, we have 
 4x-\-3y-\-2z=3a (1) 
 
 By adding the 2d and 3d, reducing and doubling, we 
 have l0x-\-4yi~2z=4a (2) 
 
 Subtracting (1) from (2), and we have 
 
 6x-\-y=a (3) 
 
 Adding the 1st and 3d, and reducing, we have 
 
 x-{-2y—a (4) 
 
 From (3) and (4) we readily find x and y. 
 —2z =40 (1) 
 
 -\-3z =35 (2) 
 
 (10.) <> 3u +* =13 (3) 
 
 y +u -W=15 (4) 
 
 jx—y-\-3t— w =49 (5) 
 
 It is easier to eliminate t than any other letter. 
 Subtract (3) from (4), and we have 
 
 y— 2u=2 (6) 
 
 Three times (3) taken from (5), gives 
 
 3x—y—\0u=\0 (7) 
 
 Add (6) and (7) and divide the sum by 3, and 
 
 x — 4w=4 (8) 
 
 Double (6), and subtract it from (8,) and we have 
 
 x=2y (9) 
 
 Eliminate z from equations (1) and (2), and we have 
 
 4#-Hlt/=190 
 But4a?=8#. Then 19t/=190, or y=10. 
 
 C2x +y 
 4y — a: 
 
 < 3 
 
EQUATIONS. 15 
 
 PROBLEMS PRODUCING SIMPLE EQUATIONS, INVOLVING TWO 
 OR MORE UNKNOWN QUANTITIES. 
 
 (1.) Let a?, y, and z represent the num 
 
 a^=600 
 
 a?z=300 
 
 1/2=200 
 Multiply (1) and (2), and divide the product by 
 3, and we have x 2 =900 ar=30. 
 
 (2.) Let x, y, and z represent the numbers. Then, per 
 question 
 
 70 (2) 
 
 5 
 
 * +2/+2=l 90 (3) 
 
 Double (1) and subtract (3), gives #=50. 
 
 This problem calls the pupil's judgment into exercise. 
 He does not know in the first place which is greatest, x or 
 z ; hence he must try both suppositions, and the one that 
 verifies equation (2) is right. 
 
 (3.) Let x,y, and z represent the shares, and put a=120 
 
 y — $(x-\-z)=a 
 z— §Gr-fy)=a 
 Clearing of fractions, we have 
 
 Ix—Ay — 4z=7a (1) 
 
 — 3x+8y—3z=8a (2) 
 
 —2x— 2y-\-9z=9a (3) 
 
 Double (1) and to the product add (2,) and we have 
 
 Ha? — ll*=22a 
 
 X— z— 2a (4) 
 
16 KEY TO ALGEBRA. 
 
 Double (3), and to the product add (1), and we have 
 — llH-22z=lla 
 
 — H- 2z= a (5) 
 
 Add (4) and (5), and we have z=3a=360. 
 
 (4.) Resolved in the book. 
 
 Let x, y, u, and z represent their ages, and s their sum. 
 
 Then, *— *=18 
 
 s—u=ld 
 
 s — y=14 
 
 s — a?=12 
 
 "Rxr orl/litirm In fift . ■ . 
 
 
 , . • t=20 
 
 r>y auuiuun, os — uu • • • 
 («5.) Let #=A's shillings. 
 
 
 • • O^^AiXfrn 
 
 y=B's " 
 
 
 
 z=C's " 
 
 
 
 After the first game they will have as 
 
 here i 
 
 •epresented 
 
 x — y — z=A 
 
 
 
 2y =B 
 
 
 
 2z =C 
 
 
 
 After the second game, 
 
 
 
 2x— 2y— 2z-—k 
 
 
 
 Sy — x — z=B 
 
 
 
 4z =C 
 
 
 
 After the third game 
 
 
 
 4a? — 4y — 4z= 16 
 
 CD 
 
 
 6y—2x—2z= 16 
 
 (2) 
 
 
 Iz — x — y= 16 
 
 (3) 
 (*) 
 
 
 Sum, x-\- y-\- 2r=3'16 
 
 
 Add (3) and (4) and we have 
 
 
 
 8z=4 16 
 
 
 . . . z=8. 
 
 (6.) This problem is resolved in the book, by equation 
 7, Art. 53. 
 
EQUATIONS. 17 
 
 (7.) Let x represent the better horse, and y the poorer. 
 
 ar+10=H(y+ ls ) 
 Therefore, J(y+10H|(y+15)+5 
 Reduced gives i/==50. 
 
 (8.) Let #= the price of the sherry. 
 
 y= brandy. 
 
 Put a=78. 
 
 2a?-}- y=3a 
 
 7x+2y=9a+9 
 
 
 3a? =3a-f-9 
 
 
 x = a+3=. . 81a 
 
 (».) Let x=X's time. y=B's time. 
 
 Then, 
 
 1 , 
 -=the part that A can do in one day. 
 
 X 
 
 And 
 
 1 
 
 _=the part that B can do in one day. 
 
 y 
 
 4 4 4 1 
 x y 16~~4 
 
 
 
 36 3 
 
 — =- Hence 
 
 y 4 
 
 (10.) 
 
 2a? 2 x-\-2 3 
 
 
 y+7 3 2y 5 
 3a?=y-f-7 5a?+10=6y. 
 
 (11.) 
 
 , 2 y , 3 * 
 
 (12.) : 
 
 Let x= the greater, and (24 — a?)= less. 
 
 
 a? 24 — x 
 
 ..4.1 
 
 
 24 — a? a? 
 
 
 a? 2 : (24— a:) 2 : : 4 : 1 
 
 By evolution, x : 24 — x : : 2 : 1. 
 
 ♦ 2 
 
 
 y=48. 
 
18 KEY TO ALGEBRA. 
 
 (13.) Let a?= the number of persons. 
 
 yz= what each had to pay. 
 Then, xy= the amount of the bill. 
 
 Put (x+4)(y— 1)= the bill. 
 
 Also, (as— 3)(y+l)= the bill. 
 
 xy + 4i/ — a? — 4 = #?/ 
 
 4y — x — 4=0 
 — 3y-{-x— 3=0 
 
 By addition, y — 7=0 
 
 (14.) \0x-\-y=4x-\-4y or, y=2x. 
 
 \0x+y+27=l0y-\-x 
 or, 9x +27= 9y 
 
 x + 3= y=2x hence, a?=3. 
 (15.) Let rr= the digits in the place of 100's. 
 y= " in the place of 10's. 
 z= «* the units. 
 x-\-y+2=ll z=2x 
 100.r+10y+2r+297=100z+10i/+a: 
 99^+297 = 99* 
 x-\-3=z=2x Hence x=3. 
 
 x— -40 #—20 , #—10 
 
 (16.) Let , , and represent 
 
 2 3 4 
 
 the parts. Then 
 
 x — 40 x — 20 a?— 10 
 
 1 1 =90 #=100. 
 
 2 3 4 
 
 (17.) Let x represent the part at 5 per cent, and (a — x) 
 
 the part at 4 per cent. Then 
 
 bx 4a— 4x . 
 
 + =6 
 
 100 100 
 
 Hence x=100&— 4a. 
 
EQUATIONS. 10 
 
 (18.) To avoid high numerals, and of course a tedious 
 
 operation, Put a=5000 ; then 2a=10000, 3«=15000, 
 
 3a 16a 
 
 — =1500, and =800. 
 
 10 100 
 
 Put #= A's capital, and r — 1 = A's rate. 
 
 x-\-2a— B's " r= B's rate. 
 
 a?-j-3a= C's " r+l= C's rate. 
 
 ~rx — x 16a rx-\-2ar 
 
 By conditions, 
 
 100 100 100 
 rx — x 3a rx-\-3ar-\-x-^3a 
 
 100 10 . 100 
 
 Reducing (1), gives a?=(16 — 2r)a 
 '27— 3r 
 
 (1) 
 
 (2) 
 
 /27 — 3r\ 
 
 " (2), " *=(— — y 
 
 Hence, 32 — 4r=27 — 3r, or r=*=5. 
 
 (19.) Put a=1000, x and y to represent the tw< parts, 
 and r and t the rates expressed in decimals, 
 
 Cx+y=13a (i) 
 
 Then by conditions,^ y ' " ^ ' 
 ' ] te=360 (3) 
 
 t ry=490 (4) 
 
 Divide (3) by (4), and we have 
 
 36 
 
 (5) 
 
 C-)Q J 
 
 49 
 
 x t 
 From (2) we have -=- 
 
 y r 
 
 Substitute the value of - in equation (5), and 
 
 T 
 
 x* 36 6v 
 
 --= — or x= — 
 
 y* 49 7 
 
 6v 
 By returning to equation (1) we have -— +2/=13e*. 
 
 •13y=13a*7 or. y=7a. 
 
20 KEY TO ALGEBRA. 
 
 (20.) Let x, y, and z, represent their respective ages. 
 Then by conditions given, x — y= z 
 5y+2z— x=U7 
 x+y+z= 96 
 (21.) Let x, y, and z, represent the respective property 
 of each, and put s=their sum. 
 
 rx+3y+3z=47a a=100. 
 
 Conditions, J y _|_ 4 X +4z=58a 
 Lz+5x-\-5y=63a 
 Add 2x to the 1st equation, 3y to the 2d, and iz to the 
 3d, observing that x-\-y-\-z=s ; then we shall have 
 
 3s=47a-f-2;z (1) 
 
 4s=58a+3y (2) 
 
 5s=63a-f4z . (3) 
 
 3s— 47a 
 
 or, 
 
 2 
 4s — 58a 
 
 5s — 63a 
 
 z= 
 
 4 
 
 3s — 47a 4s— 58a 5s— 63a 
 
 By addition, s= h H 
 
 \ 2 3 4 
 
 Hence, s=19a. 
 
 This value of s, put in equation (1), gives #=5a=500. 
 
 (22.) Taken from the book we have 
 
 Is — p ms — a ns — r 
 
 s=-—f-+ 1+ 
 
 / — 1 m — 1 n — 1 
 
 or , ,~(±.y.r+(JL Y__i_ +(JL y__ L. 
 
 V/-1/ /_i \ot-1/ m-1 \nr-l/ n-l 
 
 (I m n \ p q r 
 — + -f- 1 )s=-i--r— H — -f 
 
EQUATIONS. 21 
 
 4-1- ?■■!■■.»- 
 
 / — 1 m — 1 w — 1 
 
 Hence, s= =k 
 
 I m n 
 
 I — 1 m — 1 n — 1 
 
 As s is now known, we call its value k. 
 
 lk —P 
 Then, #= -- 
 
 mk — q 
 
 nk — r 
 r— 1 
 (23.) Let x, y, and z represent the respective sums. 
 
 x +r a (i) 
 
 y+5=« (2) 
 
 *+i=« (3) 
 
 2x-\-y=2a 
 
 3y-\-z=3a or, 4z-\-\2y—\2a 
 
 4z-\-x=4a or, 4z± x= 4« 
 
 — x+l2y= 8a 
 From the 1st 24x-\-l2y=24a 
 
 25# =16a 
 
 (24.) This problem is resolved in the work, by the 13th 
 example, page 80, (Art 51.) 
 
 (25.) Let x= the greater, and y the less. 
 5#-Hy=13 
 jx — 5y=0 or, 2x=3y. 
 
22 KEY TO ALGEBRA. 
 
 (26.) #-K(y+z)=a=51 
 
 y+*{x-\-z)=a 
 z-\-k(x+y)=a 
 x+(x-\-y+z)=2a 
 2y+{x+y+z)=3a 
 3z-}- (x-\-y-\-z) =4a 
 
 x—2a — s (1) 
 
 y =i(3a— s) (2) 
 
 *=i(4a— 8 ) (3) 
 
 s= 2a — s-f-£(3a — s)-H(a — s) 
 6«= 12a— 6s+9a-— Ss-\-Sa—2s 
 I7s=29a or 5=29-3=87. 
 
 Now equations (1), (2), and (3), will readily give x, y, 
 and z. 
 
 (27.) Let #=A's, y B's, and z=C's sheep. 
 Then by the conditions, 
 
 x+8 — 4=y-fz— 8 
 
 5(y+8)="ar+ar— 8 
 
 *(*+8)=ar+y— 8 
 
 *4-12=» y+* (1) 
 
 y+24=2x+2z (2) 
 
 z-\-32=3x-\-3y (3) 
 
 Add (1) and (3), and we have x-\-44—3x-\-4y. 
 Double (1), and subtract (2), and we have 
 
 2x — y—2y — 2a? or, 4ar=3y 
 
 But 2#-f4y=44 
 
 4#+8i/=44-2 
 
 lly=44-2 or y=8. 
 
 (28.) ;r+l_l x _l 
 
 "J 3 y+l~4 
 
 (29.) a?+2__5 a: _1 
 
 "^ 7 ^"S 
 
EQUATIONS. S3 
 
 (30.) This is a repetition of the 10th example, page 89, 
 inserted here by oversight. 
 
 (31.) Let #=A's money, and 2/=B's. 
 
 *-5=i(y+5) (1) 
 
 ar-r-5=3y — 15 (2) 
 
 Subtract (1) from (2), and we have 
 
 10=3y— 15— or y=ll. 
 
 2 Z 
 
 (32.) Let a?= the number of bushels of wheat flour. 
 
 And y= barley " 
 
 Then the cost of the whole will be expressed by 
 
 iOx-\-4y 
 The sale at 11 shillings will be llic-f-lly 
 Now by the conditions, 
 
 10aH-4y : ll#-f-lty : : 100 : 143| 
 Multiply the last two terms by 4, and 
 
 \0x-\-4y : llx-\-lly : : 400 : 575 
 Divide the two last terms by 25, and 
 
 10aH-4y : ll^-flly : : 16 : 23 
 
 bx+2y : llz-fllt/ : : 8 : 23 
 
 115#-}-46y=88;r-f88i/ 
 27^=42y 
 9x=-\4y 
 These co-efficients, 9 and 14, give the lowest proportion 
 in whole numbers. The proportion was only required. 
 (33.) Let lOx -jry represent the number. 
 
 Kito+sO-e-H 
 
 i(io*-h/)=<M-i 
 
 Now the question gives us Q =2x 
 
 And Q'^ty 
 
 . i(10*-f-y)=2ar-H or y=l. 
 
24 KEY TO ALGEBRA. 
 
 INTERPRETATION OF NEGATIVE VALUES. 
 
 (Art. 55.) 
 
 (4.) Let x represent the years to elapse. 
 
 Then 30-r-ar=3(15+a?) x=— lh 
 
 To make this equation true, the years required must be 
 taken subtr actively. 
 
 (5.) Let x= the man's daily wages, and y— the son's. 
 
 7x+3y=22 (1) 
 
 5#-f- */=18 (2) 
 
 12aH-4?/=40 
 Sx-{- y—lS #=4, y= — 2. 
 
 (6.) Let x— man's wages, y= wifes, and z~ the son'? 
 
 10#+ 8y-\- 6^ = 1030 cts (1) 
 
 \2x-\-l0y-{- 4z=1320cts (2) 
 
 \5x+l0y-{- 12sr=1385cts (3) 
 
 Subtract (2) from (3), and we have 
 
 3x+8z=65 (4) 
 
 Multiply (1) by (5), and (2) by 4, and take their differ- 
 ence, and we have 2x-\-\4z= 130 
 
 x -{- 7z=— 65 (5) 
 
 Sx+21z=— 3-65 
 (4) Sx-\- Sz= 65 
 
 By subtraction, 13z= — 4*65 
 
 z=: _4.5=— 20 
 As z comes out with a minus value, it shows that the 
 eon had no wages, but the reverse of it, he was on expense. 
 
 (7.) 10;H-4y-r-3z=1150 (l) 
 
 9£-r-82/-f 6z=1200 (2) 
 
 7x+6y-{-4z= 900 (3) 
 
 Double (1), and subtract (2), and 
 
 11^=1100 a?=100cts. 
 
EQUATIONS. 25 
 
 (8.) x+ l_3 x __5 
 
 ~~^ 5 y+f" 
 
 5 X -\-5=3y (1) 
 
 7x =5y+5 (2) 
 
 Add 12a; =8y 
 
 3^ =2y (3) 
 
 Subtract (2) from (1) and we have 
 
 —2^+5=— 27—5 (4) 
 
 #=__10 by adding (3) and (4). 
 
 y=— 15 
 The result coming out minus, shows that there is m 
 such arithmetical fraction. Algebraically, however, »~tj 
 will answer the conditions. 
 
 SECTION III. 
 
 We pass over the first three chapters of this section, as 
 requiring no aid from a key. 
 
 We commence at the examples in approximate cube root 
 *t page 126. 
 
 (1.) What is the approximate cube root of 120 1 
 
 125 has a root of 5. 
 2 
 
 250 240 
 
 120 125 
 
 370 : 365 : : 5 : root required. 
 
 (2.) What is the approximate cube root of 8.5 ? 
 8 has a cube root of 2. 
 
 16 17 
 
 8.5 8 
 
 24.5 : 25 : : 2 : root required. 
 
26 KEY TO ALGEBRA. 
 
 (3.) What is the approximate cube root of 63 ? 
 64 has a cube root of 4. 
 128 126 
 
 63 64 
 
 191 : 190 : : 4 : root required, 
 (4.) What is the approximate cube root of 515? 
 512 has a cube root of 8. 
 
 1024 1030 
 
 515 512 
 
 or 
 
 1539 : 154 
 1539 : 3 
 
 sur: 
 J980* = 
 
 3 j54x A = 
 4 3 > /708 = 
 
 \2 : : 8 : cube required. 
 : : 8 : correction. 
 
 DS IN GENERAL. 
 
 (2.) 
 
 : ^/49a 2 
 
 X2 
 
 = 7aj2 
 
 (3.) 
 
 = JWXZy 
 
 =2xjSy 
 
 (4.) 
 
 :y27£ 3 : 
 
 K2x 
 
 =3x*j2x 
 
 (*•) 
 
 =4 727. 
 
 X4 
 
 = 12 V4 
 
 (6.) 
 
 Jx? — aV= 
 
 3 JS2a 3 = 
 
 J2$a 3 x 2 = 
 
 = Jx 2 (x- 
 
 -a*) 
 
 —xjx — a* 
 
 (*•) 
 
 = V80 3 
 
 X4 i 
 
 =2a V4 
 
 (8.) 
 
 = t j4a 2 x 2 
 
 X7a 
 
 =2aav7a 
 
 (9.) Performed in the work. 
 
 V 64 V 64 V 64 4 V 
 
 (12.) /50 /25-2 /25-6 5 _ 
 
PURE EQUATIONS. $7 
 
 (i3.) * v^^6" 2 =v^(TT6T=« vh^ 
 
 (14.) /2a /T 1 — 
 
 3 9 3 V 
 
 (Art. 81.) Requires no aid from a key. 
 (Art. 82.) 
 
 (I.) 5^5X3/8 = 15^X2X4=30710 
 (2.) 4712x3^2-12724=24^/6 
 (3.) 372 X 278=6716=24 
 (*•) 2 yuX3 3 74=6 3 77X8=12 8 77 
 
 (5.) 275 x 27X0=4750=2072 
 
 (1.) Here to multiply (a-f-^) 3 by (a-f-6) we must sim- 
 ply add the exponents. 
 
 (2.) (7)^X7^=7^^=7^ 
 
 (3.) 2(3)* X3(4)*=2(3)^X3(4)*=2(27)*X 3(16)*== 
 6(27X16)*=6(432)~« 
 
 (4.) (15)*X(10)*=(15)*X(10)*=(225)*X(1000)*= 
 (225000)* 
 
 PURE EQUATIONS. 
 
 (14.) Multiply both members by Ja-\-x % and 
 t Jax-Jrx 2 -\-a-{-x=i=2a 
 jjax+x^a — x 
 ax-\-x*=a 2 — 2ax J rx % 
 3ax=a* x*=la. 
 
KEY TO ALGEBRA. 
 
 (15.) Multiply by Jtf+x 2 , then 
 
 xJa 2 +x 2J r a 2 -\-x 2 =2a i 
 xJa 2 -\-x 2 =a 2 — x 2 
 By squaring both members and reducing we have 
 Sa 2 x 2 =a 4 or x=±ajl, 
 
 (16.) Square both members, and 
 
 x 2 +2ax+a 2 =a 2 +xj¥+x* 
 Drop a 2 and divide by x, and 
 
 x-\-2a=Jb 2 -\-x 2 Square, &c. 
 
 (17.) Divide the numerator by its denominator, in each 
 
 member, and we have 
 
 4 15 
 1, =1 
 
 v/6a?-r-2 4vQx-\-Q 
 Drop 1 and change signs, and clear of fractions, and 
 16^6^-1-24=1576^+30 Hence x=6. 
 
 (18.) Cube both members, and 
 
 (4-\-xf 
 
 64-f-z 2 — Sx=- -=(4+x )2 
 
 (4+*) K T J 
 Hence, 64-\-x 2 —8x=16-\-8x-\-x 2 or x=3, 
 
 (19.) By clearing of fractions, f>-\-x-\- J x 2 -\-5x=\5 
 By reduction, 1 Jx 2 -{-5x=lO — x Square &c. 
 
 A/ x-f-Jx — \/ x — Jx= -—= 
 
 V x+Jx 
 
 M iltiply by \f x-\- Jx, and we have 
 
 
 
 X \ w X i^ Ju *L 2 /^ X 
 
 2a:+ 2 Jx — 2 Jx 2 —x =3jx 
 
 2x — Jx==2 s /x 2 — x 
 
 4X 2 — 4x,Jx-{-x=4x 2 — 4a? *=H* 
 
PURE EQUATIONS. 29 
 
 (21.) Resolved in the work. 
 
 (22.) Resolved the same as 17. 
 
 2b 7b 
 1 =1 
 
 a 
 
 \/ax-\-b 3Jax*\-5b 
 
 _ _ 96* 
 
 Hence, 6Jax+l0b=7jax-{-7b or x- 
 
 ( U. 136. ) ( & 136. ) 
 
 (23.) Square both members and we have 
 ^+a^ 2 -f-12= / i+2;r+z* 
 l Jx 2 -\-\2=2-\-x Square, &c. . x=2. 
 
 (24.) Multiply numerator and denominator by the nu- 
 merator, and we have 
 
 i 
 
 Take square root, and transpose ,j4x, and 
 N /17-Fl=3— J4x 
 Square, 4a?+l=9 — 6,j4x-{-4x Hence, #=£. 
 
 (25.) Square root gives a — x=Jb or, x=a — Jb. 
 (26.) Clearing of fractions and we have 
 
 271—^=73 
 
 4— 4;z 2 =3 x 2 =% 
 
 (27.) Take the square root of both members, and 
 
 2 1 
 =- or 4=a? — 1 
 
 x—l 2 
 
 (28.) Resolved the same as the 21st. 
 (29.) Clearing of fractions we have 
 A /r i ^M-a>-9=36 
 
30 KEY TO ALGEBRA 
 
 Jx 2 — 9#=45 — x 
 
 x 2 — 9^=452—90^+^ 
 
 81z=45-45 
 
 9-9#=5-9-5-9 #=5-5=25. 
 
 (30.) Resolved the same as (17) and (22.) 
 
 Dividing numerators by denominators, we have 
 
 10 10-5 
 3 =3 
 
 jx+2 V^-Ho 
 
 Drop 3 from both sides, change signs, and divide by &, 
 and clear of fractions, and 
 
 2 > /#-f80=21 > /x+42 Hence x=4. 
 
 (31.) Multiply numerator and denominator of the first 
 member by (Jx+Jx — a) 
 
 Then, y+</~* 
 
 a x — a 
 
 Multiply by «, and take square root, and 
 
 an 
 
 Jx+ Jx—a= 
 
 Jx — a 
 
 Jx 2 ~ax-\-x — a=an 
 Jx 2 — ax— (n+ 1 )a — x 
 x i—ax={n+\) 2 a 2 —2a(n+\)x+x % 
 Drop x 2 , and divide by a, and 
 
 — x = (n -f- 1 ) 2 a — 2nx — 2a? 
 (l+2n)x=(n-t-l)*a 
 _(n+l) 2 « 
 
 X l+2ra 
 (32.) Resolved in the work. 
 (Art. 90.) 
 
