Question

Solve the simultaneous equation and find the value of y

$\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x+y-12}{11}$

Answer 1

Consider two equations,

$\frac{x+y-8}{2}=\frac{3x+y-12}{11}$ ............(1)

$\frac{x+y-8}{2}=\frac{x+2y-14}{3}$ .............(2)

After solving (1) we get,

$5x+9y=64$ ..............(3)
After solving (2) we get,
$x-y=-4$
or
$9x-9y=-36$ ...............(4)
Adding (3) and (4) we get,
$14x=28$
$\therefore x=2$
Re substituting $x=2$ in (3) we get,
$y=6$