-\frac { 1} { 2} x + \frac { 1} { 5} y = 9 −12x+15y=9----------(1)
and
7x - \frac { 1} { 3} y = - \frac { 2} { 3}7x−13y=−23------------(2)
Multiplying equation (1) by 10 and equation (2) by 3 to eliminate the fractional part:
- 5 x + 2 y = 90−5x+2y=90--------------(1') and
21x - 1 y = - 221x−1y=−2 -----------(2')
Eliminate any one variable (say yy):
(1') + (2') x 2 =>⇒
- 5 x + (2xx 21x) + 2 y + (2xx -1y) = 90+ (2xx -2)−5x+(2×21x)+2y+(2×−1y)=90+(2×−2)
=> -5x +42x +2y -2y = 90 - 4⇒−5x+42x+2y−2y=90−4
=> 37x = 86⇒37x=86
=> x = 86/37 = 2 12/37⇒x=8637=21237
Substitute this value of xx in (1') or (2') to get the value of yy
(2')=> 21 xx(86/37) - 1 y = - 2⇒21×(8637)−1y=−2
=> 1806/ 37 -y = -2 ⇒180637−y=−2
=> - y = -2 - 1806/37⇒−y=−2−180637
-y = (-74-1806)/37−y=−74−180637
y = 1880/37y=188037
Cross check the validity of x and yxandy by substituting in left hand side (LHS)of any of (1) and (2):
LHS of (1)=>⇒
-\frac { 1} { 2} x + \frac { 1} { 5} y = -1/2 (86/37) + 1/5 xx (1880/37)−12x+15y=−12(8637)+15×(188037)
= -43/37 + 376/ 37 = 333/37 =−4337+37637=33337
= 9 =9 = RHS (right hand side ) of equation (1)
therefore x= 86/37 and y = 1880/37 are the correct values.