 (4.) Observe 180=9*20 189=921. Put a=9. 
 *fy+xy*=2Qa sHy=21a 
 
PURE EQUATIONS. 31 
 
 Multiply the first equation by 3, and add it to the second, 
 
 and x 3 +3x 2 y+3xy 2 -\-y ;i =81a=a! i 
 
 cube root, a?-f-y=a=9 
 
 The rest of the operation is obvious 
 
 (5.) Divide the first equation by (x+y) and 
 x 2 —xy+y 2 =xy 
 or x* — 2xy-\-y 2 =0 or x — y=0. 
 
 Hence x=2 y=2 
 
 (6.) x+y : x : : 7 : 5 xy+y 2 =l2& 
 
 5x-\-by—7x 
 
 5y=2x or a?=§i/. 
 Put this value of x in the second equation, and 
 
 7y 2 = 126-2 
 ^=18-2=36 . . .y=±6. 
 (7.) From the first equation we have 
 
 5x— 5y=fy 5x—9y 
 
 x z -{-4y 2 =lSl 
 1^+4^=181 
 8h/ 2 -r-100t/ 2 =181-25 
 
 181# 2 =181-25 or i/ 2 =25. 
 
 (8.) From the proportion we have 
 
 Jx+Jy^Jx— ±jy 
 
 §Jy—%jx or 25#=9#. 
 
 The rest of the operation is obvious. 
 
 (9.) Extract square root and 
 
 |jp-H=3 or • x=-lh 
 
 (10.) From the first proportion 
 
 x-\-y=Zx — dy or 4y=2x 
 
 Hence 8y>=x*. 
 
 8 V 3 — 1/ 3 =56 y=8 y=2. 
 
32 KEY TO ALGEBRA. 
 
 (11) (12) and (13) resolved in the work. 
 
 U4.) or 2 — v 8 
 
 -=6 or a?+v=6 
 
 x—y * 
 
 From the second, #y=5. 
 
 (15.) Divide the 1st equation by (x-\-y), and 
 x 2 — xy-\-y t —2xy 
 
 x 2 — 2xy-\-y 2 =xy=lQ (1) 
 
 Add 4xy =64 
 
 a?-\-2xy+y*= 80=16-5 
 Square root, x-{-y=4 ) j5 
 
 Square root of (1) x — y=4 
 
 2x =4^5+4 
 (19.) Double the 2d equation, and add and subtract it 
 from the 1st, then 
 
 x 2 -\-2xy-\-y 2 =a-\-2b 
 x 2 — 2xy-{-y 2 —a — 26 
 z+y=Ja+2b 
 x—y=Ja — 2b 
 
 (Art. 92.) 
 
 (5.) Add the two equations and extract square root, and 
 i i 
 we have x* -\-y* =±4 (1) 
 
 Separate the first member of the first equation into fac- 
 tors, and we have # 5 Cz 5 +i/ 5 )==12 (2) 
 
 Divide (2) by (1) and a?*=d=3 x=9. 
 
 (6.) is of the same form and resolved the same as (5.) 
 (?.) Add the two equations, and extract square root, and 
 
 3 3 
 
 we have x* -\-y* = J a-\-b 
 
 \ 
 
PURE EQUATIONS. 33 
 
 But a?* Gr* +#*)=« 
 
 I a 
 ar = 
 
 Ja+b 
 
 x?— or x= ( j 
 
 (a+bf \ («+6)V 
 
 (8.) Resolved in (Art. 90,) of the Key. 
 (9.) Square the first equation, and 
 
 x -fy^ia (1) 
 
 Difference 2x*y* = 12 (2) 
 
 Subtract (2) from (1) and 
 
 x— 2x*y*+y = 1 
 
 By evolution x* — y* =±1 
 
 But x *+y* = 5 
 
 2a? 5 =6 or 4. 
 
 CHAPTER V. 
 (Art. 93.) 
 
 QUESTIONS PRODUCING PURE EQUATIONS. 
 
 (5.) Let x-\-y = the greater number, 
 And ...... x — t/= the less 
 
 Difference • . . .2^=4 Sum =2x 
 
 2a?(4#i/) = 1600 Hence. . . .#=10. 
 
 (6.) Let x-\-y= the greater number. 
 x — y= the less. 
 2y : x-—y : : 4 : 3 or x= \y. 
 
 ( x 2 — y 2 )( x—y)=504 
 
 (W) (}y) =504 Hence y=4. 
 
 3 
 
34 KEY TO ALGEBRA. 
 
 (7») Let Sx= the length of the field, and 5x= its breadth 
 
 40^ x 2 
 
 Then =— = the acres. 
 
 160 4 
 
 ?# 2 X8;r=the whole cost. 
 
 26x— the rods around the field. 
 
 13X26#=the whole cost. 
 
 Hence 2x z =lS-26x or #=13 . . . 8x=104,rfns. 
 
 (8.) Let 5a?— the length of the stack. 
 
 4x= the breadth. 
 
 7x 
 Then, — = the height. 
 
 2 
 
 Ix 
 5x'4x* — -4x= the cost in cents. 
 2 
 
 Also, 5;r-4av224= the cost in cents. 
 
 Ix 
 
 Hence, 5x*4x' — •4#=5a: , 4a: , 224 
 
 2 
 
 7a?-2ar=224 or x—4. 
 
 (9.) Put x 2 — 7= the number. 
 Then x-\-Jx J +9^=9 
 
 ^2-j_9— 9— x or x —4, 
 
 (10.) Let x represent A's eggs ; then 100 — x— B's 
 
 18 8 
 
 = A's price. -=B's price. 
 
 100— x 
 
 18# 8 1 
 
 Hence =-(100— #) 
 
 100— x x x 
 
 9^ 2 =4(100— xf 
 
 3#=2(100— x) a?=40. 
 
 (11.) Let x-\-y= the greater number, 
 x — y= the less. 
 
 2#=6 or x=S 
 
QUADRATIC EQUATIONS. 35 
 
 (x-\-yY=x 3 +3x z y+ 3xy 2 +y* 
 (x^yf=x a -^3x 2 y-{-3xy 2 —y t 
 
 2x 3 +6xy 2 =72 
 
 Divide by 2x and a*+3y 2 =12 Hence . . »y=l. 
 
 (12.) Let x= one number 
 Then a 2 x= the other. 
 
 b 
 
 aV=6 2 arr=6 x=-. 
 
 a 
 
 (13.) Let a? 3 and y 2 represent the numbers. 
 
 Then a? 2 +i/ 2 =l00 
 
 x -\-y = 14 
 
 SECTION IV. 
 
 QUADRATIC EQUATIONS. 
 
 None of the examples require the aid of a key until we 
 come to the 12th, Art. 101. 
 
 (12.) Put x=\u. Then 5x 2 = jW 2 , and the equation be 
 
 comes \u 2 -\-\u= 273 
 
 m 2 +4m=1365 
 Hence w=35 or — 39 which gives . . . x=7 or — 3 j°. 
 
 (13.) Put x=\u. Then 
 
 u 2 — 20w =224 
 
 W 2__20u-f-100=324 
 w _10=±18 
 
 «=28 or — 8 and • • x=7 or — } 
 (14.) 25# 2 — 20#=— 3. 
 Complete the square by (Art. 99.) 
 
 25a; 2 — 20a?+P=/ 2 — 3 
 5atf=— 10a? t=—2 t 2 =4 
 5x—2-=^zl x=i or \. 
 
36 KEY TO ALGEBRA. 
 
 (15.) Put x=^u. Then 
 
 « 2 — 292u= — 500-21=— 10500 
 
 w 2 — 292n4-(146) 2 =108L6 
 
 w— -146=±104 
 
 w=250 or 42 
 
 Hence »=Vt° or 2. 
 
 (Art. 103.) 
 
 EXAMPLES. 
 
 (1.) Put(a?+12)*=y. Then 
 
 2/ 2 +2/=6 2/=2 or —3. 
 
 (a?+12)*=2or — 3 
 
 a:-|-12 = 16or81 a?=4 or 69. 
 
 (2.) Add b 2 to both members and extract square root 
 re then have (rc-f-«)*+&=rt2& 
 The rest of the operation is obvious. 
 
 (3.) Put y=(9x+4f Then tf+2y=\b 
 y=3 or — 5 Hence 9aH-4=9 or 25. 
 
 (4.) Put y=(l0-\-xf. 
 
 Then y 2 — y=2 y=2 or — 1 
 
 (Art. 104.) 
 
 3 
 
 (5.) Put. . . .(x — 5) 2 =?/; then 
 y* — 3^—40 y=S or — 5. 
 
 (6.) Put (1+37— x*)*=y; then 
 
 .2 ,,— : 1 
 
 2tf 
 
 From which we obtain y equal 5 or \. 
 Hence . . l-\-x — afej- or ^. From which we obtain 
 x=4t±$jU or izt&.fm. 
 
 (7.) Put. . . . (a?+ 16)^=1/ 
 
 ^2 — 3y=io Hence y=5 or — 2. 
 
QUADRATIC EQUATIONS. 37 
 
 (§.) Puty=a? n ; then 
 
 d*f—2y=B Put y=\u 
 
 w 2 — 2w=24 w=6 or — 4 
 
 y=Z x= n J2. 
 (9.) Multiply by 4, &c. Rule 2. 
 
 4a?*+4a?M- 1=3025 
 
 2a?*+l=±55 
 
 , a?*=27 or —28 
 a? s =3 or a?=243. 
 
 (10.) Put (2a?— 4) 2 =y. 
 
 8 16 
 Then -=H 
 
 3/ y 2 
 
 8y=^»+16 
 ^— .8i/+16=0 or y— 4=0. 
 
 (11.) Multiply by 16. Rule 2. 
 
 Then . . . 64a?^+16a?*+ 1=39- 16+ 1=625 
 
 8a?i+l=25 a?*=3 ;r=729. 
 
 (12.) Add 5 to both members. 
 
 Then {x 2 — 2H-5)+6(ar 2 — 2a?+5)*=16 
 
 By subtraction, 2/ 2 +6y+9=25 y=2 or — 8. 
 
 Hence ar 2 — 2a?+5=4 a?=l. 
 
 (13.) By (Art. 99) we have 
 
 x* 12a? 
 1_, 2=/2 _32 
 
 361 19 
 a? 6a? 
 
 — 1= *=— 6 * 2 =36. 
 
 19 19 
 
 x 2 12a? 
 
 1-36=4 
 
 361 19 
 
 a? 
 
 19 
 
 -6=±2 a?=152 or 76. 
 
38 KEY TO ALGEBRA. 
 
 (14.) Observe that 81 2* and — are both squares, and 
 
 ar 
 
 if these are taken for the first and last terms of a binomial 
 
 1 
 
 square, the middle term must be 9a?*-'2^18. 
 
 x 
 
 This indicates to add 1* to both members. Then extract 
 
 1 
 square root 9x-\ — =±10 Hence. • • • x=l or — 1. 
 x 
 
 (15.) The first member of (15) is the same as (14.) 
 
 Hence, add unity to both members and extract square 
 
 1 29 
 
 root ; we then have 0x-\ — = j-4 
 
 x x 
 
 u 
 93 2 — 43=28 Put 3=- 
 
 9 
 
 w 2 — 4w=28-9=252 
 
 u — 2=±16 3=2. 
 
 (Art. 105.) 
 
 (4.) Multiply every term by 3, and 
 
 ^_83 3 +19a: 2 — 123=0 
 Operate for square root thus 
 
 a? 4 — 82?+ 19a*— 12a; (a* — ix 
 
 2x*— 4x)— 82?+ 19a? 2 
 — 83 3 +163 2 
 
 33 2 — 123 
 3(a*— 4x) 
 
 (a?— 43) 2 +3(3 2 — 43)=0 
 Divide by (2* — 4a:) and 
 
 2*— 43+3=0 3=1 or 3 
 
 But the factor 2* — 43 gives 3=0 or 4. 
 
QUADRATIC EQUATIONS. 39 
 
 (5.) # 4 — 10x 3 +35a^--50^4-24=0 ( x 2 — 5x 
 
 2x?— 5x)— 10x 3 +35x 2 jr& OY THE 
 
 UNlVEESITY 
 
 10x 2 — 50*+24 v /y 
 
 (a^— 5ar) 8 +10(a?»— 5a?)+24^^XrIpQB, , »VJ 
 If we add unity to both members, we shall 
 plete squares. Extract square root, and 
 
 x*—5x= — 4 or —6 
 From these two equations we find #=1, 2, 3, or 4. 
 
 (6.) By mere inspection we perceive that this equation 
 can take the form (a: 2 —- a?) 2 — (a? 2 — #)=132. 
 
 y2—y=l32 #=12 or — 11. 
 
 x 2 — x—Vl or — 11 
 If we take — 11, the value of x will become imaginary, 
 12 gives #=4 or — 3. 
 
 ( U. 169. ) 
 (7.) This equation may be put into this form : 
 (y»_ cy)-— 2 (y 2 —cy)=c 2 
 from which the reduction is easy. 
 
 (Art 107.) 
 
 (3.) Taken from the work we have 
 ( a + l)x 2 —a 2 x=a* 
 
 or (a-\-l)x 2 =(x+l)a 2 
 
 Both members are of exactly the same form, and of 
 course the equation could not be verified unless • . a?=«. 
 
 EXAMPLES. 
 
 (1.) # 2 + llx=80. Multiply by 4, &c. 
 4.z 2 -f-./H-ll 2 =320+l21=441 
 2#-f-ll=±21 x =5 or —16. 
 
V\ KEY TO ALGEBRA. 
 
 (2.) Drop 2x from both members, and divide by 3 ; then 
 
 x — 1 x — 2 
 
 x = 
 
 #— 3 2 
 
 Clearing of fractions and 
 
 2X 2 — 6;r— 2x-\-2=x*— 3x—2x-\-6 
 x 2 — 3*=4 Put 2a=3 
 
 Hence, (Art 106) x=4 or I 
 
 (3.) Multiply the equation by 6x ; then 
 
 6x 2 
 
 — ■ |-6tf-j-6=13a? 
 
 x-\- 1 
 
 Gx 2 
 
 _4-6=7* 
 a:+l 
 
 6x 2 +6aH-6=7;r 2 -r-7a: 
 Hence x 2 -^-x—6 x—2 or — 3. 
 
 (4.) Clearing of fractions 
 
 70#— 2 \x 2 +72^=500— 1 50a? 
 21# 2 — 292*=— 500 
 
 (5.) Put f-+y\=x. Then 
 
 a? 3 -f-a?=30 or, x=5 or — 6. 
 
 6 
 Now, — H/=5 or — 6 
 
 y 
 
 y 2 — 5y=— 6 or t/2+6?/=— 6 
 
 Ay 2 —- .#+25=25— 24 
 
 2y— 5==hl y=3 or 2. 
 
 (6.) Put x%=y; then i/ 2 +7#=44 
 Ay 2 +.#+49=225 
 2y+7=±15 2/=4or— 11. 
 
 a?=(4)f or (— 11)*. 
 
QUADRATIC EQUATIONS. 41 
 
 (7.) x 2 -{-x=42. Hence x=6 or — 7. 
 
 That is yH-1 1=36 49 or y=5 or «/38. 
 
 (8.) ar+7 a: 
 
 x—1 3 
 
 33a? — 23 1— 3a?— 2 1 =x*—-lx 
 
 x 2 — 37a?=— 252 
 
 4a? 2 — .tf-I-37 2 =1369— 1008=361 
 
 2a;— 37==t=19 *=28 or 9. 
 
 (9.) 3a? 2 — 9a?=84 
 
 12 
 
 362 2 — ^+81 = 12-84-1-81 = 1089 
 Gx— 9=±33. 
 
 (10.) Clearing of fractions we have 
 
 2x+2 t Jx=\6—x 
 
 ^x-\-2 s /x=\G 
 Multiply by 12, &c. 
 
 6^/a?-{-2=±14 a?=2or7j. 
 
 (II.) 6(2a:— li; 
 
 a? — 3 
 
 3(2a?— 11) 
 
 •4a?=26 
 
 -2a?=13 
 
 a:— 3 
 
 6a?— 33-f-2a? 2 — 6a?=13a?— 39 
 2a? 2 — 13a?=— 6 
 16a? 2 — ^-1-13 2 =169 — 48=121 
 4x— 13=±11. 
 
 (12.) Multiply by x* and we have 
 
 22a: 2 
 
 10a: — \4-\-2x— 
 
 9 
 11a: 2 
 
 6a:— 7= 
 
 9 
 
 11a? 2 — 54a: =—63 
 
42 KEY TO ALGEBRA. 
 
 u 
 
 Put x=— ; then u 2 — 54m =—693 
 11 
 
 u 2 —542i-\-21 2 =3Q s 
 
 u— 27=±6 m=33 or 21. 
 
 (13.) Clearing of fractions we have 
 
 £*_! o# 2 -f- 1=* 3 — 6^ 2 ~|-9^— 3x 2 + 1 8z— 27 
 — # 2 =27.r— 28 
 tf 2 -f-27x=28 Put 2a=27 
 a' 2 -f2a^=2«+l 
 x 2 +2ax-{-a 2 =a 2 -{-2a+l 
 a?-fa=±(«+l) • .a?=lor— 28 
 
 (14.) Given mx 2 — 2mxjn=nx 2 — rnn to find x 
 
 By transposition mx 2 — 2mxJn-\-mn=nx 2 
 Square root . . Jmx — Jmn—dcjnx 
 By transposition {Jm±Jn)x=,Jmn 
 
 Jmn 
 
 CHAPTER II. 
 
 QUADRATIC EQUATIONS CONTAINING TWO OR MORI 
 UNKNOWN QUANTITIES. 
 
 Problems 1, 2, and 3, require no aid from a key 
 (4.) Put x=vy; then the equations become 
 
 v*y*+y* = 18vy* (1) 
 
 vy+y=\2 ' • . . . (2) 
 
QUADRATIC EQUATIONS. 43 
 
 From (1) we have y— 
 
 * «M-i 
 12 
 
 From (2) we have y= 
 
 v-f-1 
 
 Hence = 
 
 v-f-1 v 3 -f-l 
 
 3v 
 Divide the denominators by (v-f-1), and 2=—- • 
 
 Or, 2v 2 — 5v= — 2 v =2 or |. 
 
 •Another Solution. 
 Cube the 2d equation and we have 
 
 x 3 -\-ij 3 -\-3xy(x+y) = l2 s 
 That is. . . . 18x'i/H-3xi/(12) = 12-12-12 
 Divide by 6, and we have 3xy-\-6xy=2'\2-l2 
 
 Or 9xy=2'3'4-34 
 
 Or ^=2-4-4=32 
 
 Now, having (x-\-y) and (xy), the rest is obvious 
 
 (5.) The first equation can be put in this form 
 
 (x+y) 2 +2(x-{-y) = l20 
 Or s 2 -f2s-f-l = 121 
 
 *+l=dbll 
 
 x-\-y=\0 or — 12 
 Or a?=10 — y or x= — y — 12 
 
 From the second equation x= — 
 
 y 
 
 8+y 2 8+v 2 
 Hence . . _=10— y or, — = — y J 2 
 
 y y 
 
 8-|-2i/ 2 =:10i/ or, 8+2?/ 2 + 12y=0 
 
 y* 5y— 4 ^ 2 -J-6l/= 4 
 
 2/ 2 +2«y=2a+l 
 y-fa=±(a+l) 
 
44 KEY TO ALGEBRA. 
 
 (6.) Put x=vy ; then the equations become 
 
 vy 2 -\-2y 2 =60 
 56 
 
 y 2 - 
 
 y 2 =- 
 
 v 2 -\-v 
 60 
 
 v+2 
 
 15 14 
 Hence = 
 
 w-f-2 v 2 +v 
 
 \5v 2 + 15v=Uv-\-28 
 
 15v 2 +v=28 
 
 4-15 2 t; 2 +^4-l =28-60-)-l = 168l 
 
 2-]5v+l = ±41 v=f or— J. 
 
 (7.) Put x=vy ; then the equations become 
 
 6i; 2 i/ 2 +27/ 2 --=5^ 2 4-12 (1) 
 
 2ut/ 2 -f3i; 2 ?/ 2 =3i/ 2 — 3 (2) 
 
 12 
 From (1), . . . y 2 = — ■ — 
 
 —3* 
 
 From (2), . . . y 
 
 2 
 
 3^2 4-2^—3 
 
 4 1 
 
 Hence . . . j =0 
 
 6u 2 — 5v+2 3?; 2 +2v— 3 
 
 12v 2 +8v— 12+6t; 2 — 5u+2=0 
 
 18v 2 +3i>— 10=0 
 
 9v 2 -t-|v=5 
 Complete the square, (Art. 99.) §v 2 -\-\v-\-t 2 =t 2 -\-5 
 
 3vt=lv or, t=% * 2 = T V 
 
 9v 2 4-Iv+T , a=T , 6+5=fi 
 
 3i;-f-*=±} v =f r— f, 
 
 (§.) Put x=vy ; then the equations become 
 
 3v 2 y 2 +vy 2 =68 (1) 
 
 4?/ 2 -f-3u2/ 2 = l60 (2) 
 
 * 
 
QUADRATIC EQUATIONS. 45 
 
 68 160 
 
 3v 2 +v " 4+3i> 
 
 40 17 
 
 4+3v 'Sv 2 -\-v 
 
 120v 2 +40v=51v+68 
 
 120v 2 — llu=68 
 4-(120) 2 ?; 2 — .tf+121=68-480+121=32761 
 2-120v— ll=rbl81 u=| or— jj. 
 
 (Art. 111.) 
 
 (2.) From the second equation we get 
 
 x*y*=6 or, 2aA/* = 12 
 
 This added to the first equation gives 
 
 Put x*-\-y*=s; then s 2 +2s=35 . . . s=5 or —7. 
 
 1 1 
 
 Now if we put a? 3 =P, and y* = Q, we shall have 
 
 P+<?=5 and PQ=6 
 
 From which P and Q are readily found, and from them 
 
 x and v. 
 
 A A 
 
 (3.) Put x*=P, and y 3 = Q. Then the equations be- 
 come P+£=8 (l x ; 
 
 /"+e"+i*G"=259 (2, 
 
 Square (1) and we have P 2 +2P£+Q 2 =64 ... (3) 
 Subtract (3) from (2), and we have P 2 # 2 — 2P#=195 
 Hence PQ=15 or —13. 
 
 Now as we have P+#=8, and PQ=-15, we have P= 
 
 1 
 5 or 3, and #=3 or 5. That is, x*=5 or 3, &c. 
 
 2. 1 
 
 (4.) Put a? 3 =P, and y 5 = #; then the equations be- 
 
 tome J F+# 2 +i>+#=26, and P#=8 
 
 2jPQ =16 
 
 (P+£) 2 +(P+£)=42 Hence . .P+£=6. 
 
46 KEY TO ALGEBRA. 
 
 i 
 
 £C* 33 3 11 
 
 (5.) Put — =u ; then u 2 -\-4u— — . . u=- or . 
 
 yk 4 2 2 
 
 The remaining operation is obvious. 
 
 (6.) Given # 2 — 8# 2 y-64, and y—2x*y*=4, to find a; 
 and y. 
 
 To both members of the first equation add 16#, and to 
 the second add x, to complete the squares ; then extract 
 square root, and we have 
 
 y— 4ar*=4(a?+4)* and y*— x* = (x+4)* 
 Four times the last equation subtracted from the prece- 
 i 
 ding gives y — 4i/ 3 =0 or y=16. 
 
 (7.) Multiply in the first equation as indicated, and sub 
 tract the second equation ; we then have 
 
 x+y+2x*y*=25 or x*-\-y*=5 
 But from the second equation we have 
 
 (•*-+y*)**y* a - 80 Hence #V= 6 - 
 
 3 3 1 
 
 (8.) Divide the first equation by y 2 , and x 3 =2y*, or 
 y =kx 3 This put in the second equation gives 
 Sxt— £a?l=l4 
 al—16#*+64=64— 28=36 
 CHAPTER III. 
 
