Question

Solve the simultaneous equations and find the value of x. $\frac{x+y}{4}-\frac{x-y}{3}=1;\frac{x+y}{2}+\frac{x-y}{6}=12$

Answer 1

$\frac{x+y}{4}-\frac{x-y}{3}=1$ ...........(1)

$\frac{x+y}{2}+\frac{x-y}{6}=12$ .........(2)
Multiply (1) by $24$ and (2) by $3$ we get,
$-2x+14y=24$..........(3) {By solving (1)}
$2x+y=36$...............(4) {By solving (2)}
Adding (3) and (4) we get,
$y=4$;
Substituting $y=4$ in (3) we get,
$x=16$