4"* 



MATHEMATICS. ALGEBRA. 



[SOLUTIONS TO THE EXERCISES. 



Jax a 3*/ax 2o 



11. i ~X~ "" <f^ Z Si' Applying tne lorogomg pnucipiu, ~ u 



Clearing, T Jax- 6Jax + 3a, .'. Vox- 3a 

 .-. ox 9oV-'. x 9a. 



12. J(f 24) Jx 2. Squaring each side, 



x 24 x 4 v/x + 4. 

 Transposing, 4 ^x - 28, .-. V x - 7, A x - 49. 



13. ^'(4x + 21) 2 ,/ x + 1. Squaring each side, 



4x + 21 4x + 4 V'x + 1. 

 Transposing, 20 4 ^x, . . Js, = 5, . . x = 25. 



14. V (2* + 3) = 3. Cubing each side, 



2x + 3 = 27, . '. 2x = 24, . . x - 12. 



15. _ " ^L* I_. The numerator of the first fraction is tho cEfforence of the squares of the 



two quantities in the denominator ; therefore, the numerator is actually divisible by the denominator, so that the 

 equation is the same as 



16 ' 



B, ..6x = 25, .'. = 5. 

 By the principle at page 468, 



SIMPLE EQUATIONS wmi Two UNKNOWN QUANTITIES. 

 EXAMPLES FOK EXERCISE. PAGE 471. 



1st. Proceeding by Rule I., we 



have from the first equation x ~ ^ ; and from the 



2 



second x = :L_y ; hence equating these two expres- 

 6 



lions for x, 



23 3y _ 10 -f 2y 

 2~ 6 



Clearing fractions, 115 15y = 20 + 4y, 

 Transposing, 95 = 19y, . . y = 5, 



6~~ '5' 



Substituting this in the first we get 



4 i 4y i 3 j._23 .'. * y - 

 6 ' 6 



Clearing, 4y + 15y = 95, . . 19y = 95, . . y = 5 ; 



__ 



5 6 



2nd. Proceeding by Rule II., we have from the second 



19, 



And since x - ^L^-f?, . . x - = 4. 

 5 5 



3rd. Proceeding by the third method, multiplying the 

 first equation by 2, tho second by 3, and then adding 

 tho results, we have 



19x = 76, .-.x = 4 



Again, multiplying the first equation by 5, tho second 

 i by 2, and subtracting tho results, wo get 

 19y 95, .'. y = 5. 



o &x + 4y 68 1 Multiplying the first equation by 3, 

 ' 3x + 7y - 67 J and the second by 6, there results 



15* + 12y - 174 

 lox + 35y = 335 



Subtracting, 



And since from tho first equation x = 1?, 

 _68 28 



Otherwise, thus: multiplying the first equation by 2, 

 we have 



3x + 7y = 67 



Subtracting, 7x+ y= 49, /. y = 49 7x 



Multiplying by 3, 21x + 3y = 147 

 First equa. 6x -j- 4y = 58 



Subtracting, 1 6r, + y = 89, .-. y = 16x 89 

 .-. 16x 89 = 49 7x, /. 23x = 138, .-. x = 0, 

 ... y = 49 ?x = 49 42 = 7. 



3. ? + 8y = 194 "I Clearing fractions, these equations 



I become 

 to -131 I + My -1652 





By adding these 



And by subtracting, 



2000, /. y+x 

 40 



504, /. v x 

 8 



65z + Goy 



_ 2G(JO 

 65 



C3y 63z 

 = 504 



63 



.'. adding and subtracting, 2y = 48, and 2x = 32, 



.-. y = 24, and x = 10. 



NOTE. The rule has been departed from in the pre- 

 ceding solution ; and you perceive that considerable 

 advantage on the score of neatness and simplicity has 

 been gained in consequence. By tho rule we should 

 have proceeded thtia : 



From the first equation x = 1552 64i/ 

 Substituting in the second, y + 64(1552 64y) 1048 ; 

 That is, 64 X 1552 1048 = (64 J l)y 

 64 X 1652 1018 _ 64x15621048 

 64 J 1 65X63 



" -21 



~6T9.77l3 



And x - 1552 6-ity = 1552 64 x 24 = 16. 

 4. ?_"hj^=:5") Clearing, these equations become 



I x + y = 15 



V x =1 f yx" 7 



7 J Adding and subtracting, wo get 



2i/ = 22, and 2x - 8, . . y - 11, x = 4. 



Sx 7y 2a; + y + i i 



it " I Clearing fractions, we have 



L 15.,; :!.->./ = 6r, + 3j/ + 3 



2 



.v 



10 x-f-y-0 