 QUESTIONS PRODUCING QUADRATIC EQUATIONS. 
 
 We pass to the sixth. 
 
 (6.) Let t= the time (hours) he traveled, and r= his rate 
 per hour; then rt=36 (1) 
 
QUADRATIC EQUATIONS. 47 
 
 But if r becomes (r-J-1), t must become (t — 3), and then 
 (r+l)(/— 3)=36 . . . . (2) 
 
 Or rt-\-t— 3r— 3=36 
 
 rt =36 
 
 *.=3(r4-l) 
 
 Hence r 2 -f-r=12, and r=3. 
 
 (7.) Let x= the number of children, 
 
 And • • • y= the original share of each. 
 
 Then . . . a?y=46800 (1) 
 
 (a?— -2)(i/+1950)=46800 (2) 
 
 a?*/-f 1950a?— 2y— 2- 1950=46800 
 1950(*— 2)=2i/ 
 
 Or 975(a?— 2)*=a? f */=46809 
 
 By division, 3* — 2a?=48 #=8. 
 
 (8.) Let a?= the number of pieces. 
 
 675 
 
 Then = the cost of each piece. 
 
 x 
 
 675 
 
 48^. =675 
 
 x 
 48a? 2 — 675a*=675 
 16a? 2 — 225a?=225 
 
 ^9.j Let a?=the purchase money. 
 
 104a? 104a? 
 
 Then = the cost, and 390 = his whole gain. 
 
 100 100 6 
 
 104a? 104a? a? 
 
 Then : 390 : : 100 : — 
 
 100 100 12 
 
 Product of extremes and means 
 
 26a? 2 
 
 =39000— 104a? 
 
 300 
 
 2a? 2 
 
 =3000—8a? 
 
 300 
 
48 KEY TO ALGEBRA. 
 
 Put a=300 and divide by 2 ; then 
 
 # 2 
 
 — =5a — 4a? 
 a 
 
 # 2 -J-4a#=5a 2 
 x 2 +4ax-\-4a 2 =9a* 
 #-J-2a=3« #=0=300. 
 
 (lO.) Put x-\-y— the greater part, 
 
 And • • . x — y = the less part. 
 
 Then 2#=60, #=30, and x 2 — i/ 2 =704. 
 
 (11.) Let x— the cost ; then 39 — x= the whole gain. 
 x : 39— x : : 100 : x Ans. a?=10. 
 
 (12.) Let {x — 20)= the persons relieved by B. 
 Then . • . #-f-20 = the persons ....... A. 
 
 1200 1200 
 
 + 5: 
 
 ar-f-20 #—20 
 
 Divide by 5, and put «=240 ; then 
 
 a a 
 
 1=- 
 
 #+29 x— 20 
 o #__20a-f a?— 400 =ax-\-20a 
 # 2 =4O0+4OO=4O(«-|-1O)=4O-25O 
 Or # 2 =400-25 #=20-5=100. 
 
 Hence 80 is B's number, and 120 A's. 
 
 (13.) Let #= the price of a dozen sherry, 
 And y= the price of a dozen claret. 
 
 7x-\-l2y=50 (1) 
 
 10 
 
 — = the number of dozen of sherry for ^10 
 # 
 
 6 
 
 -=the number of dozen claret for £6, 
 
 y 
 
 10 6 
 Then . . . — =3-f- * (2) 
 
 # y 
 
Or 
 
 QUADRATIC EQUATIONS. 4tf 
 
 10 10y 
 
 70y 
 By substitution, --- {-12y=50 
 
 70t/+36?/ 2 +72?/=150?/+50'6 
 
 36*/ 2 — 8t/=300 
 
 9# 2 — 2y=75 Hence y=3. 
 
 (14.) Let 19#= the whole journey. 
 Then x= B's days, also his rate per day. 
 Or x 2 = B's distance. 
 Also, 7#-{-32 = A's distance. 
 
 x 2 -|-7aH-32 = 19# 
 tf2__12#=— 32 
 
 Hence #=8 or 4. 
 
 And 19a?=152 or 76. 
 
 If we put x for the whole journey, we shall obtain the 
 13th equation, (Art. 104.) 
 
 (15.) Let x= the bushels of wheat, 
 
 And . . . #-f-16= the bushels of barley. 
 
 24_ 24 1 
 
 x a?+16 4 
 
 x 2 + \§x 
 
 24*-f 16-24=24aH 
 
 4 
 
 a? 2 -f-16*=16-96=16-16-6 
 Put 2a=16. Then 2a-2«-6=24a 2 
 
 x*+2ax=24a 2 
 
 x+a=±:5a #=4a=32. 
 
 (16.) A put in 4 horses, and B put in x horses. 
 
 18 
 Then — =the rate per head. 
 x 
 
 4 
 
50 KEY TO ALGEBRA. 
 
 4-18 
 
 J— 18== the price of the pasture. 
 
 x 
 
 4-20 
 
 __— }-20= the price of the pasture. 
 
 x~\~2 
 
 4-18 4-20 
 Hence .... = \-2 
 
 x a?-f-2 
 
 36 40 
 
 — =-3-:+l *=6 
 
 x x-\-2 
 
 (17.) Let 4x= the price per yard, 
 And . . . 9x= the number of yards. 
 
 36a? 2 =324 x =3. 
 
 (18.) Let \0x-\-y= the number. 
 
 Then Z£=2 (1) 
 
 xy 
 
 And 10a?+y-r-27==10y-r-a? (2) 
 
 From (1). . . . 10a?=(2a?— \)y 
 From (2) • • • • x-\-3=y 
 
 ^ ,. . . l0x 
 
 By division, . . =2x — 1 
 
 ir-f-3 
 
 1 0a7— 2a? 2 + 6a?— <r— 3 
 
 2* 2 — 5a=3 ^3. 
 
 (19.) Let (.r— y), a?, and (a*-}-*/— 6) represent the num- 
 bers. Then 3a? — 6=33, or a?=13. 
 (x— y) a —x 2 — 2xy-\-y* 
 x 2 =x 2 
 (x+y— 8) 2 =x 2 +2xy-\-y 2 — 12a?— 12i/+36 
 
 3a? 2 +2?/ 2 — 12r— 12i/+36=441 
 
 By subtracting the value of 3a? 2 — 12a?+36, we have 
 2y 2 —l2y=54: Hence i/=9. 
 
 (20.) Let x and y be the numbers. 
 
QUADRATIC EQUATIONS. 51 
 
 Then x+y^xy (1) 
 
 X +y= X 2 +y 2 . .' (2 ) 
 
 Put x=vy, and the equations become 
 
 vy+y=vy 2 (3) 
 
 vy+y=v 2 y 2 -\-y 2 • • . (4) 
 
 v 2 + l 
 Divide (4) by (3), and =1 
 
 This gives v==h±h J — 3 
 
 But from equation (3) we have i>y=v-H=«=J±$«/ — 3 
 
 (21.) Let x-\-y= the greater number, 
 
 And .... x — y= the less. 
 
 ^^ 7 ^24 (1) 
 
 2x+2x 2 +ty 2 =62 (2) 
 
 Or x+x 2 +y 2 =3l 
 
 Add (1) x 2 — y 2 =24 
 
 2x 2 -\-x =55 #=5, y=l. 
 
 (22.) Let x-\-y= the greater number, 
 And . - . x — y= the less. 
 
 **— 2/ 2 +2x=47 (1) 
 
 2x 2 +2y 2 — 2#=62 (2) 
 
 2x 2 — 2?/ 2 -r-4a?=94 
 
 4^-f2x= 156 
 
 2x*+ x= 78 x=6, y=l. 
 
 (23.) Let (a?+y)= one number, 
 And .... (x — y)= the other. 
 
 2z=27 (1) 
 
 x*+3x*y+3xy*+y* 
 X s — 3x*y -f- 3.ri/ 2 — t y 3 
 
 2x 3 +6^ 2 =5103 (2) 
 
 Divide (2) by (1), and we have **-f-3i/ 2 =189 
 
52 KEY TO ALGEBRA. 
 
 3 2 -9 2 
 
 \-3y 2 =9-2l 
 
 3*81 
 
 Divide by 3, and f-y=3-21 
 
 4 
 
 3-81-j-4t/ 2 =3-84 
 
 4?/ 2 =3-3 or 2y=3 y==|. 
 
 *-h/=v+f=i5, &c. 
 
 (24.) Let #+y= one number, 
 And .... x — y= the less. 
 
 2x= 9 
 
 x 4 -\- 4x*y-\- 6xy-\-4xf-\-y* 
 x 4 — 4# 3 2/+6#y— 4ay-r-y 4 
 
 2a? 4 +12a?y +2^=2417 
 By resolving this we shall find y=\. 
 
 (25.) Let #+?/= the greater, and x — y= the less. 
 
 Then O^ 2 — 3/ 2 ) (2# 2 -f 2*/ 2 )=1248 (1) 
 
 Or x 4 — -y 4 =624 
 
 Also 4a??/=20 (2) 
 
 Or y=- 
 
 x 
 
 x 
 625 
 
 625 
 
 a? 4 — =624 
 
 «4 
 
 X* 
 
 # 8 — 624a? 4 =625 Put 2a=624 
 
 Then x*—2ax 4 -\-a 2 =a 2 -\-2a+l 
 
 x 4 — «=±(«+l) 
 
 a? 4 =2«-f-l or — I a?=5 
 
 y=i. 
 (26.) Let a?= the days required by one, 
 And .... #+10= the days required by the other. 
 
ARITHMETICAL PROGRESSION. 53 
 
 Then . • . -= one day's work of the .first. 
 x 
 
 = one day's work of' the second. 
 
 #+10 J 
 
 x #+10 12 
 
 SECTION V. 
 
 CHAPTER I. 
 ARITHMETICAL PROGRESSION. 
 (Art. 116.) 
 
 (4.) Z=l+3kZ (A) 
 
 280=16(24-31)^ (B) 
 
 (B) reduced gives (/=£ ; then (A) gives Z=16£. 
 
 (5.) Here a=|, L=h n=5. 
 
 *=l+4d (A) 
 
 *=(*+«i (2*) 
 
 From {A) we have d=£ T 
 
 5+2^=5= the first mean, &c. 
 
 (6.) Here a=9, Z=109, n=ll. 
 
 J09=9-f-10rf </=10, Am. 
 
 (7.) Z=1+364X2 (A) 
 
 s=730X365X£=(J5), Ans. 
 
 (8.) Z=20+(n— l)3=17+3n (A) 
 
 s =(37-f-3n)irc=438 (#) 
 
 (Art. 117.) 
 
 (2*) Represent the numbers by x — y, x, and tf+y. 
 3#=18 a?=6 
 
W KEY TO ALGEBRA. 
 
 x*=x 2 
 (*4-y) 2 =* 2 4-2ay-f-*/ 2 
 
 3x 2 +2/=158 
 
 (3.) Let a?— 3y, a;—?/, ar+y, and a?-f-3y represent the 
 numbers ; then 2?/ =4. 
 
 The product of the 1st and 4th is 
 
 a? 2 -— 9y 2 ; of the 2d and 3d is (a?—v*). 
 x*—y* v * ; 
 
 x 4 — 9#y 
 
 — a-y+ y 
 
 a? 4 —10a:y+9y 4 = 176985 
 
 9^= 144 
 
 x 4 — 40a? 2 =176841 
 
 (4.) The same notation as in last example. 
 
 2#=8 x=4 #2_y2=i5 
 
 (5.) Let n= the number of days. 
 
 Then Z=l + (rc— i)i=n 
 
 *===(l-f»)in= the whole distance. 
 
 Also (n— 6)15 = the whole distance. 
 
 n 2 -frc=30rc— 180 
 n 2 — 29n=— 180 .... n=9 or 20 
 
 6 6 
 
 3 14 
 (6.) The first day he must pay 1+t; % representing 
 the interest of one dollar for one day. 
 
 First day 1 -j- { 
 
 2d day 1 -f- 2i 
 
 3d day 1+ 3i 
 
 Last day l+60i 
 
 (2+61 i )30= the whole sum 
 
GEOMETRICAL PROGRESSION. 55 
 
 to be paid ; but as this sum is to be paid in 60 equal pay- 
 ments, each payment must be 
 
 1-1 — Ans. $1 and f of a cent, nearly. 
 
 z 
 
 ($.) Let x — 3^, x — y, x-{-y, and x-{-3y represent the 
 numbers; then 2x 2 -\-lSy 2 =50 
 
 2x 2 -{- 2# 2 =34 
 
 16*/ 2 =16 y=l. 
 
 GEOMETRICAL PROGRESSION AND HARMONICAL PROPORTION. 
 
 (Art 124.) 
 
 (1.) Let x represent the mean sought. 
 
 2-6-12 
 
 x= =8 
 
 18 
 
 (2.) Let a?= the number sought. Then, by harmonical 
 
 proportion 234 : x : : 90 : 144 — x 
 
 90#=234-144— 234# 
 
 324^=234-144 Hence #=104. 
 
 (3.) Let x= the number sought. 
 
 Then .... 24 : x : : 8 : 4— x 
 
 Or 3 : x : : 1 : 4— x . • #=3. 
 
 (4.) Let x= the second. 
 
 Then. . . 16 : 2 : : 16— a? : 1 . • . .#=8. 
 
 (5.) Let x= the first number, and y= ratio. 
 
 Then • • x+xy-\-xy 2 =2l0 (1) 
 
 xy 2 —x= 90 (2) 
 
 By subtraction .... 2x-{-xy=l20 or x—. 
 
 120 
 
 90 
 From (2) we have x= 
 
56 KEY TO ALGEBRA. 
 
 4 3 
 
 — ; — = or, Ay 2 — 3v=10 v=2. 
 
 2+y y*—l * y y 
 
 (5.) Let x, xy, xy 2 , and xy* represent the numbers. 
 
 mi xy 3 y 2 4 
 
 Then . . . - — ==-*_=_ 
 
 xy-\-xy 2 l-\-y 3 
 
 From this equation we perceive at once that y=2 ; then 
 
 #-r-2#-f-4.r-|-8;r=15#=30 x=2. 
 
 (6.) Let x, xy, xy 2 and xy* represent the numbers. 
 
 x+xy 2 =l48 (1) 
 
 xy-\-xy 2 =888 (2) 
 
 Or x(l+y 2 )=4-37 (3) 
 
 xy(i+y 2 )=4'222 (4) 
 
 Divide (4) by (3), and .... y=6. 
 
 (T.) Let x, Jxy, and y represent the numbers ; then 
 
 x-\-,Jxy+y=14 (1) 
 
 And x 2 +xy+y 2 =84 (2) 
 
 Put x-{-y=s, and Jxy—p ; 
 
 Then • • a?-\-xy-{-y*=8 2 — p 2 , and equations (1) and (2) 
 
 become s-\-p=\4 (3) 
 
 s 2 —p 2 =84 (4) 
 
 Divide (4) by (3), and we have s — p=6 (5) 
 
 Add (3) to (5), and divide by 2, and 
 
 s=10. Hence p=4. 
 (8.) Let x, xy, xy 2 , and xy* represent the numbers ; 
 
 Then xy 3 — xy=24 
 
 xy z -\-x : xy 2 -\-xy : : 7 : 3 
 Or . . . y«+l : y 2 + y : : 7 : 3 
 Divide the first couplet by (y-\-l), and we have 
 2/-2/-H : y : : 7 : 3 
 3y 2 —3y-\-3=7y or 3y 2 —l0y=—3 
 From this equation we have y=3, the ratio. 
 
GEOMETRICAL PROGRESSION. 57 
 
 (9.) Let x, xy, xy 2 , and xy z represent the numbers ; 
 
 Then *(i+y+yHy)=y+i 
 
 And a?= T y Put (y+\)=Jl. 
 
 Then T V(^+% 2 )=^ 
 
 A+Ay*. = 10./? rfy 2 =9A or i/=3. 
 
 Hence y'^, T 3 o, &c. are the numbers. 
 
 (10.) Let x, , and y represent the numbers; then 
 
 x+y 
 
 x+—±+y=2G 1 
 
 x+y 
 
 And xy=72 
 
 Put x+y—s; then equation (1) becomes 
 
 144 
 
 s-f- =26 or 5 2 —26s=— 144. .. .3=18. 
 
 5 
 
 (11.) Let x, xy, and xy 2 represent the numbers ; 
 
 Then «y=216 (1) 
 
 x 2 +x a y 4 =328 (2) 
 
 From (I) xy = G, or x =— 
 
 J/ 
 328 
 
 From ( 2 ) *~7+y 
 
 36 328 9 82 
 
 or 
 
 y' l+y* y 2 l+y* 
 
 9^4 — 82y 2 ^ — 9 Hence y=3. 
 
 (12.) Let x, Jxy, and y represent the numbers ; then 
 
 x+J^y+y=i3 (i) 
 
 (*+y)J*y=*o_ (2) 
 
 x+y=\3—Jxy (3) 
 
 *+y— ^ W 
 
 Jxy 
 
 _ 30 — 
 
 13 — Jxy=—= Hence . . Jxy=3 
 
 Jxy 
 
58 KEY TO ALGEBRA. 
 
 2xy 
 (13.) Let x, , and y represent the numbers ; then 
 
 x+y 
 
 *+</=18 (1) 
 
 — -=576 (2) 
 
 18 V ; 
 
 xy 
 
 — =24 xy=72 (3) 
 
 3 
 
 From (1) and (3) we find x and y. 
 
 (14.) Let x, xy, and xy 2 represent the numbers ; then 
 {xy 2 — xy) {xy — x) are the 1st differences, and 
 xy 2 — 2xy-\-x= 6 
 xy 2 -\- xy-\~x=42 
 
 Difference Sxy =36 .... . xy=\2. 
 
 2ry 
 
 (15.) Let x, , and y represent the numbers. If y 
 
 x+y 
 
 (2xy \ / 2xy \ 
 y ) ( x ) 
 
 4xy 
 
 are the 1st differences, and y \-x=2 the 2d diff. 
 
 x+y 
 xy=72 Put (x-{-y)=s; 
 
 4-72 
 
 Then s =2 
 
 s 
 
 s 2 —2s+ 1=289 
 
 5 —l = 17. . . .s=x+y=18. 
 
 (lY.) Let x 2 , xy, and y 2 represent the numbers ; then 
 x 2 -\-xy-\-y 2 =3\, and x 2 +y 2 =26 
 
 (18.) Let x, xy, xy 2 , xy 3 , xy 4 , and xy 6 represent the 
 numbers. Then, by the conditions, we have 
 
 x-\-xy-\-xy 2 -\-xy*-\-xy 4 -\- xy 5 =lS9=a • • (1) 
 And xy-t-xy*=54=:b (2) 
 
PROPORTION. 59 
 
 But equation (1) may be put into this fofm 
 
 a 
 Or x-\-xy 3 = 
 
 Multiply this last equation by ?/,and its first member will 
 be the same as the first member of equation (2) ; therefore 
 
 ay 
 
 i+y+3/ 2 
 
 ratio. 
 
 •■=& ; a quadratic from which we obtain y, the 
 
 (19.) Take the same notation as for (18) ; then we have 
 
 {x+xy)+(x+xy)y A = \S$— 36=153=«. . (1) 
 And \x+xy)y*=ZQ=b (2) 
 
 Divide (1) by (2), and we have 
 
 1+V 4 153 51 tt 
 
 £_= =_ Hence y=2. 
 
 y* 36 12 * 
 
 CHAPTER III. 
 PROPORTION. 
 
 (5.) Let x and y represent the numbers ; then 
 x — y : x-\-y : : 2 : 9 
 x-\-y : xy : : 18 : 77 
 
 From the first, 2x : 2y : : 11 : 7or,a?=yy. 
 
 ISy U^ 
 
 7 7 
 
 18 : 77 
 
 y : Uy* : : 1 : 77 y=7. 
 (6.) Let x and y represent the numbers. 
 
 x+4 : y+4 : : 3 : 4 .... (1) 
 #_4 : 2/_4 : : I : 4 .... (2) 
 
 From (2) we have 4a? — 16=y — 4, or, y=4x — 12. This 
 value of y put in (1) gives 
 
60 KEY TO ALGEBRA. 
 
 x-\-4 : 4a;— 8 : : 3 : 4 
 x-\-4 : x — 2 : : 3 : 1 
 
 x+4=3x— 6 x=5. 
 
 (7.) Let x and y represent the numbers ; 
 
 Then x-hy=lG 
 
 And xy : x*-\-y 2 : : 15 : 34 
 
 Double the first and third terms, then add and subtract 
 (Theorem 4), and 2xy : x 2 -\-y* : : 30 : 34 
 x*i-2xy-\-y 2 : x 2 — 2xy+y 2 : : 64 : 4 
 x-{-y : x — y : : 8 : 2 
 
 16 : x — y : : 4 : 1 or, a: — y=4. 
 
 (8.) Let #= the gallons of rum, 
 
 And . . . y= the gallons of brandy. 
 
 x — y : y : : 100 : x 
 x — y : x : : 4 : y 
 
 Product (x — y) 2 : xy : : 400 : xy 
 Dividing the second and fourth by xy, and 
 
 (x—y) 2 : 1 : : 400 : 1 
 
 x — y : 1 : : 20 : 1 or, a;— y=20. 
 (9.) Let x-\-y= the greater number, 
 And . . • x — y= the less. 
 
 Then x 9 —y 2 =320 (1) 
 
 {x+yf^+Sxty+Sxy'+y* 
 (x — y ) 3 = x s — 3x*y + 3 xy* — y 3 
 
 6# ? 3/+2?/ 3 =diff. of the cubes. 
 2y= difference. Cube of (2?/)= 8I/ 3 
 §x 2 y+2y* : 8y s : : 61 : 1 
 3X 2 -f- y 2 : 4y 2 : : 61 : 1 
 3x 2 +y 2 =244y 2 3a*=243y 2 
 
 ^= Sly 2 
 This value of x 2 put in equation (1), gives 
 
 80t/ 2 =32Q or y=2. 
 
PROPORTION. 61 
 
 (10.) Let a? and y represent the numbers ; 
 
 Then a?-f-t/=60 
 
 And . . . . xy : x 2 -\-y 2 : : 2 : 5 
 Double the first and third terms, aud 
 
 2xy : x 2 -\-y 2 : : 4 : 5 
 Add and subtract and 
 
 x 3 -{-2xy-\-y 2 : x 2 — 2xy-\-y 2 : : 9 : 1 
 
 60 : x—y : : 3 : 1 x—y=20 
 
 But a?+y=60 
 
 (11.) Let 3a? and 2a? represent the numbers; 
 
 3a?-}-6 : 2a?— 6 : : 3 : 1 Hence, x=8 
 
 (12.) Let 16a? and 9a? represent the numbers ; 
 
 Then . . 16a? : 24 : : 24 : 9a? Hence, a?=2 
 
 (13.) Let a? and y represent the numbers ; 
 Then x-\-y : x — y : : 4 : 1 Hence, 5y=3x 
 
 This reduced equation will be true if we take y=3 and 
 a?=5, or, if we take any multiples of 5 and 3, as 10 and 6, 
 15 and 9, 20 and 12, &c. Hence the answer in the book 
 is correct, but many other answers are equally correct, and 
 the problem is therefore indeterminate. 
 
 Again let a? and y represent the numbers i 
 
 And a? 2 -ft/ 3 : x : : 102 : 5 
 
 Or, a? 2 +2 9 ^ x 2 : a? : : 102 : 5 Hence a?= 15 
 
 (14.) Let a? and y represent the two numbers 
 Then, x-\-y=20, and a? : y : : 9 : 1 
 
62 KEY TO ALGEBRA. 
 
 APPLICATION OF THE BINOMIAL THEOREAL 
 
 (3.) To expand (a — £>)~',we take formula (3). Then 
 a=a, x= — 6, and m— — 1. 
 
