SOLUTIONS TO THE EXERCISED] 



MATHEMATICS. ALGEBRA. 



Or, transposing, 9x 38y = 3 ... (1) 

 x y = 10 ... (2) 

 ... Q x 9,j = 90 ... (3) 

 Subtracting (1) from (3), 29y = 87, .'. y = 3 

 . . x 10 + y = 13. 



I O "^ 



6. _ + 8i/ = 31 Clearing the fractions, we have 



Or, x + 24y = 91 



4Oe + v = 763. 

 Multiplying the first by 40, 



Subtracting, 



959y = 2877,.-.y="g 7 = 3 



763-y 760 _ 

 *" ~40~ = 40 - 



7. ax -f- by = 1 1 Equalising the coefficients of z, we 

 a'x + fc'i/ = 1 J have 



aa'x + a'by = a 

 aa'x + db'y = a 



Subtracting, (ab r a'% = a a' 

 a a' 



' y= 5/~ 



In like manner equalising the coefficients of y, 

 ab'x + b'by = I' 

 a'bx -r b'iy = b 



Subtracting, (a'6 a&> = 6 6' 



6 '' 

 "* 



8 ax + l>y = e \ Equalising the coefficients of z, we 

 a'x + fc'y=c'; have 



aa x + a^y = a'c 

 aa'x -j- ai'i/ = ac* 



Subtracting, (ab' a'6)y = ac' a'c 

 at? a'c 



i/=a 6 cr ( ,v 



lu like manner equalising the coefficients of y, wo have 

 ab'x + b'by = b'c 

 afbx + b'by = W 



Subtracting, (a'6 aV\x ~W l'c 



be b'c 

 *-?6_,. 



' ~* 12 Multiplying the first equation 

 iby 3, ami then 

 second, we have, 



Multiplying the hrst equation 

 >by 3, ana then subtracting the 



'J 

 _ 1 ,= 36 2? = 33f 



y 



Multiplying by 1y, 56 = 233y, .'. V = 233' 



1 ^.L 1 12 233 _ 439 x - 50 

 Also, - - 12 + - = 12-- ^ - w , . 



10 x 2 10-* y 10 = ol _, 



10. ^ j Clearing frac- 



> tions, these equa- 

 : tions become, 



m 24 200 + 20/ lay + 150 = 

 10i/ + 32 Gx 3y 6x 78 = 0. 

 Transposing, 32x 15y = 74 



. 



Multiplying tho first by 3. and the second by 8, we have, 

 46i = 222 



368 



Adding, ' = 590, .'.y = 10; 



and since 13y 46 = 12*, . . 130 46 = 12*, 



7. 



11. 



x+6_y+2 



ff i ^ y 2 I Applying to those fractions tho 



^..3 y 7 r general principle at page 468, wo 



..- = ^5 have 



x ~|- x y J-O j 



^=|,.-. 2x + 10 = y 



^3_V-10 ... 3x + 9 = y-10. 



1 3 



Equating the two expressions for y, we have 

 Zx. + 19 = 2x + 10, .-. x = 9 

 . . y = 2x + 10 = 18 + 10 = 8. 

 12. x + y = 191 Dividing the second equation by 

 y? y 2 = 95) the first, we have 



x y = 5 

 Also, s + y = 19 



.-. adding and subtracting, 2x = 24, and 2y = 14, 

 .-. x = 12, andy = 7. 



QUESTIONS rs SIMPLE EQUATIONS. PAGE 472. 



1. Find a number such that if it be increased by one- 

 half, one-third, and one-fourth of itself, the sum shall 

 be GO. 



Let x be the number ; then by the question, 



+T+f+T-* 8a 



Multiplying by 12, to clear the fractions, 

 12x + Gx + 4* + 3* = 600, 

 that is, 25-e = COO, . . x = 24. 



Othertcise. To avoid fractions, put 12x for the num- 

 ber ; then by tho question, 



12z + Gx + 4x + 3x = 50, 

 that is, 25x = 50, . . x = 2, .'. 12x = 24. 



2. There is a fraction such that if 4 be added to tho 

 denominator, tho value is } ; and if 3 be added to tho 

 numerator the value is J : required tho fraction. 



x x 



Let tho fraction bo - ; then by the question, . ^- 



y + 4 



- = 



4' ~^ -- 4' 



therefore clearing the fractions, 4j; 



Subtracting 



12 2y ! , 



Also ae = | 



3, /. tho fraction is 



3. What number is that, such, that if it bo increased 

 by 7, the square root of the sum shall bo equal to tho 

 square root of tho number itself and 1 more ? 



Let x be the number, then the condition is that 



Squaring, x + 7 - 



Transposing 6 = 2V x, .'. 3= J x, .'. 9 = x ; 



hence the number is 9. 



4. Fifty labourers are engaged to remove an obstruc- 

 tion on a railway ; some of them are, by agreement, 

 to receive ninepence each, and tho others fifteen pence. 

 Just 2 are paid to them ; but, no memorandum having 

 been made, it is required to find how many worked for 

 Od., and how many for 15ti. 



Suppose there are x workmen at Qd., and y at 15rf , 

 then the number of pence received by tho former is 9x, 

 and the number received by the latter I5y ; and since 

 by the question the number of pence paid altogether 

 is 2 X 20 X 12 = 480, wo must have the equation 



9j; 4. 15y = 480 



Moreover, x + y = 50 by the question. 

 .-. 9^ + 9y = 450 



Subtracting, 6i/ = 30, /. y = 5, /. x = 45. 

 Consequently, 5 of the labourers received 15 d., and 45 

 received 9d. 



Othtrivise. Let * be the number of labourers at 9<( , 

 then by the question 50 x was tho number at 15/ ; 

 consequently the former received 9.6 penco, and the 