 Hence, 
 a m +ma m -'b-\-mPj a m ~ 2 b 2 , &,c.=a~ x +a--6+a" 3 6 2 , &c. 
 " 1 b b 2 
 
 Or -+-7+-T' &c - 
 
 a a z cr 
 
 b 2 
 (4.) Take formula (1), and put x= — . Then 
 
 a 2 
 
 K J \ 2 2 22 3 / 
 
 / b 2 b* 36 8 \ 
 
 a ( H H &c. ) 
 
 \ 2a 2 2-4a 4 2-46a 6 / 
 
 (5.) Expand = -=-( H ) 
 
 («•+*•)* c(n-gy A c2) 
 
 The numerical coefficients of this series must be the 
 
 same as the last, because m is numerically the same ; the 
 
 sign of the second and every alternate term will be minus. 
 
 d/ x 2 3x 4 3-5x 6 3-5-7.T 8 \ 
 
 -( 1 H -f- Ac. ) 
 
 c\ 2c 2 2-4c 4 2-4-6c 6 2-4-6'8c 8 / 
 
 , X3 / « 2 \? |/ # 2 \f a 2 / a*\f 
 
 («•) (*)'0— ) =<<'—) =7 ( 1 ^) 
 
 3 a; x 2 
 Here m= 
 
 4 a a 2 
 x m — 1 x 2 
 
 Formula (2) 1+m — \- m- &c. = 
 
 w a % a 2 
 
 / 3^ 3 a* 4 3 5* 6 \ 
 
 ( 1 <fcc. ) This multiplied by 
 
 V 4 a 2 4-8 a 4 4-8- 12c^ / * , 
 
 a 2 /l/ 3^ 3x 4 5x 6 \ 
 
 the factor will give \/ -l a 2 &c. 1 
 
 j a a\ 2 2 2 5 a 2 2 7 c 4 / 
 
COMPOUND INTEREST. 63 
 
 (7.) (a+y) m +=a"+mar-'y-\-m'?i±a m - 2 y 2 J &c, and 
 il m== — 4, the series will be 
 
 a--^— 4a-^/-r-10a-y— 20a-y'-f- &c. 
 
 1 4?/ l(ty 3 20 w 3 
 Or -H ■ &c. 
 
 a 3 1 / 6 3 V' 
 
 (8.) = =( H } The cube root of this 
 
 V ' a 3 +6 3 IP \ a 3 / 
 
 a 3 
 
 is . . .(l-h— ) =1 1 &c. 
 
 V a / 3a 3 3*6a 6 3-6-90 9 
 
 6 3 26 6 146 9 
 Or =1 h &c. 
 
 3a 3 9a 6 81a 9 
 
 COMPOUND INTEREST. 
 
 (4.) The general equation is pJi n =a, and for this exam- 
 ple jo=5, c#=(1.05) a=9, and n is unknown. 
 Hence 5(l-05) n =9, or (i*05) n =l-8 
 
 log. 1.8 0.25527 
 
 Or • . . . n= = =12.04 years, nearly. 
 
 log. (1-05) 0.02119 J ' 
 
 (5.) 1000(l+r) 6 =1800 or (l+r) 6 =1.8 
 
 By logarithms ... 6 log.(l+r) = log. (1.8) 
 
 0*25527 
 
 Hence . . . -log. (l-fr) = =0.04254 
 
 6 
 
 By the tables we find l-{-r= 1*103, or, r=.10 T \, nearly. 
 
 (6.) In this example the general formula is 
 
 jo(1.04) 4 =350.9575. By logarithms we have 
 log. p+4 log. (1.04)= log. 350.9575 
 
 Or log. /H-0.06812=2. 54525 
 
 By reduction log. jo=2. 47713 
 
 Hence jo=300, Ana. 
 
64 KEY TO ALGEBRA. 
 
 The general equation applied to this problem is 
 
 3600(1.05)"=5000(1.04) 1 2 
 
 By reduction . . . . (1.05) 7, = |°(1.04) 12 
 
 n. log. (1.05) = log. fl+12 log. (1.04) 
 
 w(0.02119)=0.14267+0.19236 
 
 0.33503 
 
 n= =16 nearly. 
 
 0.02119 J 
 
 ANNUITIES. 
 
 (6.) That is, the rent is an annuity to continue forever ; 
 
 what sum of money will purchase it ? 
 
 mi p 3000 
 
 The general equation is P=-= =100000. 
 
 r .03 
 
 (7.) The general equation applied to this problem is 
 
 350[(1.04) 8 — 1] 
 JSt'= — -=8750[ (1.04) 8 — 1] 
 
 Log. J2'= log. 8750+ log. ( (1.04) 8 — 1) 
 But (1.04) 8 — 1=.3685 
 
 Log. .#'=3.94201— 1. +0.56644=3.50845 
 
 A' 
 But the answer to the question is P= — (Art. 155.) 
 
 That is by log. (log. P.) =3.50845—0. 13624=3.37221 
 Or P=2356.46 Jlns. 
 
 (8.) If no interest were required, the sum to be paid at 
 each annual payment would be t3 °° dollars ; but this must 
 be paid and compound interest on the same for 1, 2, 3, &c. 
 years, up to 7. Call —. 7 — =/?. 
 
 Let ^=(l+r) = 1.04. Then by (Art. 153), 
 
 The sum to be paid the first year must be pA 
 
 ' second year pJP 
 
 • third year pJP 
 
 • last year p^F 
 
EQUATIONS. 65 
 
 This is a geometrical series, and its sum (Art. 120) is 
 
 A — 1 r r ' 
 
 1200 1.04/ N 300.104 
 Or, 5= X ( (1.04) 7 — 1 )= X.3158 
 
 7 .04 V ' J 7 
 
 But, if this sum is to be paid by seven equal payments, 
 
 each payment must be 
 
 300X104 312 
 
 X.3158= X31.58=$201-l- dns. 
 
 49 49 
 
 p 250 
 (9.) Here P=-= =$35712 Jlns. 
 
 v ' r .07 7 
 
 SOLUTIONS OF EQUATIONS OF THE HIGHER DEGREES. 
 NEWTON'S METHOD OF APPROXIMATION. 
 
 (1.) Given x 3 -{-2x 2 — 23# ^70, to find one value of x. 
 
 By trial we find that one value of x is between 5 and 6, 
 nearer 5 than 6 ; therefore, let a=5 and y— the remaining 
 part of the root. Then x=a-\-y. 
 
 Expand, neglecting all the terms containing the powers 
 of y after the first, and we shall have 
 
 x z = a 3 -f3a 2 2/-f &c. 
 2x 2 =2a 2 -\-4ay+ &c. 
 — 23x=— 23«— 23y 
 
 By addition, 
 
 ar4-2;c 2 — 23#=«H-2« 2 — 23a-f-(3a 2 +40— 23)y=70 
 
 In this last equation we observe that a has the same 
 powers and coefficients as x, and the coefficients to y may 
 be found by the following 
 
 Rule. Multiply each coefficient of x by its exponent, 
 diminish each exponent by unity, and change x to a. 
 
66 KEY TO ALGEBRA. 
 
 70-|-23a— 2a 2 — a 3 
 
 Now y — — Giving a its value, 5, we 
 
 3 Sa 2 -\-4a— 23 6 
 
 have y=s||a=.l4- Now make a=5.1, and substitute 
 
 again in the preceding formula, we have a new value of y. 
 
 2.629 
 
 Thus y= =.03 Now make a^5. 13, and substitute 
 
 u 75.43 
 
 again, and our new value of y will be .004578-}- Hence 
 
 a+y or #=5.134578-{- 
 
 (2.) Given x 4 — 3a?*-f 75^=10000, to find one value of x. 
 
 By trial we find x must be near 10. Hence put a=10 
 
 and x=a-\-y. Then by the preceding rule 
 
 10000— 75a-{-3a 2 — a 4 —450 
 
 v = = =—.11 
 
 9 4ft 3 — 6a-r-75 4015 
 
 Now make «=10 — .11 = 9.89. If we have the patience 
 to substitute this value for a in the equation, we shall have 
 a new value to */, true to 6 or 7 places of decimals, and of 
 course a value to x to the same degree of exactness. 
 
 (3.) Given 3x 4 — 35x 3 — 11a.- 2 — 14#-f-30=0 to find one 
 value of x. 
 
 By trial we find that x must be near 12. Let a=12, 
 and x=a-\-y. Then by the rule 
 
 — 30+14H-lla 2 +35a 3 — 3a 4 —6 
 
 y- 
 
 -00112 
 
 12a 3 — 105a 2 — 22a— 14 5338 
 
 Hence z=12— .00112=11.99888. 
 
 (4.) Given 5x? — 3x 2 — 2.r=1560, to find x. 
 We find by trial that one value of x is more than 7. Put 
 x—a-\-y, and a=7. Then by the rule 
 
 1560+2a4-3a 2 — 5a 3 6 
 
 y = ! = =.00867+ 
 
 * 15a 2 — 6a— 2 689 
 
 Hence #=7.00867+ 
 
EQUATIONS. 67 
 
 CHAPTER in. 
 YOUNG'S METHOD OF RESOLVING THE HIGHER EQUATIONS. 
 
 This is really Horner's method, but extracted from Young. 
 
 (6.) Given ^+7^=1194, to find the values of x. 
 We perceive by inspection that one value of x must be 
 more than 31 ; therefore put r=31. 
 
 «-f-r= . . . . 
 
 • 38 
 
 1194 (31.2311-f &c. 
 
 r+s 
 
 31.2 
 
 1178 
 
 a-{-2r-\-a 
 
 69.2 
 
 1600 
 
 s-\-t 
 
 23 
 69.43 
 
 1384 
 
 
 21600 
 
 
 31 
 69.461 
 
 20829 
 
 
 77100 
 
 
 &c. 
 
 69461 
 
 &c. 
 As the sum of the two roots is equal to — 7, the other 
 root must be — 38.23 11 -f- 
 
 (7.) Not necessary to have place in a key. 
 
 (8.) Given x 2 — 21^=214591760730, to find one value 
 of x. 
 
 In this example the pupil might be at a loss as to the 
 most expeditious manner of finding r by trial. 
 
 Conceive — 21a? not to exist ; then the value of a? will be 
 the square root of the absolute term ; but this term has six 
 periods of two figures each, and the superior period is 21. 
 The greatest square in this is 16, root 4 ; hence r must be 
 at least 400000, six places ; try this number. 
 
68 
 
 KEY TO ALGEBRA. 
 
 — a-\-r .... 
 r-f-a • • • 
 
 =399979 
 . 460000 
 
 214591760730 
 1599916 
 
 ( 400000=r 
 60000=* 
 
 ~a+2r-|-s. . . 
 
 • 859979 
 . . 63000 
 
 5460016 
 5159874 
 
 3000=* 
 200 =u 
 
 a+2r+2s+t 
 
 . 922979 
 3200 
 
 3001420 
 2768937 
 
 50=v 
 l=tt> 
 
 
 926179 
 250 
 
 2324837 
 1852358 
 
 
 « 
 
 926429 
 51 
 
 4724793 
 4632145 
 
 #=463251. 
 
 926480 926480 
 
 926480 
 As the algebraic sum of the two roots must make 21, 
 (Art. 156 in the work,) therefore the other root must be 
 —463230. 
 
 (9.) Given 7x 2 — 3#— 375, to find one value of x. 
 Or # 2 — ?#= 3 I 5 . Putx=\y. (Art. 166.) 
 
 Then 
 
 y 2 Sy 375 
 
 Or I/ 2 — 3y=375X 7=2625. 
 
 49 49 7 
 
 In this equation we perceive that y must be more than 
 
 the square root of 2625, that is, more than 50. Hence put 
 
 r=50. 
 
 rs t 
 
 — a-\- r . . . . 
 
 .47 
 
 2625 ( 52 
 
 r-f-s 
 
 52 
 
 235 
 
 — a+2r-fs. . 
 
 . 99 
 
 275 
 
 s+t 
 
 2.7 
 
 198 
 
 
 1017 
 
 7700 u 
 
 
 75 
 
 7119 H 
 
 
 10245 
 
 58100 
 51225 
 
 6875 
 
 52.756-f- „ , 
 Hence x= =7.+ 
 
EQUATIONS. 69 
 
 (10.) Given 2x* — lla?= — 7|, to find one value of x. 
 Orx 2 - 'Jar==— y Put a?=£y. 
 Then . . . ly 2 — y y=— y or. . .# 2 — lli/= — 15. 
 In this equation one value of y is between 9 and 10 ; 
 therefore put r—9. 
 
 — a+ r —2 —15 ( 9.405124838+ 
 
 r-\-s 9.4 —18 
 
 — a-f-2r-f-s 7.4 300 
 
 *+* 4 296 
 
 7.805 40000 
 
 51 39025 
 
 78101 97500 
 
 12 78101 
 
 781022 1939900 
 
 24 1562044 
 
 7810244 377856 
 
 The division is not carried out for the last three figures. 
 
 Hence one value of y is 9.405124838-J-, and as the two 
 
 values must make 11, (Art. 166), the other value must be 
 
 1.594875161. But x=iy, therefore 
 
 #=.797437580+ or, #=4.7025624194- Am. 
 (11.) Given |# 2 +|a?= T 7 T , to find one value of x. 
 Or # 2 -f-ia:=!f or, ar 2 +.8#=0.848484848+ 
 
 Here it is obvious that x cannot be 1 ; by trial we find 
 
 it must be near .6 ; therefore r=6. 
 
 a+r 1.4 0.8484848484 &c. ( 0.604233+ 
 
 r+s 60 84 
 
 2004 8484 
 42 8016 
 
 20082 46884 
 23 40164 
 
 200843 672084 
 
70 KEY TO ALGEBRA. 
 
 We omit several examples as they present no difficulty. 
 
 CUBIC EQUATIONS. 
 
 (3.) Given x*-{-2x a — 23#=70, to find one value of x. 
 
 By trial we find x must be a little over 5; therefore 
 r=5, .#=2, ,#=—23, N=70. 
 
 B —23 
 
 r(r+A) 35") 
 
 1st Divisor 12 > 70 ( 5.134 
 
 r 2 . . 25J 60 
 
 B' 72 10000 
 
 *(a+3r+^) 17n 7371 
 
 2d Divisor 7371 ^ 2629000 
 
 s 2 1J 2276697 
 
 B" 7543 352303000 
 
 *(3Q+t)t 4599 305649104 
 
 3d Divisor 758899 46653896 
 
 f 9 
 
 B" 763507 
 
 61576 
 
 4th Divisor 76412276 
 
 16 
 
 Common division will give three or four more figures to 
 perfect accuracy. 
 
 (4.) Given x s — 17^+42^=185, to find one value of x. 
 Here .#=—17, 5=42, JV=185, and we find by trial 
 that x must be between 15 and 16 ; therefore r=15. 
 
 * Q represents the root as far as previously determined. 
 
CUBIC EQUATIONS. 71 
 
 B 42 
 
 r(r+Ji) —30^ 
 
 I r st 
 
 1st Divisor ... 12 > 185 ( 15.02 
 
 r 2 . . . 225J 180 
 
 B' 207 5000000 
 
 *(s-f3r-M) .... 0") 4154008 
 
 2d Divisor 207 f 2077 ) 845992 ( 407 
 
 s 2 OJ 8298 
 
 B" 207 16192 
 
 t(3Q-{-t) 7004 14539 
 
 3d Divisor 2077004 1653.0 
 
 Hence x=15.02407-|- 
 
 (5.) Given x 3 -{-x i -^500 i to find one value of x. 
 
 Here Aml % B=0, r=7. 
 B 
 
 r(r-\-Jl) 56 
 
 1st Divisor .... 56 500 ( 7.61 
 
 r 2 49 392 
 
 B' 161 108 
 
 (3r+*-M)s . ■ • • 1356 104736 
 
 2d Divisor .... 17456 ■ 3264 
 
 * a 36 1887181 
 
 18848 )1376819 
 
 2381 
 Continue by common division. 
 
 3d Divisor .... 1887181 
 1 
 
 1889563 
 
 (6.) Given # 3 +10x 2 -j-5tf=2600, to find one value of x. 
 
72 KEY TO ALGEBRA. 
 
 Here ^=10, B=5, r=ll. 
 
 B 5 2600 ( 11.006 
 
 r(r+^). ... 23 n 2596 
 
 1st Divisor . . ■ 236 V 4 
 
 r 2 121J 3529188216 
 
 B' 588 470811784 
 
 (3/?+tt)w. • • • 198036 
 
 . , t. . . ~ nntnnn ^ n Continue by common division. 
 
 4th Divisor . . . 588198036 J 
 
 The four miscellaneous examples on page 262, the last 
 
 page of the school edition. 
 
 (1.) Let x-\-y represent the greatest extreme, and x — y 
 
 x 2 — y 2 
 the least extreme ; then — — = the harmonical mean. 
 x 
 By the conditions of the problem we have 
 
 , x 2 — v 2 (x 2 — V 2 ) 
 
 2x+ -=26. • . .(1) and * —=576 . . . (2) 
 
 x x 
 
 By reducing (1), and taking the square root of equation 
 
 (2), we have 2x 2 +x 2 — y 2 =2Qx (3) 
 
 x 2 —y 2 =24 ) Jx~ (4) 
 
 By subtraction . . . 2x 2 = 2Qx — 24 Jx 
 
 If we put jjx=v, and reduce the resulting equation, we 
 shall have »*— 13t>=— 12 
 
 By (Art. 163) we find v=3. -Hence a?=9, &c. 
 
 (2.) Let x-\-y= the greater number, and x — y= the less. 
 Then by the problem we have x=5 
 
 And 8^1/4- 8.n/ 3 =1040 
 
 Or y*+25y=26 Hence y=l 
 
 (3.) Let x, y, and z represent the numbers ; then by 
 
 the problem x-\-y-\-z=2S (1) 
 
 x+yz=5l (2) 
 
 y+xz=S7 (3) 
 
EQUATIONS. 73 
 
 By adding (2) and (3) we have (x+y)-\-(x+y)z=l38 
 But by equation (1), x-\-y =28 — z ; therefore 
 
 28— s+28z— z 2 =138 or z=5 
 
 (4.) Let x, y, and z represent the numbers ; then 
 
 ^+2/ 2 +z s = 195 (1) 
 
 x 9 +y 9 +z 3 =l799 (2) 
 
 xyz= 385 (3) 
 
 The form of these equations will not be changed by ta- 
 king x for z or x for y. A full and formal solution of these 
 equations would become tedious, and to avoid this, we may 
 venture a solution by inspection. As 195 and 385 both 
 have 5 in the unit's place, it is more than probable that the 
 value of one of the symbols is 5. Therefore assume z=5. 
 This gives ic 2 -f-i/ 2 =170, and #*/=77, from which we find 
 a?=7 or 11, and y=ll or 7, and as 5, 7, and 11 will verify 
 equation (2), this is a true solution. 
 The following examples are in the University Edition only. 
 
 (6.) Find one value of x from bx*— 6x 2 -\-3x=— 85. 
 
 As the result is negative, we will change the second and 
 every alternate sign of the equation, (Art. 178), and find a 
 value of x from the equation 5r J -f-6x 2 -r-3a;=85 
 
 Use the formula of (Art. 194.) c=5, J3=6 y J5=3, and 
 by trial we find r=2. r s 
 
 B 3 85 (2.1 
 
 (cr-f-^)r .... 32V 70 
 
 1st divisor . . 35 f 15 
 
 cr* 20J 9.065 
 
 87 5 935 
 
 * ' * * Continuing this we shall find the 
 
 2d Divisor . • • 90.65 value of x to be 2.16399-f , and its 
 
 cs _^ sign changed will be the value of x 
 
 94.35 in the original equation. 
 
74 KEY TO ALGEBRA. 
 
 (7.) Find x from the equation \2x*-{-x 2 — 5#=330 
 
 Here c=12, J2=l, J5=— 5. r=3. 
 
 £ —5 
 
 (cr+A)r lin 
 
 I r »< 
 
 1st Divisor .... 106 ^ 330 ( 3.036 
 
 cr* 108J 318 
 
 B' 325 "~12 
 
 {3cE-\-ct)t. . . • 11208 9783624 
 
 3d Divisor. . . . 3261208 2216376 
 
 ct 2 108 
 
 Continue thus. 
 
 3272524 
 
 In the same manner perform (8) and (9.) 
 
 ( Art. 195. ) Page 323. 
 
 (3.) Extract the cube root of 1-352-605-460-594-688. 
 
 For the sake of brevity, take r=ll, in place of 1. 
 1st Divisor. • -121 r st 
 
 JS'^r 2 . • 363 1-352-605-460-594-688 (110592 
 
 (3B-\-t)t. • 16525 1 331 
 
 2d Divisor 3646525 21 605 460 
 
 25 18 232 625 
 
 3663075 3 372 835 594 
 
 (3R+u)u • • 298431 3 299 453 379 
 
 3d Divisor 366605931 73 382 215 688 
 
 81 73 382 215 688 
 
 366904443 
 (3i?-f-v)u • ■ 663544 
 
 36691108844 
 
 (4.) By a table of cubes which run to 8000, we perceive 
 at once that r in this example is 17. 
 
EQUATIONS 
 
 1st Divisor 289 
 
 Sr 2 =B' 867 
 
 (3#-fs)s 2575 
 
 2d Divisor 89275 
 
 25 
 
 (3B+t)t 
 
 9J875 
 10504 
 
 rs t 
 5382674 ( 175.2 
 4913 
 
 469674 
 446375 
 
 23299000 
 18396008 
 
 75 
 
 3d Divisor 9198004 4902992 
 
 4 Complete another divisor, 
 9208512 t * ien contmue as m simple di- 
 
 It is not important to show a solution to the remaining 
 examples under this article. 
 
 Page 324. 
 
 (»•) 
 
 Change 
 
 all the signs, 
 
 then 
 
 
 a*+a*+3?— £=500 
 
 
 r=4 
 
 
 r\ i 
 
 
 1 
 
 —1 
 
 500 (4 
 
 
 4 
 
 
 20 
 
 84 
 
 332 
 
 
 5 
 
 
 2l 
 
 83 
 
 1~68 
 
 c 
 
 4 
 
 
 36 
 
 228 
 
 
 
 9 
 
 
 57 
 
 311 
 
 
 go 
 
 4 
 
 
 52 
 
 
 
 b 
 
 Ti 
 
 
 To9 
 
 
 
 • 
 
 4 
 
 
 
 
 _ 8 
 
 1 17 
 
 109 
 
 311 
 
 168 (.4 <fcc. 
 
 (2.) Resolved in the same manner as (1.) 
 
 (3.) The signs of all the terms must be changed, then 
 
76 
 
 '1 —9 
 
 .1 
 
 KEY TO 
 
 —11 
 —.89 
 
 ALGEBRA. 
 
 —20 
 —1.189 
 
 — 4(.l 
 —2.1189 
 
 03 
 
 cJ 
 
 S-l 
 
 -•J 
 
 | 
 
 fa 
 
 —8.9 
 
 1 
 
 —11.89 
 
 — 88 
 
 —21.189 
 — 1277 
 
 —1.8811 
 
 —8.8 
 
 1 
 
 —12.77 
 
 — 87 
 
 —22.466 
 
 
 
 —8.7 
 1 
 
 —13.64 
 
 
 1 — 
 
 -8.6 - 
 
 -13.64 
 
 —22.466 — 1 
 
 .8811 (.07 &c. 
 
 (4.) x 5 = 
 
 5000. 
 
 Here all the coefficients 
 
 are zero ex- 
 
 cept the first, and r- 
 
 = 5. 
 
 
 r 
 
 1 
 
 
 
 
 
 
 
 =5000 ( 5 
 
 5 
 
 25 
 
 125 
 
 625 
 
 3125 
 
 5 
 
 25 
 
 125 
 
 625 
 
 1875 
 
 5 
 
 50 
 
 375 
 
 2500 
 
 
 10 
 
 75 
 
 500 
 
 3125 
 
 
 5 
 
 75 
 
 750 
 
 
 
 15 
 
 150 
 
 1250 
 
 
 
 5 
 
 100 
 
 
 
 
 20 
 
 250 
 
 
 
 
 5 
 
 
 
 
 
 25 
 
 
 
 
 s 
 
 1 25 
 
 250 
 
 1250 
 
 3125 
 
 = 1875 (.4&c 
 
 (5.) X*: 
 
 64.Z 2 
 
 = or 
 
 z 4 +2r4-l 
 
 ar s +2» 8 +s= 
 
 =64 
 
 r 1 
 
 1 
 
 2 
 
 
 
 1 
 
 =64(2 
 
 2 
 
 4 
 
 12 
 
 24 
 
 50 
 
 2 
 
 6 
 
 75 
 
 25 
 
 14 
 
 2 
 
 8 
 
 28 
 
 80 
 
 
 4 
 
 14 
 
 40 
 
 105 
 
 
 Continue as in last example. 
 
EQUATIONS. 77 
 
 In the editions of Algebra which we're published prior 
 to 1857, some few important problems were passed over 
 in the Key without a solution. 
 
 They were purposely omitted, as an experiment, to dis- 
 cover whether there would be any special call upon the 
 author for solutions; and the experiment is now over. The 
 call for their solution has been very frequent and earnest, 
 from all parts of the country: and this demonstrates the 
 importance of a Key. 
 
 We now insert them, and some others which have been 
 added to the enlarged edition, first published in 1858. 
 
 A more general Key to mathematical science is pub- 
 lished in our Operations. In that work, the advanced 
 student and the amateur mathematician will find many 
 curious and useful problems, not merely in Algebra, but 
 in Geometry and the higher sciences. 
 
 (Art. 107.) Example 18. 
 
 a 
 
 Given .... x — l=24-__ 
 Jx 
 
 Place . Jx=y. Then . y 2 — 1=2+? 
 
 y(y»— l)=*(y+l) 
 
 Divide by (y+1) then y(y— 1)=2 or y 2 — y=2. 
 Whence, y=2 or 1. 
 
 (Art. 114.) Example 17. 
 
 Let x and y represent the number. Then, by the given 
 conditions, we have: 
 
 x+y=xy ( 1 ) 
 
 and xy=x 2 — y 2 (2) 
 
 Assume x=vy. Then vy-\-y—vy 2 , or y=l-| — (3) 
 and vy 2 —v 2 y 2 — y 2 
 
78 KEY TO ALGEBRA. 
 
 Whence, . y=ldn _=_J_ 
 
 But, . . . •« f -^J • L^ =H 3±^) 
 
 y l + ^ 5 1-V5 -~i-4 ^ 1 +^;- 
 
 Example 31. (Page 189.) 
 Let x and y represent the number. Then, by the given 
 conditions: 
 
 xy—x 2 — y a (1) 
 
 x*+y a =x*— y 3 (2) 
 Assume x=vy. And this value of x, placed in (I) and 
 (2), produce 
 
 vy 2 =v 2 y 2 — y 2 (3) 
 
 and v 2 y 2 -\-y 2 =v 3 y 3 — y 3 (4) 
 Dividing (3) and (4) each by y 2 , and we obtain 
 v=v 2 — 1 (5) 
 and v 2 +\ = (v 3 —l)y (6) 
 Observe that (5) is a quadratic, and its solution gives 
 
 2*>=l± > /5 (7) 
 Again: Add 2 to each member of (5), and we have 
 
 v-\-2=v 2 -\-l. 
 Also, multiply (5) by v, and 
 
 v 2 =v 3 — v y or # 2 ~j-z;=fl 3 . 
 
 But v 2 =v-\-\\ 
 
 Whence, . 2v-\-\—v 3 , and 2v=v 3 — 1. 
 Now, Equation (6) becomes 
 2vy=v-\-2 y 
 or 4vy=2v-^-4. 
 Substituting the value of 4v and 2v, as found in (7), we 
 have 
 
 2(\±j5)y=5±j5 
 
 2y= 5/y±T =±V5 ******* Ans ' 
 
 Lastly, x^vy=^ (\d-j5)^ s /5=\ (J5±5) Arts. 
 
EQUATIONS. 79 
 
 Another solution may be found in our Mathematical 
 Operations, pages 115 and 116. 
 
 (Art. 204.) Example 1. 
 
 x*—5x 3 +5x 2 —l=0. 
 Here the sum of the coefficients is zero, therefore one 
 root is -f-l> and the equation is divisible by x — 1. 
 Dividing by (x — 1), the quotient is 
 
 x *-\-x 3 —4z 2 -\-x-{-l=:0. 
 Here, again, the sum of the coefficients is zero, show- 
 ing that another root is +1; and this, again, is divisible 
 by (x — 1). The next quotient is 
 
 x 3+2x 2 — 2x— 1=0. 
 Here, again, the sum of the coefficients is zero, show- 
 ing a third root equal to -(-1; and hence, we divide again 
 bj x — 1. The quotient is 
 
 s*_|-3;z+l=0. Whence, x= — \ (3±J5). 
 N". B. — We might have taken x*-\-x 3 — 4# 2 -|-£-|-l=0, 
 and solved it by the rule (under Art. 203). 
 
 2. x*+5x 3 ~\-2x 2 ±5x-{-l=0. 
 By the rule (under Art. 203) we have 
 
 x< -j_5z 3 -f( 2+V )x a -j-5^-1-1 = yz 2 ; 
 Extracting square root, and 
 
 x 2 =— 1, or x=±J—l; 
 Or, x 2 +5x=— 1 *=H— 5±J2l~ 
 
 3. x*— 4x 3 -\-4x— 1=0. 
 (a.*— 1)— 4.r(z 2 — 1)=0. 
 
 Dividing by (x 2 — 1), and x 2 -\-\ — 4^=0, or x 2 — 1=0, 
 Or . . #=d=l; or x 2 — 4.r-)-4=3. #=2± A /3. 
 
 4. x*—^x 3 -{-2x 2 — \x -(-1=0. 
 
 By the rule (Art. 203), add \\x 2 to each member— 
 then *«— lx 3 +\\x 2 — fz-f^ffz 2 
 
 Square root, > x 2 — |-ar+l ==fc|ar, 
 
80 KEY TO ALGEBRA. 
 
 Whence, . . # 2 -|-l=0, or x 2 — fa--|-l=0 
 
 x=±z,J — I, or x=2 or £. 
 
 5. 5x*-\-Zx 3 +9x 2 +8a--f-5=0. 
 
 Add .... 2a-- 2 =-|-U>a; 2 
 
 and # 4 +fa: 3 +2a: 2 +fav}-l = ?L 
 
 Add ISfl =1^1 
 
 25 25_ 
 
 Square root, x 2 -\-$x-\-l=±xj2\ a quadratic. 
 
 Place .... 4 ~-^ 21 =2fl 4— 721=10a, 
 
 5 
 
 Then .... s 2 +2az+a 2 =a 2 — 1 
 
 a?= — a-±ija 2 — 1 
 
 Because a 2 = — _5l — / a 2 j is imaginary, and, 
 
 therefore, the values of x are imaginary. 
 
 6. 4a* 6 — 24a? 5 +67a: 4 — 73a- 3 +57a- 2 — 24a-+4=0. 
 Divide by x 3 ; then 
 
 4a: 3 — 24ar 2 +57a;— 73+^— ^4--i-==0 (1) 
 
 XXX 
 
 1 ^7 
 
 Assume a--|--=y, Then 57a;+ — =57y, 
 
 a; a; 
 
 a: a +2+-L=y 2 — 24z 2 — — =48— 24y s 
 
 a; 2 a; 2 
 
 Also, a- 3 +-U-3 (x-\S\^y\ or a-^*^ 3 — 3y, 
 # 3 \ a?/ * 
 
 4 
 
 and 4# 3 4- — =4y 3 — 12y. 
 x 3 
 These values placed in ( 1 ) 
 And 4y 3 — 12y+48— 24y 2 +57y— 73=0 
 
EQUATIONS. 81 
 
 Or 4y 3 — 24y 2 -f-45y— 25=0, (2) 
 
 Here the sum of the coefficients is 'zero. Therefore, 
 one value of y is 1. 
 
 Then &4-!=l. Whence x= l±z ^~^- 
 
 1 x 2 
 
 Dividing (2) by y — 1, and we obtain 
 4y 2 — 20y+25=0 
 Square root 2y — 5=0, or y=f. 
 
 Whence #-!-_=§, and #=2 or £. 
 x 
 
 7. 4j? 4 +3« 3 — 8a; 2 — 3ar+4=0. (1) 
 Here the sum of the coefficients is zero. Therefore, 
 
 #=1 for one root. Again, if we change the second and 
 every alternate sign, we shall have 
 
 and the sum of the coefficients is still zero; therefore, 
 x=l in this equation, which corresponds to — 1 in the 
 given equation. 
 
 Therefore, (1) has two roots, x=\ and x= — 1: hence, 
 that equation is divisible x 2 — 1. 
 
 The quotient is, 4# 2 -|-3a; — 4=0, 
 
 Whence x=~ 3±z 'J 13 
 
 8 
 
 8. x*+24x 3 — 2z 2 — 24z-fl=0. 
 
 This, like the preceding, and for the same reason, is 
 divisible by x 2 — 1. 
 
 The quotient is, # 2 -f24.r — 1=0, 
 Whence *==■— IZdzjUB. 
 
 9. x *—2x*—7x 2 —Sx+\6=0. (1) 
 Here the sum of the coefficients is zero. Hence, a?=l 
 
 for one root, and the equation can be depressed to 
 
 x 3 — x 2 — 8z— 16=0. (2) 
 
 Assume x=nP. Then 
 
 n s p*—ri> p3_ 8w p_! 6==0 
 
 Letrc=4; then 64P 3 — 16P 2 — 32P— 16=0 
 Divide by 16, 4P 3 — P 2 — 2P— 1=0. 
 
82 KEY TO ALGEBRA. 
 
 Here the sum of the coefficients is zero. Therefore, 
 P=l. But x—?iP, and n=4, P=\. Whence, rr=4, for 
 another root of the equation. 
 
 Now (2), divided by x — 4, produces 
 ar 2 -j-3:r-J-4=:0, 
 
 "Whence x=— — -V Imaginary. 
 
 Another Solution: 
 
 The attempt to extract square root results as follows: 
 #4_2z 3 — lx 2 — 8z-j-16 (x 2 — x 
 x* 
 
 ^2x^~—7x 2 
 2x 2 —x — 2x 3 -\- x 2 
 
 — 8x 2 — Sx 
 If the remainder were -\-8x 2 in place of — 8# a , this ex- 
 pression would be a square. 
 
 It will be 8x 2 , if we add -{-\6x 2 to each member. Then 
 we shall have 
 
 x x — 2x 3 -\-x 2 -\-8x 2 — 8x+16=\6x 2 , 
 Or . . . (x 2 — x) 2 +8(x 2 — x)+\6 = 16x 2 , 
 
 Square root x 2 — x-\-4=zh4x. 
 
 10. x*+2x 3 —3x 2 —4z+4--=0. (1) 
 Here one value of # is 1. Dividing by x — 1, we obtain 
 
 x 3 +3x 2 — 4=0. 
 Here, again, x=l; and another division produces 
 
 Square root . . (#+2) (z+2)=0, x=— 2 z=— 2. 
 
 11. x*— 2x 3 — 25.c 2 +26.r-f-120=0. (1) 
 Assume x=nP. Then 
 
 n i P<—2n 3 P 3 — 25w 2 P 2 -f2GttP+ 120=0 
 p4 _2P3_25 yj2+ 26 p+ i20 ===()! 
 n n 2 n 3 n l 
 
 Letw=3. Then P^—fpz — ^pz+^P+^^O, 
 Or 27P 4 — 18/ J3 — 75P 2 +26P+40=0. 
 
 Here the sum of the coefficients is zero; therefore, P=\. 
 But n=o, whence x=3. 
 
EQUATIONS. 83 
 
 If in (3) we assume n—5, the coefficients will again be 
 zero, showing that another root is 5. 
 
 Now, equation (1) divided by x — 3 or x — 5, or thei r 
 product, will produce a quadratic whicb^fl^gi\]e r towp\ 
 other roots, #= — 2 and x= — 4. Jy^^ of the ^ 
 
 12. r 4 — 2x 3 +2x 2 — x^ofm N I V E EjCSDX T Y 
 
 (x 3 — \)x=2x 2 (x — 1V^ - ^ 
 
 As x is a common factor, x=Q; an 3^ft^Jl 7^(^60^ ■»**• 
 common factor, therefore' x=\. Dividing 
 tors, we obtain x 2 -]~x-\-l = 2x, 
 
 Or x *—x=—\. Whence x^^^" 3 
 
 2 
 
 1 3. x 4 —4x 3 -\-8x 2 — 32a-=0. ( 1 ) 
 
 Or (x— 4)x 3 +{x— 4)8#=0, Whence z=0, 
 
 Or . . x=4, or # 2 -|-#=0, or x=dtzj — 8. 
 
 14. x 3 +5x 2 +3a?— 9=0. (1) 
 Here the sum of the coefficients is zero; therefore, 
 
 x=l. And by division we obtain 
 
 £2_|_6:r+l=0, Whence #=—3^78. 
 
 15. o;3_|_ 6 , r 3_ 7x _ 60=0. (1) 
 Assume x=nP. Then 
 
 n 3 P 3 +6n a P 2 —7nP—60=0. 
 
 Suppose n=S. Then 
 
 3)27P 3 +54P 2 — 21P— 60=0 
 9P 3 -{-\QP 2 — IP— 20=0. 
 
 Here the sum of the coefficients is zero; therefore, 
 x=S, for one of the roots o^pthe equation. The other 
 roots are — 4 and — 5. 
 
 16. ^ 3 -[-8^ 2 + 17^4-10=0. (1) 
 Change the 2d and each alternate sign, then 
 
 x 3—8x 2 _f_17#— 10=0. 
 Here the sum of the coefficients is zero. Hence, x=\ 
 for the last equation; or, x= — 1 for the given equation. 
 Now divide the given equation by *-|-l, and the quotient 
 will be x 2 4-7x-\-\0=0, x=—2 or —5. 
 
84 KEY TO ALGEBRA. 
 
 17. x 9 —- 29x 2 
 
 -f 198z— 360=0. 
 
 Place x=?iP. Then 
 
 
 p 3 4,V [ . 
 
 198 P 360_ Q 
 
 JT — | 
 
 n 
 
 n 2 n* 
 
 Assume w=3. Then 
 
 
 (') 
 
 p* _?£p 2 J_22P —12=0 
 
 3 3 
 
 3/?3_29 J p2^_ 66 p_ 40=0 
 
 Here the sum of the coefficients is 0. Therefore, x=3; 
 and, dividing the equation by (x — 3), we obtain 
 x 2 — 26*+ 120=0, 
 
 x=6 or 20. 
 
 18. 4x 3 — 112z 2 + 109ar— 27=0. (1) 
 
 Here the absolute term — 27 is not numerically large 
 enough to make the sum of the coefficients zero; there- 
 fore, we take an operation which will increase it: 
 p 
 Place x= — Then 
 
 2 
 
 4P 3 112P 2 , 109P 
 
 •27=0 
 
 8 4 ' 2 
 
 Multiply by 2, and P 3 — 56P 2 +109P— 54=0. (2) 
 
 Here the sum of the coefficients is zero. Therefore, 
 P=\. Whence x=\. 
 
 Dividing (2) by P — 1, we obtain 
 
 P 2 —55P4-54==0. 
 Here, again, P—\; and, of course, another root of the 
 original equation is \, and it has two equal roots. 
 Dividing (1) by x — \, a#d we obtain 
 4x 2 — 110^+54=0, 
 Whence .... #=27 
 
ON INDETERMINATE EQUATIONS. 
 
 For the complete solution of a problem, we must have as 
 many independent equations as unknown quantities to be 
 determined. 
 
 When this is not the case, the problem is indeterminate. 
 For example, x-\-y—20. x may be one or two or three, or 
 5, or any other number, whole or fractional, under 20, and 
 y will take the remaining part of 20, and the equation is 
 indeterminate in the strictest sense of the term. 
 
 If, however, we restrict the values of x and y, to whole or 
 integer numbers in the equation rc-f-?/=20, x cannot have 
 more than 20 different values, when, without this restriction 
 x might take an infinite number of values, and still preserve 
 the equation x-\-i/=20. 
 
 In some cases, the number of answers to an equation 
 may be infinite, and the particular values restricted to inte- 
 gers. The following is a general case of the kind 
 
 ax — by—c. This givi s x= —£— : , where y may be any 
 
 a 
 
 value whatever, that will give ' , a whole number, but 
 
 Cb 
 
 numberless such values of y can be found, as c-\-by is a 
 magnitude that can rise higher and higher, without limit, 
 according to the assumed value of y. 
 
86 KEY TO ALGEBRA. 
 
 But take y what we will, and the equation still exists, 
 and therefore, the member of ansivers for x is unlimited or 
 infinite. In such equations, if the least values of x and y 
 are required, we have definite problems. 
 
 In equations like the following, ax-\-by—c, the number of 
 answers in integer numbers, may be very limited, may be 
 only one, or may be impossible. The equation gives 
 
 a 
 Now, if c is very large, and b and a small, y may take 
 many different values before c — by is so small that we ean- 
 not divide it by a, and obtain an integer quotient. 
 
 When c is not large in reference to b and &, we may ob- 
 tain only one value of y and #, and if by making y=l, we 
 
 c 
 
 find — -, a proper fraction, the problem is impossible. 
 a 
 
 3x-\-5y=\3. a ?= ' Q ° y . If we take y—l. x=-^, 
 
 not an integer, therefore, y must not be taken equal to one. 
 Take y=2, then a?=l, both integer values, and the only 
 integer values that will answer the conditions of the equa- 
 tion. 3#-)-5^=6, is an equation in which it is impossible 
 to give integer values to both x and y, because 3— 1-5 is more 
 
 c—b . , . 
 
 than 6, or a-\-b greater than c or is a proper fraction. 
 
 The equation az-\-b?/=c. is always possible in integers, 
 when c is greater than (ab — a — b). and a and b prime to 
 each other. The equation is sometimes possible, when c is 
 not greater than {ab — a — h.) 
 
 The equation lx-\-l3i/==7\, is impossible in integers, for 
 both x and y. But the equation lx-\-\3y=27. is possible, 
 a? =2 and y=l. Here a and b are the same in both 
 equations. , In the first equation, e=71 ; c is 
 
 not large enough to make the equation possible, for c, 71, ia 
 
 y 
 
INDETERMINATE' EQUATIONS. 87 
 
 not greater than (7*13 — 20), but if c was any number greater 
 than 71, the equation would be possible -in integers, and it 
 is possible, with some numbers less than 71. 
 
 If a and b are not prime to each other in the equation 
 ax-^-by=c, c must be divisible by the same number that di- 
 vides a and b, or the equation is impossible in integers for x 
 and y. 
 
 Examples. 
 
 1. Giv en 6a-f9y=32. O r, 2a?-f3//=y. 
 
 But if x is a whole number, 2x will be a whole number, 
 and by the same considerations, 3y must be a whole number, 
 and two or more whole numbers added together can never 
 make a fraction, therefore the equation 2x-^-3y= 3 f , or 
 6r-|~9?/=32, is impossible in integers. 
 
 In cases where solutions are possible, our rules of opera- 
 tion rest entirely on these considerations: 
 
 1st. A whole number added to a whole number, the sum is 
 a whole number. 
 
 2d. A whole number taken from a whole number, the 
 remainder is a whole, number. 
 1 3d. Multiply a whole number by a whole number, and the 
 ■ product is a whole number. 
 
 For instance, if x is a whole number, 2.r, 3a?, 4x, or any 
 integral number of x is a whole number. 
 
 2. Given 3 a?-f»5y=35, to find x and y in whole num- 
 
 **»■ i — | I W r * "" 
 
 bers. a?= Q But as x is a whole number, its equal, 
 
 o 
 
 35— 5» 
 
 or — q~^i must equal some whole number. 
 
 Now 11 — y being a whole number take it away, and 
 
 * the remainder — -— , must also be a whole number. 
 o 
 
88 KEY TO ALGEBRA. 
 
 Sv 
 
 But y being a whole number, -- is a whole number, 
 
 3 
 
 ml 3y , 2—2;/ ?/4-2 . . 
 
 Therefore, - -j — -= — =— ^— is a whole number, which 
 
 number call ;?. And - '- ==p. Or, y=3p — 2. 
 
 For the least value of y make p=l, and ?/ will equal 1, 
 
 35 5„ 
 
 anda?= — = — =10. Make p=2, then ?/=4, and #=5. 
 3 
 
 Make ^=3, then ?/=7, and a?=0. 
 
 Hence, ?/=l or 4, and z=10 or 5, are the only results 
 this equation admits of 
 
 In operating on the fractional expression — - — *, it was 
 
 3 
 our object to work down the coefficient of y to 1 . To accom- 
 plish ti>is object, we cast out whole numbers, add and sub- 
 tract whole numbers in the shape of fractions, &c, only- 
 taking care to keep expressions that are equal to integers, 
 until the coefficient of y becomes one. 
 
 3. Given 35a? — 24?/= 68, to find the least values of # and 
 y m integers. 
 
 The number of values in this case is unlimited. 
 _68+24?/_ 33+24?/ 
 *- 35 - 1+ ~35" 
 
 Hence, — ~J~- — - = some whole number ; but -~ is also 
 3d 3d 
 
 Soy 33-4-24// Ut/ — 33 
 
 a whole number, -p- ' - • = -^ — =an integer. 
 
 3d 3d 3d ° 
 
 33 j /-99_33//-29 
 
 Ur ' ""35 ~ 35~ 
 
 35?/ 33?/— 29 27/-J-29 _ .■ . 
 
 -h^ qcl~ = qc — a wnoIe number, multiply by 18, 
 
 3d 3d 3d 
 
 . 36?/+18-29 , % . , y+32 . . 
 
 and — ' ■ — =3/-f-14-f-- - ' - = a whole number. 
 3d 3d 
 
 Therefore V —^-=p. Or, y=Sbp— 32. 
 
INDETERMINATE EQUATIONS gg 
 
 Take p=l for the least value of y. and y—3. Therefore 
 *=4. 
 
 4. A man wishes to lay out $500 for cows and sheep: 
 the cows at 17 dollars per head, and the sheep at 2 dollars. 
 How many of each did he purchase 1 
 
 Let z== the number of cows, and y the number of sheep. 
 Then 17#-|-2?/=500. We know this equation is restricted 
 to whole numbers ; because the man could not have part of 
 a cow, or part of a sheep. 
 
 To find the least number of cows, transpose 17a?. 
 
 Then y = 250— 8*— ^. 
 
 * 2 
 
 Now as y must be a whole number, and 250 — Sx must 
 
 x 
 also equal some whole number, must be a whole number. 
 
 That is, the number of cows must be an even number ; be- 
 cause the number must be divisible by 2. 
 
 x 
 Hence, -^=P- Or, x=2p. Makep=l, and a?=2, the 
 
 least number of cows. Then #=233, the corresponding 
 number of sheep. 
 
 Now if the man wished to purchase as few sheep, and as 
 many cows as possible, we should transpose the other term, 
 
 500— 2y on , 7—2?/ 
 thus: * = ~ 17 = + ~T7 ' 
 
 7 2?/ 
 
 Therefore, — =— -- = a whole number. Multiply by 8, 
 
 and — — — - = a whole number, to which add — -±- and we 
 17 17 
 
 have — ~^=3-| — ^. Drop off the whole number 3, then 
 
 5 I y 
 
 -J^L=zp. Or, y=l7p — 5. Making p=l, gives #=12, 
 
 the smallest number of sheep. This gives #=28, the cor- 
 responding number of cows. 
 
 The number of cows or a?, may be any one of the even 
 numbers from 2 to 28. a 
 
90 KEY TO ALGEBRA. 
 
 5. A man wished to spend 100 dollars in cows, sheep, and 
 geese ; cows at 10 dollars a piece, sheep at 2 dollars, and 
 geese at 25cts., and the aggregate number of animals to be 
 100. How many must he purchase of each ? 
 
 Let x=. the number of cows, y the sheep, and z the geese. 
 
 Then - - - 1 Qx+2y+ | = 100. (1) 
 And - - - - x-\- ?/-|-* =100. (2) 
 Clear equation (1) of fractions, and 
 
 40*+8#- r -;?=40G. 
 , T _j_ y _j_ 2 r == l00. 
 
 39x+7y =300. 
 
 300—7?/ „ , 27—7?/ , , . 
 
 xz= — ' '— H ^- 1 =a whole number. 
 
 Or, 5( S9 j^= — 3 ^— - = a whole number, add -^ 
 
 , 4y+135 40* -f- 1350 . ?/+24 , . 
 
 and Kg- , or ^^— = ?/+34+ tjjj- = a whole 
 
 number. 
 
 Therefore, 8tt-~ ^ Or, ?/=39^— 24=15. 
 
 This value of y, gives a?=5. Hence, z=80. 
 
 If we take p=2, we shall have y=54; then a? will come 
 a minus quantity an inadmissible circumstance in any prob- 
 lem like this. Therefore, 5 cows, 15 sheep, and 80 geese, 
 is the only solution. 
 
 6 A person spent 28 shillings in clucks and geese; for the 
 geese he paid 4s. 4d. a piece, and for the ducks, 2s. 6d. a 
 piece. What number had he of each? 
 
 Let #=the number of geese, and ?/ the number of ducks. 
 
 Then. 52.r-f30.?/=28-i 2. Or, 26*4-15?/= 168. 
 
 We will now show another operation to reduce this kind 
 of equations. 
 
 Take the lowest coefficient, (in this example it is 15,) and 
 observe whether it will divide the other numbers or not. If 
 
INDETERMINATE EQUATIONS. yi 
 
 it will divide, reserve the whole numbers, and omit the frac- 
 tions. In this case, 15 will not divide 26, and will divide 
 168. The quotient will be 11, disregard the remainder. 
 
 Now assume p=x-\-y — 11. Then, since x and y are 
 whole numbers, p must be a whole number. Multiply by 
 15, and transpose, and we have 
 
 15j9 — 15a; — 15y= — 165. 
 2llr+15?/= 168. 
 
 By addition, 15;>{-lla? =3. Assume q=x-]-p. 
 
 Then - - - - l\q— Up— Ux=0. 
 
 Sum 11<H~ ^P =«*• Assume r=p-\- < Zq- 
 
 Or, 4r— 4p— 8q=0. 
 
 Sum 4r -(- 3q='3. Assume s=r-\-q — 1 
 
 Or, 3s— 3/— 3?=— 3. 
 
 Sum, 3s+ r =0~ ~Or, r=—3s. 
 
 Now having worked down to unity for a coefficient, the 
 problem is essentially reduced. We can make 5=0, hence 
 r=0. Then in the last assumed equation q=l, and in the 
 equation, r=p-^-2q, gives p= — 2 ; and in the preceding 
 assumed equation, q=x-\-p, that is #=3, and the first as- 
 sumed equation gives y=Q. 
 
 Hence 3 geese and 6 ducks is the answer, and no other 
 numbers will do. 
 
 7. Divide the number 100 into two such parts, that one 
 of them may be divisible by 7, the other by 11. 
 Let 7x= one part, and ll^=the other. 
 
 Then 7a;-|-lly=100, and x and y must be whole num- 
 bers. Assume p=x-\-y — 14. (1) 
 
 Then 7p—7x— ly=— 98. 
 
 But 7x+ Uy= 100. 
 
 By addition, - - 7p +4^=2 
 
 Assume q=p-\-y. (2) 
 
 Then 4q — ip — 4y=^0. 
 
 Add, and - - - - 4? 4. 3^=2. 
 
92 KEY TO ALGEBRA. 
 
 4?+ 3p=2. 
 
 Assume r—q^-p. (3) 
 Then 3r— 3q— 3p=0. 
 
 By addition - - - 3r-\- q =&" Or, q=2—3r. 
 
 Take r=0, then <7=2, and p= — 2. And from equation 
 (2), 2= — 2-|-#, or #=4, and lly=44, one of the numbers, 
 and of course 56 is the other, 
 
 8. Find a number which being divided by 6, shall leave 
 the remainder 2, and the same number divided by 13 shall 
 leave the remainder 3. 
 
 Consider that in division, the divisor and quotient multi- 
 plied together, and the remainder added, gives the number 
 divided. 
 
 Let N represent the number divided, x and y the quotients. 
 
 Then 6*4-2= N, and 13y+3=N. 
 
 Consequently, 6a? — 13//=1, an equation in which x andi/ 
 must be whole numbers, because they represent the whole 
 numbers of the division. 
 
 Assume p=x — 2y. We take 2y because 6 is contained 
 in 13, twice. 
 
 Then 6p— 6*4-12?/=0. 
 
 And ...... 6o? — 13?/==1. 
 
 Add, and - « - - 6p — //=1. Or, y=6p — 1. 
 
 For the smallest value of y we must take p=\. Then 
 7=5, and 13?/-|-3=68, the answer. 
 
 9. What number is that which being divided by 1 1, leaves 
 d remainder of 3, divided by 19, leaves a remainder of 5, 
 and divided by 29, shall leave a remainder of 10. 
 
 Let N be the required number, and x, y, and z the several 
 quotients, and of course they must be whole numbers. 
 Then 1 1*4-3= N, and \9y-\- 5=N, and 29;:4-10=N. 
 
 Hence, *J?^t?, and *J2t&. 19^=29*4-5. 
 
INDETERMINATE EQUATIONS. 93 
 
 Or, 19y— 29*=5. 
 
 Assume p=y — %. (1) 
 Then \9p— \9y+\ 9z=0. 
 
 Add, and - - 19^ — 10z=5. 
 
 Assume q=p — z. (2) 
 Then - - - - 10 g — 10p+10z=0, 
 
 By addition - lO^-j- 9p =5. 
 
 Assume r=q-\-p. (3) 
 Then - - - - 9r— 9q— 9p=0. 
 
 Add, and - - - 9r+ ? =5. Or, ?=5— 9r. 
 
 By returning to equations (3) and (2), we find p=\Or — 5, 
 and z=19r — 10. Not only must z be a whole number, 
 
 but to make x a whole number, - "£- must be a whole num- 
 ber. Substituting the value of z in this last expression, and 
 
 . 551r— 283 , . . CA oc , r— 8 
 
 we nave ■ — a whole number, or 50r — 25-| — — — a 
 
 whole number. Therefore, — — - must be a whole number, 
 
 which call i\ then r = 1 1 *-|-8. Let *=0, then r=8, z=z 1 42, 
 and 29z-f-10=N=4148, the number. 
 
 10. Required the least number that can be divided by each 
 of the nine digits without remainders. 
 
 Let x— the number. 
 
 _,. xxxxxxxx „ . 
 
 The " 2' 3' V 5' 6' 7' 8' 9' mUSt aU be 
 
 whole numbers. 
 
 Now if we make - a whole number ~j and — , the 
 
 double, and quadruple will be whole numbers of course. 
 
 x . x 
 
 Also if is a whole number, will be a whole number, 
 y o 
 
 Therefore we have only to find such a value of x as will 
 
 flf* fp ffi ff* ft* f 
 
 make -, 5 , -j — , - whole numbers. — , may also 
 
 y 8 7 o 5 
 
 be cast out, on consideration that 6=2-3, and 2 3 are factors, 
 one cf 9, the other of 8, in preceding expressions. 
 
94 KEY TO ALGEBRA. 
 
 X X X X 
 
 Hence we have only to make - , — , — , — , whole 
 
 9 8 7 5 
 numbers. Put x=9p. 
 
 Then ^ = jp_[_^. Hence, p=8q. Then z=9p=72?. 
 
 — -= a whole number. - = a whole number, or -=a4- 
 7 7 7 * ' 
 
 . Make -J[ =r, or <?=7r. In the same way we find 
 
 r=5s. Take 5=1, then r=5. ?=35. a?= 72-35 =2520. 
 
 N. B. By the aid of the Indeterminate analysis, combined 
 with some well known properties of numbers, we may some- 
 times solve miscellaneous problems in a very summary 
 manner. For example, we refer to the following. 
 
 11. The product of five numbers in arithmetical progres- 
 sion is 10395, and their sum is 35. What are the numbers 1 
 
 As the number of terms is odd, let x represent the mid- 
 dle term, and y the common difference. Then (x — 2y), 
 (x — y), x, (x-\-y), (x-\-2y), will be the numbers, and their 
 sum 5#=35, or x=7. 
 
 Now as 10395 is the product of all the numbers, 7 must 
 be one of its factors. Therefore divide by 7, and we have 
 1485 for the product of the four remaining terms, but as 
 this number ends in 5, it can be divided by 5, and it is more 
 than probable that 5 is one of the numbers, and the one 
 preceding 7, therefore, 2 is the common difference, and 3, 
 5, 7, 9, and 11, the numbers. 
 
 12. Given the sum of the squares of three numbers =195, 
 the sum of their cubes =1799, and their continued product 
 =385, to find the numbers. 
 
 By the common rules, without considering any circum- 
 stances, this problem would produce equations of high order, 
 and difficult of solution. But let us call to mind the fact, 
 that the numbers cannot be fractional, if they were, their 
 squares, cubes, and product would not probably come whole 
 numbers. Also, some of the numbers must be under 10; 
 
INDETERMINATE EQUATIONS. 95 
 
 if even two of them were over 10, the cubes and product 
 must be larger than the numbers mentioned. From consid- 
 erations like these, we decide that the answer must be whole 
 numbers, and some of them under 10. Noav as the product 
 of the three is 385, and the sum of their squares is 195, 
 both numbers ending in 5, it is so probable that one of the 
 numbers is 5, that we shall so consider it, and let x and y 
 be the other two numbers, then a: 2 -|-?/ 2 -{-25 =195, and hxy 
 =385. Equations from which we readily find one number 
 to be 7, the other 11. 
 
 Now by trial we find 5 3 +7~+lP=1799, and therefore 
 these are in fact the numbers. 
 
 Section XX. 
 
 To determine the number of solutions an equation 
 may admit. 
 
 It has already been observed that an equation in the form 
 of ax — by=c admits of an infinite number of solutions, 
 
 as x= ' * a quantity all positive, and the only restriction is 
 
 to assume y of such a value that the numerator may be 
 divided by a. 
 
 But equations in the form of ax-\-by=c. Then x= - 
 
 and the numbers of solutions depends on the relative values 
 of c, 5, and a. 
 
 If c be very large in relation to b and a, as we have 
 before observed, the equation may have many solutions, 
 otherwise not. 
 
 We come now in a general manner to determine the num 
 ber of solutions an equation of this form may have. 
 
 Let ax-\-by-=c. Assume ax! — by'=l. Which equation 
 
96 KEY TO ALGEBRA. 
 
 is always possible, and from which x' and y' can be known 
 in integers. 
 
 Multiply the assumed equation by c, and acx' — bcy'=c. 
 Put the two values of c equal to each other, and 
 ax-\-by=acx' — bey' 
 x=cx' — b(y-\-cy'). Or, x=cx' — bm. (1) 
 
 y=;a( C ^j^)— cy'. Or, y=am—cy'. (2) 
 
 In these theoretical equations, (1) and (2), m has different 
 values, it being an arbitrary number taken at pleasure, so 
 that ex' may be greater than bm, and am greater than cy' to 
 render x and y positive. 
 
 But if no such value of m can be found, it is proof that 
 
 values of x and y do not exist in positive integers, and on 
 
 the contrary as many suitable values of m as can be found, 
 
 so many solutions will the equation admit of, and no more. 
 
 Now as - - - mb <^ ex. and am ^> cy' 
 
 r\ ■ ^ cx ' A -^ C V' 
 
 Or, m<—-. and m > — 
 
 ^ b ' ^ a 
 
 That is m at the same time is found to be greater and less 
 than known quantities, therefore its limit or range is found. 
 
 For instance, if m must be greater than 30, and less than 
 40, we conclude that it may be any number between 30 and 
 40, and the number o£ different values it can take is 9. 
 
 We perceive that the difference between the integral parts 
 
 of — — and — will express the range of m, and the number 
 
 of different solutions which the equation admits of, (except 
 in certain cases ; ) as m is more than one of these fractions 
 and less than the other, the difference between the expressions 
 
 — and — is sometimes one more than the number of dif- 
 o a 
 
 ex 
 ferent values of m, such is the case when — is an integer, 
 
 in such cases, subtract one from the difference of these quan- 
 tities for the range of m, but this case very seldom occurs. 
 
INDETERMINATE EQUATIONS. 97 
 
 N. B. In making use of the expression — and — care 
 
 ' Cb 
 
 must be taken, not to take their difference as factional ex- 
 pressions, their absolute difference is not wanted, it is the 
 
 ex' cii f 
 difference between the integral farts of — and- 1 -. 
 
 ° r b a 
 
 Example. 
 Required the number of integral solutions to the equation 
 
 7*4-%= 100. 
 Find the least value to a?', y' in the equation^' — 9/=l. 
 
 lx' — 9y'=l. Assume p— x' — y' 
 Then - - lp— 7x'+7y'=0. 
 
 Add, and lp — 2y'=l. Assume q=3p — y' 
 Then - - 2q— 6p+2i/=0. 
 
 Add, and 2q-\- p=l. Or, p=l — 2q. 
 
 Take £=0, then p=l. y'—% # =4. 
 
 mL ex' 100-4 j cy 100-3 
 Then — =— — - and -£ = — — . 
 b 9 a 7 
 
 That is ^=441 and ^-=42« 
 b 9 a 1 
 
 Disregarding the fractions, the difference of the integral 
 
 parts is 2, that is there are two integral solutions to the 
 
 equation. 
 
 400 300 
 
 9 7 
 
 .... 2800 2700 100 13 _ 
 fractional form, thus: — ^ .— - =—=137. 
 
 Here the integral difference is one, which without this 
 caution might be taken for the number of solutions. The 
 integral difference in this case is not the difference of the 
 integrals. 
 
 Observe in this example 44£ and 42|, the fractional part 
 
 of — is |, and the fractional part of — is f , the former is 
 o a 
 
 less than the latter* in such cases the integral parts must be 
 taken separately. 
 
 If we had taken the difference between — — , in a 
 
98 KEY TO ALGEBRA. 
 
 ex 
 But when the fractional part of — - is not less than the 
 
 b 
 
 cv 
 fractional part of -— , but equal or greater than it, we may- 
 find the number of solutions by taking the difference of the 
 
 ex' cy' _ , 
 expressions — . Reduce to a common denominator, 
 
 and take the difference of the numerators, and we will have 
 
 — - — — ; but ax' — by'= 1. Therefore, we have — 
 
 ab ab 
 
 for the number of solutions, at once. 
 
 Example. 
 
 What number of integral solutions will the equation 
 1 9,z-f-l 3j/ =2000 admit of? Ans. 17. 
 
 ~9 X 13=117) 2000(17. 
 But as we cannot know whether the fractional part of 
 
 C11 CI J 
 
 ~y~ is not less than the fractional part of — we cannot be 
 b r a 
 
 sure that dividing c by ab will give the true number of solu- 
 tions. It either will be the true number or one less. 
 
 The equation 5#-|-9#=40 admits of no solution in whole 
 numbers, c=40 will not be divided by aZ>=45. 
 
 Now take the equation hx' — 9#'=1. ' And we find x'=2, 
 
 and y'=l. 
 
 ex' 80 cy' 40 
 
 Therefore, — = n =8£, and — = - =8. Now as there 
 b 9 y a b 
 
 is no difference between these integral parts, it indicates as it 
 should, that there is no solution. 
 
 But let us take 5#-f-9//=37, the same equation, except a 
 smaller value of c. If c would not divide by ab before, 
 much less will it now. Yet in this last equation we have 
 
 ex' 74 
 
 a solution, a?' =2, y'=l, as before, and c=37. — =-^-=8|- 
 
 and — = _ =7|. Here the difference of the integrals is 1, 
 a 5 
 
 and indicates one solution. x=2, y=3. 
 
INDETERMINATE EQUATIONS. 99 
 
 How many solutions will the equation 2a:-)-5^=40 admit 
 of? 
 
 The auxiliary equation 2a?' — 5/=l, gives #'=3, y'=l. 
 
 «*[_. 24. ^=20. Or, 4 solutions. 
 
 C2?' 
 
 But observe that ~r in this case, is a complete integral, 
 
 24 ; agreeable then, to previous considerations we must de- 
 duct one, and the number of solutions are but 3, as follows: 
 #=5. 10. 15. #=6. 4. 2, and no other solution can be 
 found. 
 
 What number of solutions. in whole numbers can be found 
 for the equation 3x-\-5y-\-lz=\00. 
 
 As x and y each cannot be less than one, z cannot be 
 
 greater than = =131. That is, z cannot be greater 
 
 than 13, in whole numbers. Now suppose 2=1, and the 
 equation becomes 3x-\-oy=93. 
 
 The number of solutions for this equation, found as pre- 
 viously directed is 6. rm, of - J a? = 26. 21. 16. 11. 6. 1. 
 inatis^ ?/== 3 6 9 ]2 15 1Q 
 
 Now x and y can make these 6 changes, and z be con- 
 stantly equal to 1, and satisfy the primitive equation. 
 
 Take z=2, and the equation becomes 3#-j-5 < y=86. 
 
 This equation has also 6 solutions, z being through all the 
 changes of x and y equal to 2. 
 
 Now take z=3, then the original equation is 3z-|-5y=79. 
 This equation has five solutions. 
 
 Now take z=4, then 3.r4-5y=72. This equation has 
 four solutions. 
 
 Take z=5, then 3#-|-5y=65. This equation has four 
 solutions. 
 
 Take 2=6, then 3z-[-5y= 58. This equation has four 
 solutions. 
 
 In this manner, by taking z equal to all the integers up 
 to 12 in succession, we find 41 solutions. 
 
100 KEY TO ALGEbKA. 
 
 THE DIOPHANTINE ANALYSIS. 
 
 The Diophantine Analysis teaches hew to find square 
 and cube numbers under given conditions, or having given 
 relations to each other. 
 
 Examples. 
 
 Case 1st. Find such a value of x as will make the ex- 
 pression ax-\-b a square. 
 
 Put ax-\-b=n 2 , n 2 being any square, it is, therefore, an 
 indefinite problem. ' 
 
 _ . . n 2 — b . - 
 
 From the equation x= ; take n equal to any num- 
 ber whatever, and a and b being known, x becomes known. 
 
 Eight times a certain number added to 9, makes a square. 
 What is the number ? 
 
 Let a? = the number. Then Sx-\-9=n 2 , that is any 
 
 o i n a n2 — b n2 ~ 9 A 
 
 square. a=o, o=9, and x= = — — — . Assume n 
 
 a o 
 
 =7, then #=5, the required number. But there are many 
 other numbers that will answer the condition according as 
 we assume n more or less. 
 
 Find x, such that the following expressions shall be square 
 numbers. 
 
 9*4-9. 7#+2. 3x— 5. a?+|. 
 
 All these correspond to the general expression ax-\-b. 
 
 Case 2d. Any algebraic expression in the general form 
 of ax 2 -\-bx, may be made a square by supposing its square 
 root equal px; x must be in some part of the root, because 
 the expression contains a: 2 , that is some function of a?. Now 
 if px is the root, ax 2 -\-bx=p 2 x 2 . Divide by*, &q and we 
 
 have %=--£ — • p may be any assumed value whose square 
 
 is greater than a, 
 
DIOPHANTINE ANALYSIS. 101 
 
 Examples. 
 
 Six times the square of a certain number, added to five 
 times the number is a square. What is the number ? 
 
 5 
 
 6x 2 -\-5x=p 2 x 2 . Or, x= -• 
 
 p 2 — 6 
 
 Here it is obvious that p 2 must be greater than 6, other- 
 wise it is unlimited. Take p=3, then a?= f = If. 
 
 x 2 
 Find the value of x to make — -\-3x a square. 
 
 Ans. #=6. 
 
 Case 3d. Any algebraic expression in the general form 
 of x 2 ±bx-\-c, can be made a square, by putting xdzp equal 
 its square root. 
 
 We can if we please take x — p for the root in all such 
 cases. Then if p is less than x, the square is diminished, if 
 greater, the whole root will be essentially minus, but the 
 square will be plus, and may rise to any amount. Therefore 
 x — p is far more general than x-\-p. 
 
 Case 4th. Any expression in the form of ax 2 dzbx-\-c 2 , 
 can be made a square, by taking its root equal to cdzpx. 
 
 It will be observed that x must be in the root of the pre- 
 vious expression, because it has x 2 , and c must be in the root 
 of this last expression, because it contains c 2 . 
 
 In the first we have x 2 ±bx-\-c=x 2 ±2px-{-p 2 . Or, 
 p 2 — c 
 
 In the second, x 2 ±bx-\- c 2 —c 2 ±: 2cpx-{-p 2 x 2 . Or, 
 
 x±:b= ±3cp-\-p 2 x. Or, x= - ^~- In both cases as- 
 
 Idap 2 
 
 sume p of any convenient value to render x positive, and as 
 
 small as possible. 
 
 Find a number such, that if it be increased by 2 and 5 
 separately, the product of the sums shall be a square. 
 Let z=the number, then (a?4-2) (#+5) =a? 2 +73+10, 
 
102 KEY TO ALGEBRA. 
 
 must be a square. b=l, c=10. General solution, x=J—^ — 
 
 2p±7 
 
 Now p 2 must be more than 10 ; hence, take j?=4, and 
 *= T 6 5 = |, the least number that will answer the conditions. 
 
 Case 5th. An expression in the form of ax 2 -\-bx -\-c, 
 where neither the first nor the last terms of the expression 
 are squares, neither branch of the root can be directly found, 
 and the expression cannot be made a square, unless we can 
 separate it into two rational factors, or unless we can first 
 subtract from it some simple binomial square, and can then 
 divide the remainder into two rational factors. 
 
 By reminding one of the nature of quadratic equations, 
 all may perceive that the expression ax 2 -\-bx-\-c must be 
 the product of two factors, but whether rational factors or 
 not is the subject of inquiry. 
 
 To find the factors which make the product ax 2 -\-bx-\-c, 
 put this expression equal to 0, and work out the values of x 
 thus, ax 2 -\-bx-\-c=0. Or, ax 2 -\-bx= — c. Complete the 
 square, and 4a 2 x 2 -\-\abx-\-b 2 = b 2 — \ac. 
 Or, 2ax-\-b= ±J \b 2 —4ac.\ 
 
 We now perceive that the values of x must be rational, 
 provided J \b 2 — 4ac\ is a complete square. If it be so, let 
 
 la-J \^-iac\-l=n h and -I J («•-*«}- £•* 
 
 Then the two values of x are x = m and x=n, and 
 (a? — m) (x — n), are the factors which will give the expression 
 ax 2 -\-bx-^-c. 
 
 Examples. 
 
 1. Find such a value of x as will make 6a? 2 -|-13x-|-6 a 
 square. 
 
 Here «=6, £=13, e=6. £ 2 =169, 4ac=144, Z> 2 — lac 
 =s25 and J \ b 2 —4ac } =5. Now 12o:+l3=±5. *=--§ , 
 Or, *=— |. Or, 3z+2=0, and 2*4-3=0. 
 
DIOPHANTINE ANALYSIS. 102 
 
 That is, (3x-\-2) (2#-|-3) will produce the expression 
 6x 2 +13tf+6. 
 
 Now to make the expression a square, put 
 (3*+2) {2x+3)=p 2 {3x+2) 2 
 
 Then 2x+3=p 2 (3z+2.) And »=-?£_. 
 
 Take p=l, and x=l. We might have seen at first, 
 that in this particular expression, the value of x being 1, 
 would make it a square, as 6-j-13-|-6=25, a square. 
 
 That is in all cases when the sum of the coefficients 
 make a square number, the value of x may be one. 
 
 2. Find such a value of x as shall render the expression 
 13x 2 -\-]5x-\-7*a. square. 
 
 Here as neither the first nor last terms are squares, nor 
 b 2 — 4^ a square, the expression cannot be made a square, 
 unless we can separate the remainder into factors after taking 
 away some simple square. But in this case, \ac is greater 
 than b 2 , we must then, in subtracting our square, diminish 
 a and c and increase b. 
 
 To accomplish this end we will subtract the square of 
 * — 1, not x-\-\. 
 
 That is 13z 2 +15z+7. 
 
 Subtract x 2 — 2.r+l. 
 
 Remainder,- - - - l2x 2 -\-\lx-\-Q. 
 
 In this last expression, a=12, £=17, cz=6. Hence, 
 b 2 — 4ac=289 — 288=1, a square ; we are now sure ra- 
 tional factors can be found to produce the expression 
 12z 2 +17a?-|-6. 
 
 By assuming 12 2 -}-17a;-|-6=0, and finding the values of 
 s by the quadratic, (merely to get the factors,) we find 
 «ass— |, and a?== — |. Or, 3; +2=0, and 4*+3=0. 
 
 And (3z+2)(4z+3)=12* 2 +17z+6.* 
 
 ♦The values of x used to obtain these factors have no connection with tat 
 values of z to render the original expression a square. 
 
104: KEY T0 ALGEBRA. 
 
 Now the original expression is the same as (a? — l) 2 -f» 
 (3a?-f 2) (4a?+3.) Put its root equal to (x— l)+^(3a?+ 2 )) 
 and square this root, and 
 
 (a?— l) 2 +(3a?+2) (4a?-f-3)=(a?— 1) 2 + 2^(a?— 1) (3a?+ 2)+ 
 /(3a?+2) 2 . By reduction, 4a?+3 = 2p{x—l)+f[3x+2. 
 Or, (— 4+2p+3/)a?=2p+3— 2p 2 . 
 
 Therefore *- 2 ^+ 3 — 2 ^ 
 inerelore, ^^—-^ 
 
 Take^=l, then a;=3, and 13:r-J-15a?-}-7=169a square, 
 Ryan makes a; in this example equal i and the expression 
 
 equal *§ l a square, but an integer number is always more 
 
 satisfactory. 
 
 3. Find such a value of a? as will render 14a? 2 -j-5a? — 39, 
 a square. 
 
 After a few trials this expression is found to be the same 
 as (2a? — l) 2 -|-(5a? — 8)(2a?-|-5.) Assuming its root to be 
 2a? — l-\-p(5x — 8 ) Then by squaring the root, making it 
 
 equal to its power, and reducing, we find a? = , 7T A < / 
 
 Assuming p=l, a?= 1 T 5 , and the expression equals 36, a 
 square. Other values can be found, by assuming different 
 values to p. 
 
 4. Find such a value of x as shall make 2a?*-(-21a?-|-28 
 a square. 
 
 After a little inspection we find this expression equal to 
 (a;+4) 2 +(a?+l) (a?+12.) Now if we make 
 
 (z+4) 2 +(*+l) (a?+12)= {(*f4>-p(*+l.)J» 
 
 After reduction, we shall find a?= , ^ 1 • 
 
 pr — 2/? — 1 
 
 Assume p=4, then a?=4, and the original expression is 
 144 a square. 
 
 If (a?+ 4) 2 +(a?+l)(a?+12)={(a?+4)-^(a?+12)} 2 , we 
 shall find a?= ^~ P ~ . If we take^=l, a?=f . If we 
 
DIOPHANTINE ANALYSIS. 105 
 
 take 2?=!, a?=8, and we might find many other numbers 
 that would answer the conditions of the expression. 
 
 Case 6th. When we have an expression in the form of 
 a^x^-^-bx^-^-cx'-^-dx-j-e, we can assign a value to x that will 
 make the whole expression a square, if we can extract 
 three terms of its root. 
 
 Assume such terms as the whole root, square the root so 
 assumed, making it equal to the given expression, and by 
 reducing, we shall have a value of x which will make the 
 original expression a square. 
 
 Example. 
 
 Find such a value of x as shall make 4x 4 -\-4x 3 -\-4x 2 -\- 
 2x — 6 a square. 
 
 We commence by extracting the square root as far as 
 three terms, and find them to be 2# 2 -|-#-|-£. Now assume 
 4z +4x 3 -\-4x-+2x— 6=(2z 2 +#+!.) 2 
 
 3a? 9 
 
 Expanding and reducing, we have 2x — 6=— — |-— . 
 
 And x=\3\. 
 
 Essentially the same method must be pursued in other 
 problems of the like kind. 
 
 Case 7th. Find such a value of x as shall render 2ar 2 -f-2 
 a square. 
 
 Expressions of this kind, when neither a nor c are squares, 
 nor b 2 — \ac a square, and which cannot be resolved into 
 factors, presents an impossible case, unless we can first find 
 by inspection, some simple value of x that will answer the 
 condition. 
 
 In the present example, it is obvious that if #=1, the 
 expression is a square. But we wish to find other values than 
 1 that will render this expression a square, and having found 
 that one will answer, we are now enabled to find other val- 
 ues, thus : 
 
 Let a?= l±y. Then x 2 =lzk2y+tf. 
 
106 KEY TO ALGEBRA. 
 
 And 2x 2 -|-2=4d=%-{-2y z . Here the original expression 
 is transformed into another expression, having a square for 
 its first t^rm. 
 
 Now we must find such a value of y as shall make 
 4-j-4?/-|-2y 2 a square. 
 
 Assume 4-}-4^-|-2^ 2 = (2 — my) 2 — 4 — 4;my-\-m 2 y z . Or, 
 
 a i r> /I i \ T1 4(m-f-l) 
 
 4+~j/= — lm-\-m-y. Hence #=s -^f- 9 i m mav De an y 
 
 number greater than one. Put m=2. Then #=6, and 
 5c=l-(— y=7, and the original expression 2;r 2 -f-2=98-}-2= 
 100, a square. 
 
 N. B. It often occurs incidentally in the solution of prob- 
 lems, that we must make a square of two other squares. 
 This can be done thus : Let it be required to make x 2 -\-y 2 
 a square. Assume x=p 2 — q 2 , and i/=2pq. 
 
 Then, x 2 =p* — 2p 2 q 2 -\-q\ 
 
 And, y' 2 = 4p*q 3 . 
 
 Add, and - - x 2 -\-y-=p -\-~p 2 q 2 -{-q 4 , which is evidently 
 a square, whatever be the values of p and q. We can, 
 therefore, assume p and q at pleasure, provided p be greater 
 than q. 
 
 Double and Triple Equalities. 
 
 In the preceding section it was only necessary to find such 
 a value of the unknown term as to render a single expres- 
 sion a square. But there are problems where it becomes 
 requisite to find such a value of the unknown term as to 
 render several different expressions squares at the same time. 
 And this is called double and triple equalities. 
 
 Case 1st. As a general equation for double equality, let 
 it be required to find such a value of x that ax-\-b may be 
 
D10PHANTINE ANALYSIS. 107 
 
 a square, and the same value of x give cx-\-d a square. 
 
 f2 Q 
 
 Let ax-{-b=t 2 . Then x. 
 
 And cx-^-d=p' 1 . Then x= 
 
 a 
 p 2 — d 
 
 Therefore = — . Or, ct 2 — cb = ap 2 — ad. 
 
 a c $ 
 
 Tranpose cb and multiply by c and we have 
 c 2 t 2 -=acp 2 -\-c 2 d — acd. 
 
 As the left hand side of this equation is a square, what- 
 ever may be the values of c and t, it is now only requisite 
 to find such a value of p 2 as shall render the other side a 
 square, which can be done by some one of the artifices in 
 the preceding section. 
 
 To illustrate this we give the following definite problem. 
 
 The double of a certain number increased by 4, makes 
 a square, and five times the same number plus one, also 
 makes a square. What is the number * 
 Let x represent the number. 
 
 Then 2^+4= * 2 ] 
 
 J> From which < 
 And 5a?-j-l=p 2 
 
 x= — 
 
 5 
 
 Hence 5* 2 — 20=2p 2 —2. Or, 5* 2 =2j9 2 +18, multiply 
 by 5, and 25^ = 10;? 2 +90. 
 
 The left hand side of this last equation being a square, 
 whatever be the value of t, it is now only necessary to 
 find such a value of p 2 as to make 10^ 2 -f-90 a square, an 
 expression which corresponds to case 7 of the last section. 
 We therefore cannot proceed unless we find by trial, by 
 observation, by intuition as it were, some simple value of 
 p that will make 10/> 2 -|-90 a square, and we do perceive 
 that if p= 1, the expression will become 100, a square. 
 
 Now if ;?=1, it will give a definite and positive value to 
 *, and the problem is solved. If not we must find other 
 values of p. 
 
108 KEY TO ALGEBRA. 
 
 But we have found that #= — - — , and if p=l, £=0, 
 
 and the original expressions 2z-|-4, and 5#-|-l, become 4 
 and 1, squares it is true, which answer the technicalities, but 
 not the spirit of the question. 
 
 ^To find another value of p. Put p = 1 -|- q. Then 
 10^ 2 +90 = 100*-f20;?+2 ? 2 . To make this a square 
 assume 100-\-%0q+2q 2 = {l0— nq) 2 = 100— %0nq-\-n 2 q 2 . 
 
 By reduction, q= - —}— •". . Now n must be so taken, 
 
 J a n 2 — 10 7 
 
 that n 2 is more than 10: take ti=5 and ^=8, ^=9, then 
 #=16 and the original expressions 2#-f- 4 = 36, a square, 
 and 5#-|-l=81j a square. 
 
 Case 2d. A double equality in the form of ax 2 -\-bx=Q 
 and cx 2 -\-dx, also equals a square, may be resolved by 
 
 making *=— , tlten the expressions will become — (a-\-by) 
 
 and — - (c-\-dy,) which must be made squares. 
 
 But if we multiply a square by a square, or divide a 
 
 square by a square, the product or quotient will be square. 
 
 Now as each of the preceding expressions are to be 
 
 squares, and as they obviously have a square factor — it is 
 
 only necessary to make a-\-by, and c-^-dy squares, as in the 
 first case. 
 
 We may also take another course and assume ax 2 -\-bx 
 
 =p 2 x 2 , which gives x= - , which value put in the other 
 
 expression, and we have c (^^) +\^ZZ^) ==D ' 
 
 Multiplying this by the square (p 2 — a) 2 , and the expres- 
 sion becomes cb 2 — dbd-\-abp 2 = some square, from which 
 the value of p can be found and afterwards x. 
 
y 
 
 And 
 
 DIOPHANTINE ANALYSIS. IQg 
 
 Example. 
 
 A certain number added to its square, the sum is a square, 
 and the number subtracted from its square, the remainder is 
 a square. What is the number? 
 Let x = the number. 
 
 Then x 2 -{-x=Q. And x 2 — #== some other square. 
 
 Assume x=— • Then — ( \-\-y ) = Q. 
 
 The problem will be solved if we can find such a value 
 of y, as will at the same time make \-\-y and 1 — y squares. 
 
 Therefore put l-^-y=p 2 , and 1 — y=q 2 - 
 
 From the first, y=p 2 — 1. And y=l — q 2 . 
 
 Therefore, q 2 =2 — p 2 . As q 2 is a square, we have only 
 to find such a value of p 2 , as shall render 2 — p 2 a square. 
 But this cannot be done unless we can find some simple 
 value of p by inspection, and we do observe it must be one. 
 But p being equal to one, gives y=0, which will not answer 
 the conditions. Therefore, let p=l-\-t. 
 
 Then 2— p 2 = \— 2t— t 2 ={\— ut) 2 = l— 2ut+u 2 t*. 
 
 0r > fa ^pT' Take«=* <_} p=l+t=\. 
 y=p 2 — 1=| j- x= — • x = ||, a number that will an- 
 swer the given conditions. 
 
 Case 3d. To resolve a triple equality. 
 
 Equations in the form of ax -\- by = t 2 , ax-\-dy=* u*i 
 ez-{-fy=s 2 , can be resolved thus : 
 
 ■r> • c j dt 2 — ^ 2 
 
 By expunging ?/, we find x= — j — - — 
 
 ^2^,2 ct 2 
 
 Then by expunging x. y=-~- — — - . 
 
 Substituting these values of a; end y in the tnird equation 
 
110 KEY TO ALGEBRA. 
 
 j i. ni. ( a f— bc)u 2 +(de— cf)t 2 , 
 
 and we shall have ^ '- — P~\ J — L — =s 2 . 
 
 ad — be 
 
 Assume «=±te. Then u-=t 2 z 2 . Put this value of 
 u 2 in the above, and divide by f, and we shall have 
 (af — bc)z 2 -\-de — cf__s 2 
 ad — be ~~ t*' 
 
 The right hand side of this equation is a square, and 
 therefore all that is now requisite, is to find such a value of 
 z as shall make the other side a square, which when possi- 
 ble, can be done by case 7, section 20. 
 
 After z is found t may be assumed of any convenient 
 value whatever. Now u is known, and with t and u known 
 quantities, we know x and y. 
 
 The preceding are some of the most comprehensive and 
 general methods yet known ; but there are cases in practice 
 where no general rules will be so effectual, as the operator's 
 own judgment and penetration. 
 
 Much, very much will depend on skill and foresight dis- 
 played at the commencement of a problem, by assuming 
 convenient expressions to satisfy one or two conditions at 
 once, and the remaining conditions can be satisfied by som.6 
 one of the preceding rules. 
 
 Examples. 
 
 1. It is required to find three numbers in arithmetical 
 progression, such, that the sum of every two of them may 
 be a square. 
 
 Let x 1 x-\-y and x-\-2y represent the numbers. 
 
 Then by the general formula, 
 
 2x+y=t 2 , 2x-{-2y=u 2 , 2x-{-3i/=s 2 . 
 
 -n • • i t* — V u2 — % 
 
 By exterminating x, we have ■ — ^-'-= — jz — -• 
 
 Continuing thus after the general equations, we* find a 
 
DIOPHANTINE ANALYSIS. HJ 
 
 long and troublesome process, and in conclusion, we find the 
 numbers to be 482, 3362, and 6242. 
 
 The above is according to the common as well as the 
 general method. 
 
 The following is Mr. Young 1 s Solution. 
 
 Let x — y, a?, and x-\-y represent the numbers. Then 
 2x—y, 2a?, and 2x-\-y must be squares. 
 
 Assume 2x=m 2 -\-n 2 , and y=2mn. 
 
 Then 2x — x=m 2 — 2mn-\~n 2 , and 2x-\-y=m 2 -\-2mn-\-n* 
 are evidently squares. It therefore only remains to make 
 2x or m 2 -\-n 2 a square, and this can be done as explained at 
 the close of the last section by assuming m=r 2 — 5 2 , and 
 n=2rs. Then 2x=.m 2 -\-n 2 =^{r 2 -\-s 2 Y^ an expression in 
 which r and s can be assumed in numbers. But they must 
 be so assumed that x shall be greater than y to make x — y 
 the first number, positive and for this reason, we must give 
 the literal expressions for the numbers before taking definite 
 values for r and s. The expressions for the numbers are 
 S&— y= £(r 2 +s 2 ) 2 — 4rs (r 2 — s 2 .) x= £(r 2 -f-5 2 .) 2 
 x-\-y— ±(r 2 -\-s 2 ) 2 -\-4rs(r 2 =s 2 .) 
 
 Take r=9, 5=1, and 482, 3362, 6242, are the numbers. 
 
 Another Solution. 
 
 x 2 x 2 x 2 
 
 Let -^ y->~n an( ^ o — |-y be the numbers. 
 
 Then x 2 — y, x 2 -\-y, and x 2 must be squares. 
 
 But the last being a square, we have only to make x 2 — y 
 and x 2 -\-y, squares. 
 
 Assume y=2x — 1. Then x 2 — y=x 2 — 2#-|-l, a square, 
 and we now have only to make x 2 -\-2x — 1, a square. 
 
 Therefore make * 2 -l-2# — \=(x-\-nf—x 2 -\-2nx-\-n 2 . 
 
 X -2{l-n) 
 
 It is manifest that n must be less than one, make it .& 
 
 „, 1.64 41 _ x 2 1681 72 1440 
 
 Then ... -=- Or, - = — ^fo^W 
 
112 KEY TO ALGEBRA. 
 
 m , 482 3362 6242 
 
 Then 400' 400 ' 400" are the numbers. 
 
 Here we have the numbers expressed in fractions, but 
 the denominators are common, and is a square number, we 
 may therefore multiply all three by 400, and we shall have 
 482, 3362, and 6242 for the numbers, as in the other solu- 
 tions. 
 
 If we take n= f in this last result, we shall have 2162, 
 7442, and 9442 for the numbers. 
 
 2. Find two numbers such, that if to each, as also to their 
 sum, a given square, a 2 be added, the three sums shall all 
 be squares. 
 
 Let x 2 — a 2 and y 2 — a 2 represent the numbers : then the 
 first conditions are satisfied. 
 
 It now remains to make x 2 -\-y 2 — 2a 2 -\-a 2 a square, or, 
 ff'+y 2 — & 2 =n. Assume y 2 — a 2 =2ax-\-a 2 . This assump- 
 tion will make the expression a square, whatever be the 
 values of either x or a. But the assumed equation gives 
 y*=2&a+2& 2 , and as y 2 is a square, we must find such values 
 of x and a, as shall make 2ax-\-2a 2 : a square. Put x=na. 
 Then 2na?-\-2a 2 = □ ,or, a 2 (2?i+2)= □ . Hence it is sufficient 
 that we put 2^+2 = some square. Therefore, assume 
 2?i+2=16. Hence n=7 and x=7a. Now take a equal 
 to any number whatever. If <z=l, x=7, ^=4, and 48 
 and 15 are the numbers, add 1 to each, and we have 49 and 
 16, squares; sum, 63+1=64, a square. 
 
 3. Find three square numbers whose sum shall be a 
 square. 
 
 Let x 2 +y 2 +z 2 = □ . Assume y 2 =2xz. Then x*+2xz-\-£ 
 is a square. But 2xz=\J. Let x=uz, then 2wz 2 =D, or 
 2w=D=16, tt=8, x=8z, z=l, z=8, y=4. 
 
 Therefore 64+16+1 =81 =9 2 . 
 
 4. Find three square numbers in arithmetical progression. 
 Let x 2 — y, x 2 . and x 2 -\-y represent the numbers. Assume 
 
 x*=y 2 -\-\,then the first and last will be squares, and it only 
 
DIOPHANTINE ANALYSIS. ^3 
 
 remains to make the middle term, or y 2 +J> a square. 
 
 ©2 1 
 
 Therefore, put y 2 -^-\z=(y — py y which gives y—~ 5— i ' 
 
 tip 
 
 Take ^=1, then #=£, and y 2 -\- |=§f=# 2 . Therefore, 
 oVj If? if? are tne numbers; but we can multiply them all 
 by the same square number 64, and their arithmetical rela- 
 tion will not be changed, and they will still be squares ; 
 hence 1, 25, and 49 may be the numbers, or 4, 100, and 196, 
 
 5. Find two whole numbers, such that the sum and dif- 
 ference of their squares, when diminished by unity, shall be 
 a square. 
 
 Let #-f-l=one number, and y= the other. Then by 
 the conditions we must make squares of x 2 -\-y 2 -* r 2x, and 
 K 2 — ^2_|_2#. Assume 2x=a 2 , and y 2 =2ax, then the ex- 
 pressions become x 2 -\-2ax-\-a 2 , and x 2 — 2ax-\-a 2 , obvious 
 squares, whatever be the values of x and a. But the equa- 
 tions 2x=a 2 and y 2 =2ax must be satisfied. Take a=4, 
 then a?=8, y=S, and #-(-1=9. Therefore 9 and 8 are the 
 numbers. 
 
 6. Find three whole numbers, such, that if to the square 
 of each, the product of the other two be added, the three sums 
 shall be squares. 
 
 Let z, xy, xv : be the numbers. Then by the conditions, 
 x 2 -\-x 2 yv, x 2 y 2 -\-x 2 v, x 2 v 2 -\-z 2 y, must be squares. As 
 each term contains a square factor # 2 , it will be sufficient to 
 make l-{-yv=n 1 y 2 -\-v=D 1 and v 2 -\-y=n> 
 
 Assume i/=4v-j-4, and this will make the first and last 
 expressions squares. Substitute this value of y in the second 
 expression, and we shall have 16v 2 -j-33r-|-16, which must 
 be made a square. Hence put 16^ 2 -)-33v-|-16=(4 — fv) 2 , 
 
 which reduced gives v= 2 J L Take ^=5, then v= 7 ^ 
 
 Now take #=9, and we have 9, 73, and 328 for the numbers. 
 
 7. Find two whole numbers whose sum shall be an inte 
 
 8 
 
H4: KEY TO ALGEBRA. 
 
 gral cube, and the sum of their squares increased by thrice 
 their sum shall be an integral square. 
 
 Let x-\-y=n 3 , that is some cube. Then x 2 -\-y 2 -\-3n 3 =z 
 □ . Put 2xy=3n 3 , then x 2 -\-2xy-\-y 2 is a square, whatever 
 may be the values of x and y. But x and y must conform 
 to the equations x-{-y=n 3 , and 2xy=3n 3 . Work out the 
 value of x from these equations, on the supposition that n is 
 known, and we shall find 2x=?i 3 -\- t J \n 6 — 6ft 3 .} 
 
 Now x will be rational, provided we can find such a value 
 of n as shall render n 6 — 6n 3 a square, but if we add 9 to 
 this, we perceive it must be a square, and we have two 
 squares, which differ by 9. Therefore one must be 16, the 
 other 25, as these are the only two integral squares which 
 differ by 9. Hence n 6 — 6?i 3 +9 = 25. Or, ?i 3 —- 3 == 5. 
 n 3 =8, 7i=2, and #==6, y=2. 
 
 8. Find three numbers, such that their sums, and also the 
 sum of every two of them, may all be squares. 
 
 Let x 2 — 4a; = the first, 4x= second, and 2#-|-l= third. 
 By this notation, all the conditions will be satisfied, except 
 the sum of the last two. That is Qx-\-l must be a square, 
 but fro have three differe?it whole numbers, no square will 
 answer under 121, the square of 11. Hence put 6x-\-l = l21. 
 Or, a?=20. And the numbers will be 320, 80, and 41. 
 
 9. Find two numbers, such that their difference may be 
 equal to the difference of their squares, and the sum of their 
 squares shall be a square number. 
 
 Let x and y be the numbers. Then x — y=x 2 — y 2 . 
 Divide by # — y, and l — x-\-y. Hence a? = l — y, and 
 z 2 -f-y 2 =l — 2j/-f-2y 2 . Which last expression 1 — 2y-\-2y 2 
 must be made a square. For this purpose, put 
 
 1— %+2# 2 =(l— nyf. Hence y=- J|^ • 
 
 Take n any value to render y less than one in order to 
 
DIOPHANTINE ANALYSIS. ^5 
 
 give x a positive value. Therefore take w=3, and y=z 4 
 
 Consequently x= f , answer. 
 
 10. Find three numbers in geometrical progression, such 
 that if the mean be added to each of the extremes, the sums 
 in both cases shall be squares. Ans. 5, 20, and 80. 
 
 11. Find three numbers, such, that their product increased 
 by unity shall be a square, also the product of any two in- 
 creased by unity, shall be a square. Ans. 1, 3, and 8. 
 
 Assume 1 for the first number, and x and y for the other 
 
 12. Find two numbers, such that if the square of each bo 
 added to their product, the sums shall be both squares. 
 
 Ans. 9 and 16. 
 
 13. Find three integral square numbers in harmonica! 
 proportion. Ans. 25, 49, and 1235. 
 
 14. Find two numbers in the proportion of 8 to 15, and 
 such that the sum of their squares shall be a square number. 
 
 Ans. 136 and 255. Bonnycastle's answer, 476 and 1080. 
 
 15. Find two numbers such that if each of them be added 
 to their product, the sums shall be both square. , 
 
 Ans. i and J. 
 The above require no explanation from us. 
 
 There are many severe and tedious problems in the Diophan- 
 tine Analysis, proposed by Bonnycastle, Young, and others, 
 which require more time and practice than algebraists in 
 general ought to give for the advantage derived, as time and 
 thought may be better employed in Analytical Geometry, 
 the Calculus, or Astronomy. 
 
116 KEY TO ALGEBRA. 
 
 THE DIOPHANTINE ANALYSIS. 
 
 The Diophantine Analysis is sometimes useful in solving 
 numerical Equations, in which squares and cubes are in- 
 volved, as the following examples will show. 
 
 1. Given j ^__ _ 7 \ to ** n( * one va * ue °^ x an< * V' 
 By subtraction, and the transposition of y 2 we have 
 
 x 2 +x-\-y=y 2 (a) 
 
 As the second member of this equation is a square, the 
 first must be a square in fact, if not in form. 
 
 But, we perceive, that if we put y=x-\-\, in the first 
 member, it will be in a square form. 
 
 Put this value of y, in the first equation, and we have 
 x*-\-x=6; which gives #=2; hence y=3; values 
 which verify both equations. 
 
 N. B. This method of operation is not general. It 
 only serves to resolve particular cases. We might have 
 made the first member of equation (a), a square, by putting 
 y—3x-\-4, or 5#-f-9; but the results of these substitutions 
 would not verify the primitive equations. 
 
 2. Given \ 2x *— 3x y+ 2/ 2==4 ? to find values of x and v. 
 
 \ # 2 +3y 2 — 2xy=9 ) 
 
 As 4 and 9 are squares, the first members are square in 
 fact, though not in form. But we can make the first mem- 
 bers square in form, by assuming 
 
 2x 2 —3xy =0, and 3y 2 — 2#i/=0. 
 
 Then y 2 =4 and # 2 =9, or i/=2 and #— 3 ; va- 
 lues which verify all the equations. 
 
DIOPHANTINE ANALYSIS. 117 
 
 3. Find such integral values of x, y, and z, as wiil verify 
 the equations .... x i -\-y iJ txy=31 i 
 
 and . . x 2 -)rz i -{-xz =49. 
 If we add xy to the first equation, and xz to the second* 
 the first members will be square ; and, of course, the second 
 members will be square in fact, though not in form. 
 
 We have, then, to make 37-f-#y> and 4§-{-xz t squares, 
 *o accomplish this. 
 
 Put 37-ftft/=49, or xy=12 (1) 
 
 and 49~\-xz=64, or xz=\5 (2) 
 
 12 15 
 
 From (1), • . a?= — ; from (2) . . x=— . 
 
 y * 
 
 Hence « • 12z=15y, or, z=— . 
 
 Take y=4, then z=5, and x=3 ; values which will 
 verify the given equations. 
 
 4. Find such integral values of y and z as shall verify 
 the equation .... *y 2 -]-z 2 -\-yz=6l. 
 
 Add yz to both members, then put 6l4-y*=n2. 
 
 Now if we assume n=8, yz=3. 
 
 But yz—3 will give y-\-z=8, and these two equations 
 will not give integral values to y and z. Therefore, take 
 n=9, then w 2 =81, t/z=20, and y-f-z=9. Hence z=4 
 or 5, and y=5 or 4. 
 
 5. Find such values of x and y as will verify the equa- 
 tions xy-\-xy 2 =l2 . (1) 
 
 and. ... x +xy*=\8 (2) 
 
 12 
 
 Equation (1) may be put into this form y*— — — y. . .(3) 
 
 * # 
 18 
 
 Equation (2) into this y 8 = 1...(4) 
 
 x 
 
 The first member of equation (4) is a cube ; therefore 
 
 Put 11-13=8. Whence z=2. 
 
 x 
 
118 KEY TO ALGEBRA. 
 
 r - y-\-u-{-y 2 u 2 =^2l [to find one value of each 
 
 | x-\-v-\-2?v 2 =41 | of the symbols. 
 
 A regular solution would result in a very high and tedi- 
 ous equation ; but if the values are integral, we can soon 
 determine them as follows : 
 
 Take the first equation, and put v-\-u=s, and transpose 
 s. Then vhi 2 =l3 — s; which shows that 13 — s must be 
 some square; and if s is positive, the square cannot be 
 greater than 9. That is, 13 — #=9 or, s=4. 
 
 Then v-f-w=4, and vu=3 ; giving v=l or 3, and 
 t/=3 or 1. By taking u=l 9 in the second equation, we 
 find y=4. With the values already found we obtain x, 
 from either the third or fourth equations, v=3, w=l, 
 a?=2, y=4. 
 
 7. Given \ ^~~~ ( to find values of x and y. 
 
 { 2?/ 2 -f-3.ri/= 8 J * 
 
 Put xy=p ; transpose, &c, we have 
 
 4# 2 =12-{-2;) and 4i/ 2 =16— 6p. 
 
 Now we must find such a value of p as shall render the 
 second members of these last equations square at the same 
 time; which is p=2; this gives 4x*= 16, x=2. 
 
 ft p. C 6x 2 -\-2y 2 =5xy -f- 12 ) to find one value of 
 
 ( 3x 2 -\-2xy=3y z — 3 ) x and y. 
 This problem is under (Art. 110) of both editions. 
 Ac(^ the equations together, and reduce, and we have 
 
 9x 2 =y 2 -\-3xy-{-9. 
 The first member of this equation is a square ; therefore the 
 second member is a square, but to make it a square inform, 
 as well as in fact, we perceive it is only necessary to make 
 
MISCELLANEOUS EXAMPLES. H9 
 
 07=2. Or call y 2 -\-3xy-\-9 a binomial square, and decide the 
 value of x agreeable to section 8, This gives a?=2, the 
 answer. 
 
 q p- 5 ^ 2 — *'/= 28? to find the rational values 
 ' ° 1V ^4y 2 4-3ary=160 5 of a? and y. (Y. 139.) 
 
 Ans. x = ±4, y= ±5. 
 There are, and may be equations of the preceding forms 
 which have no rational values, such are not susceptible of 
 of this mode of treatment. 
 
 
 OF THE 
 
 \ 
 
 Section XXIIli UNI VERS IT 
 
 Miscellaneous Exampl 
 
 %IP0R^' 
 
 1. When wheat was 8 shillings a bushel and rye 5, a 
 man wished to fill his sack far the money he had in his 
 purse. Now if he bought 15 bushels of wheat and laid out 
 the rest of his money in rye, he would want 3 bushels to 
 fill the sack: but if he bought 15 bushels of rye, and then 
 filled his sack with wheat, he would have 15 shillings left. 
 How much of each must he purchase to fill his sack, and 
 lay out all his money ? 
 
 (Colburn, page 50.) Ans. 10 bushels of each. 
 
 Solution by Mr. T. J. Matthews. 
 
 Let x= the wheat, and y= the rye. Then it is evident 
 that when he buys 15 bushels of wheat he has too much, as 
 he has not money enough left to fill his sack with rye. Now 
 15 — x is the excess of the wheat purchased above what he 
 ought to have had, and this excess of quantity, multiplied by 
 the excess of a bushel of wheat above one of rye, wiJJ give 
 the deficiency of his money, or equal to 3 bushels of rye at 
 5 shillings. Consequently, 3(15 — a?) =15, or #=10. 
 
 By similar reasoning, it will appear that when 15 bushels 
 of rye are purchased, he buys too much, and the excess is 
 15 — y, which multiplied as before, will equal the excess of 
 
120 KEY TO ALGEBRA. 
 
 his money, viz: 15 shillings. Therefore 3(15 — y)=15, 
 and y= 10. 
 
 2. 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 1 (Colburn.) 
 
 Solution by T. J. Matthews. 
 Assume 5z, and 4z the sides of the Cubes. 
 Then 25a; 3 — 16^=9, by the first condition. 
 Therefore, x=h 1254=500. 64-5=320. 
 3 Or, £26. £IQ. Answer. 
 
 3. Given x-\-y-{~z=25. ary=6 #2=60, to find #, y 
 and z. 
 
 Solution. From the two latter, z—lOy. Then the first 
 becomes x-\-lly—25. Or z 2 -\-Uxyz=2ox, but from the 2d, 
 ll2?y=66. Hence ^—-25^=— 66. (A) 
 
 Assume 2«= — 25, then 6&-f-9= — 66, and equation (A) 
 becomes (z 2 -{-2ax-{-d i =a 2 -{-6a-{-9. Or, a?=3,) 
 
 4. Given a^=125#-|-300y. And y 2 — x 2 =90000, to find 
 * and y. (Young p. 146.) Ans. #=400. y=500. 
 
 Put y =px, then px 2 = 1 25#+300^#. ( 1 ) 
 
 And p 2 y 2 —x 2 =(300) 2 (2) 
 
 125 
 From equation (1) p= ~~^ (3) 
 
 (300) 2 
 From equation (2) p 2 — l = v — — 
 
 THe right hand side of this last equation being a square 
 the other side is also a square, and one accustomed to the 
 analysis will perceive that p must equal f, to make the ex- 
 pression p* — 1 = D- Others can go through the form and 
 they will find that p=l, which value put in equation (3) 
 gives a?a=400. 
 
MISCELLANEOUS EXAMPLES. 121 
 
 It is not imperative that we should resort to the Diaphan- 
 tine to solve this problem but it is very Convenient. 
 
 5. Find two numbers, such that the fifth power of one 
 may be to the cube of the other, as 972 to 125. 
 
 Ans. 6 and 10. 
 
 Let a? and nx be the numbers. 
 
 Then a; 5 : n 3 x 3 : : 972 : 125. Or, x 2 : n 3 : : 972 : 125. 
 
 Multiply the first and third terms by x, x 3 : n 3 : : 972a: 
 
 : 125. 
 
 1 25x 3 
 Therefore, 972*=— f-. 
 n 3 
 
 The right hand side of this equation is a cube, therefore, 
 972#= a cube: Or, 27*36x=a cube. Hence, 36a? must be 
 a cube, which it evidently is, when #=6, as 36 is the square 
 of 6. 
 
 6. Given x +i,+xy(x+y)+r* y 2 =85 ) fi , , 
 And xy+{x+?jy+xy(x-\-y)=97 \ V ' 
 
 Young. 145. Ans. x=Q y=l. 
 Put (x-\-y)=s xy=p, then the equations become 
 
 s-\-sp+p 2 =85 } 
 And p-\-sp-\-s 2 =97 $ 
 
 By addition (s-\-p)+s- -\-2sp-\-p 2 =\82. (1) 
 
 Assume Q,=s-{-/>, then equation (1) becomes 
 Q,2_|_ a==18 o Hence Q= 13. Or, s+p=l3. The re- 
 maining steps are obvious. 
 
 12 
 
 7. Given l Jx-\-l2= — -=, to find x. 
 
 144 
 Square x-{- 12=^= -£-. Put .r-f-5=y. 
 
 144 
 
 Then y+7= . Or, ^+7^=141 
 
 Put 2&=7. Then 18a-f 81 = 144. 
 Hence y>+2ay+a*=a2+l8a~)-8h 
 y-{-a=a-(-9. Or, — a— 9. 
 
122 KEY TO ALGEBRA. 
 
 & Given J ( a? +y)(«y+l)=18^- l to find z and 
 
 o. driven, j ( a?2 _jl y 2> ) ^2 ?/2 _^ 1)=208a , 2y 2 J y 
 
 Ans. o:=2±V 3 - 0r ?±V 3 - 
 y=7=fc4«/3. Or2d=V3. 
 
 Solution. Take x-\-y=s. xy-\-l = t and xy=p. 
 
 Thenst=lSp. (1) And (s 9 —2p)(t 2 —2p)=20Sp 2 . (2) 
 
 Multiply as indicated and take the value of s't 2 from 
 
 equation (1) and we have 324/— 2p (s 2 +2 2 )+4/=208/. 
 
 „ , . **4-* a . st 
 
 By reduction, p= — ' - — . From equation (1) p=z—^. 
 
 Therefore, ~*~ =-. Assume s—nt. 
 
 00 lo 
 
 Then this last equation becomes, by a little reduction, 
 n 2 +l » „ '" 1 
 
 HO =3- Hencew = 3 - 0r '3- 
 
 This establishes a relation between x -f- y and #y-|-L 
 
 Another Solution by Charles E. Matthews. 
 
 Multiply the original equations as indicated, and 
 
 x 2 y-\-xy - -\-x-\-y= 1 8xy. 
 
 x*y 2 -{-x 2 y A ^-x 2 -\-y 2 =2Q8x 2 y 2 . 
 
 Divide the first by xy, the 2d by x 2 y 2 y then 
 
 x + y +l +1 =ia And x2 +y*+^+}*= 208 - 
 
 Assume x-\ — =??i. And y-A — =?j. 
 ^x s ^y 
 
 Then w+w=18. And m 2 -\-n 2 =212. 
 
 From which »i and ?i are easily found, afterwards x 
 
 and y. 
 
 9. A square public green is surrounded by a street of uni- 
 form breadth. The side of the square is 3 rods less than 9 
 times the breadth of the street: and the number of square 
 rods in the street exceeds the number of rods in the perime- 
 ter of the square by 228. What is the area of the square ? 
 
 (Day 307.) Ans. 576 rods. 
 
 10. A man wishes to purchase a certain number of acres 
 
MISCELLANEOUS EXAMPLES. 1x3 
 
 of land for the money he has at his command. Cleared land 
 is worth 10 dollars per acre; uncleared land is worth $8.-— 
 He finds that if he buys 120 acres of cleared land, and lays 
 out the rest of his money for that which is not cleared, he 
 will not get the quantity of land he wants by 25 acres, but, 
 if he buys 220 acres of uncleared land, and then buys a suf- 
 ficient number of acres of cleared land to make up the num- 
 ber of acres he wants, he will have 4 dollars left. How 
 many acres of each must he buy to have the quantity he 
 wishes, and lay out all his money? (Harney page 203. 
 Ans. 20 acres cleared, 218 uncleared. 
 N. B. Call to mind problem first of this section. 
 
 In working geometrical problems algebraically much la- 
 bor may be saved by paying attention to the relation oi the 
 given numbers. 
 
 We give the following as illustrative of these remarks. 
 
 1. If the perimiter of a right angled triangle be 720, and 
 the perpendicular falling from the right angle on the hy- 
 pothenuse be 144 ; what are the lengths of the sides ? 
 
 (Day Alg. p. 305.) Ans. 300, 240 and 180. 
 
 If we use the identical numbers given 144 and 720, as 
 nine-tenths of our teachers do, they will give large and tedi- 
 ous equations, but if we compare 144 and 720, we shall 
 perceive that one is exactly 5 times the other, and consider- 
 ing the nature of similar triangles, we can work on one of 
 only 144th of the linear dimensions of the first, or a triangle 
 whose perimiter is 5, and perpendicular from the right 
 angle 1. 
 
 Solution. Let x and y be the two sides, then 5 — x — y will 
 be the hypothenuss. 
 
 And x*-\-y*=(o — x — y) ? , and xy=5 — x — y. 
 
 Each member of this last equation expresses the double 
 area of the triangle. Put x-\-y=s. xy=p. 
 
124 KEY TO ALGEBRA. 
 
 Then $ 2 _2/?=(5— s) 2 =25— 10s+s 2 , and ^=5— *. 
 Or, 105 —2^=25 
 And 25 +2j5=10 
 
 By addit'n 12s =35. _ 35 
 
 0r ' 5 =12' 
 
 But s=x-\-y, the sum of the two sides which, taken 
 
 from 5, or — >, gives— , for the hypothenuse of the small 
 
 25 
 
 triangle, hence by 144=300, the hypothenuse of the large 
 
 triangle. 
 
 2. The sum of the two sides of a plane triangle is 1155, 
 the perpendicular drawn from the angle included by these 
 sides to the base, is 300 ; the difference of the segments of 
 the base is 495, what are the length of the three sides ? 
 (Day 305.) Ans. 945, 375, 780. 
 
 Write the given members in order, thus 300, 495, 1155. 
 Divide them by 15, and their relation is 20, 33, 77. 
 
 The two latter numbers have a common factor 11, which 
 call a. Put 6=20. 
 
 Then the three given lines will be b, 3a, and la. Let x= 
 the less side, la — z=the greater side, #=the shorter seg- 
 ment, and g-[-3a—the longer segment of the base. 
 
 Then p*-\-b*=xx*. (1 
 
 And y 2 -\-Qa?j-\-9a 2 -\-b 2 =49a 2 ^Uax+x 2 . (2) 
 
 Subtract (1) from (2), drop 9a 2 from both sides, and di 
 vide by 2a, and 3y=ab — Ix. (3) 
 
 We write 2b in place of 40, after dropping 9a 2 . 
 
 From the square of (3) subtract 9 times, equation (l,)and 
 we have — 9b 2 =a 2 b 3 — 14 abx-\-2bx*. 
 
 Divide by b, afterwards by 2, recollecting that 6=20, and 
 
 we have — 90=10& 2 — lax-\-x 2 . 
 
 9a 2 
 Add —j-, to both sides to complete the square. 
 
MISCELLANEOUS EXAMPLES. 125 
 
 Then?|— 90= ® (a»-40)==®-81 = ^->tax+x>. 
 
 Extract square root- "9= — x. 
 
 Or, a?=25. Then 25. 15=375. 
 
 3. Divide the number 74 into two such parts that the 
 difference of the square roots of the parts may be 2. 
 
 Ans. 25 and 49. 
 Let x — 1 , and x-\-\ be the square roots, of the two parts. 
 This problem can also be solved by the Diophantine analysis. 
 
 4. Given £*+# 2 =45 and— -) — =£, to be solved by the 
 
 x y 
 
 Diophantine analysis. Ans. x—6. y=3. 
 
 5 Given a? 2 -f-# 2 =45 and (x-\-y)x=54 to find x and y by 
 the Diophantine analysis. Ans. x=Q. y=3. 
 
 The two preceding should also be worked by common 
 algebra. 
 
 6. A and B traveled on the same road, and at the same 
 rate, from Huntingdon to London. At the 50th mile stone 
 from London, A overtook a drove of Geese, which were 
 proceeding at the rate of 3 miles in 2 hours, he afterwards met 
 a stage wagon, which was moving at the rate of 9 miles in 
 4 hours. B overtook the same drove of Geese at the 45th 
 mile stone, and met the same stage wagon exactly forty mi- 
 nutes before he came to the 31st mrle stone. Where was B 
 when A reached London 1 
 
 Solution. Let #= miles traveled by each per hour, and 
 y= distance B was behind A, then 50 — 2x= the distance 
 from London where A met the wagon. 
 
 Also,~- J g__ — — time elapsed between the meeting 
 
 9y 
 of the wagon with A and B, therefore -.—- ?-x= distance tra- 
 
 4 l r-f-9 
 
 veled by the wagon during this time, consequently 
 
 9?/ 
 50 — %x-\- d _. =distance of the wagon from London,when 
 
126 KEY TO ALGEBRA. 
 
 2a? 
 met by B. But this distance is also =31-}- ~ therefore 
 
 
 
 50_2x+ 4 -^=31+| : . Again y+5 : 5 : : » : |, 
 
 whence 3y-|-15=10a? and y= , substituting and 
 
 reducing, we get the quadratic 
 
 . 123a: 378 xktL 
 
 x 2 = -s— . Whence a?=9, consequently y=25. 
 
 7. Given a? 2 -f-a?y=77, and xy — y 2 = 12, to find a? and y, 
 by the Diophantine analysis. Ans. x—7. y=4:. 
 
 8. Three equal circles touch each other externally, and 
 enclose between the points of contact a acres of ground, 
 what are the radii of the circles ? 
 
 A Ko,i6ik) i 
 
 9. A person has £21 65. in guineas and crown pieces, out of 
 which he pays a debt of £14 17s., and finds that he has ex- 
 actly as many guineas left as he has paid crowns away ; 
 and as many crowns as he has paid away guineas ; how 
 many of each had he at first ? 
 
 Ans. 9 crowns paid away ; 12 guineas paid away. 
 Suppose a?= the guineas paid away. 
 And y=- the crowns paid away. 
 Then 21a?-|-5#=297= amount paid out { . 
 
 And 5a? +21^=249= amount on hand \ P er <l uestlon - 
 Add these equations and divide by 26, &c. 
 
 10. The sum of three numbers in harmonical proportion is 
 191, and the product of the first and third is 4032. What 
 are the numbers ? Ans. 72, 63, 56. 
 
 11. Is it possible to pay £50 by means of guineas and 
 three shilling pieces only. Ans. Impossible. 
 
 12. A merchant drew every year upon the stock he had in 
 trade, the sum of a dollars for the expense of his family. 
 
 His profits each year, were the nth part of what remain- 
 
10. Given \ V( 5 V*+5V</)+V*+Vy=l<> ) to find 
 
 MISCELLANEOUS EXAMPLES. 127 
 
 ed after this reduction, but at the end of the 3d year he finds 
 
 his stock exhausted; how much had he at the beginning; 
 
 a(3n 2 +3n+l) 
 
 Ans. . 
 
 (n+1) 2 
 
 And a?*+y*=s275 ) x and V* 
 
 Put Jipjx+bjyj^n. Then the first equation becomes 
 
 n 2 
 
 .-- \-n=\0. Which equation gives n=5. 
 
 Whence Jx-\-,Jy=5. 
 
 From this last and the 2d equation, we find x=9, and 
 t/=4. 
 
 The following are from Bland's Problems, and involve 
 
 equations only of the second degree. They are too severe 
 
 for learners, but we are tempted to leave one or two of them, 
 
 without solution, for the benefit of those who deem keys 
 
 unnecessary. More of like character might be given. 
 
 18 7 
 
 x = . Ans. #=16 or 1. 
 
 * Jx— 2 
 
 2x*(x*-\-cc i y=2x i (x-\-2a)-\-a 2 (x— a). 
 
 Ans. x=\a or — a. 
 
 *-f(«V— n+2*)* =7+2^- y* ; 
 
 (3i/ — #+7 )* = ^. Ans. *=4, y=2, 
 x—y 
 
 Double the first equation, transpose — 2y 2 , and subtract 1 1 
 from both members, then we have 
 
 2y 2 +2x— ll+2(3i/ 2 — ll+2#)^=3+4y. 
 Add y l to both members, and conceive the terms in the vin- 
 culum to be P ; then 
 
 P 2 +2P=3+4y+y«; 
 Or, P+l=2+y. 
 
128 KEY TO ALGEBRA. 
 
 By restoring the value of P, and reducing, we have 
 a?=6-f-2/ — y 2 . Put this in the 2d eq., &c. 
 i 
 4. Given x 2 y — 4=4x*y — |?/ 3 , 
 
 and x 2 — 3=#y ( x a — y* J, to find x and y. 
 
 Ans. fl?=l, y=4. 
 
 Put # =P, y a =Q, and we have 
 
 P4Q2__ 4 = 4PQ2_|Q6 1 (!) 
 
 P3_ 3 =PQ(P— Q), (2) 
 
 Now put P=nQ, and eq. (1) becomes 
 (4rt 4 -r-l)Q 6 — 16rcQ 3 =16. 
 Conceive n to be a known quantity, then the last equa- 
 tion is quadratic, and a solution gives 
 
 4(2tt 2 -r-2tt-r-l) _ 4 _ 4 
 
 4n 4 -p-l 2m 2 — 2/i-f-l 2n*— 2n-f2— 1* 
 
 But from (2), Q 3 = 3 3 2 , =-— L-— 
 
 Put the two values of Q 3 equal, and put n 2 — n-f-l=R, (3) 
 
 Th6n 2^-I=i Whence 2 „= 6 |=? (4) 
 
 But from (3), resolved as a quadratic, 
 
 2n=l±V4R^3 (5) 
 
 From (4) and (5), 2R±2R^/4R~ 3~=6R— 3 ; 
 Or ±2R74R— 3=4R— 3. 
 
 Put V4R— 3=S, 
 
 Then S 2 ±2RS=0, Or, S(S=b2R)=0. 
 
 This last equation may be verified by taking either factor 
 equal to zero ; and as the first factor only gives a rational 
 quantity, we take that which gives R=f . 
 
 By retracing, we easily find x and y. 
 
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