Question

If $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$, then the values of $x$ and $y$ is

• A. $x=1,y=3$
• B. $x=5,y=2$
• C. $x=3,y=3$
• D. $x=2,y=6$

Answer 1

The correct option is D $x=2,y=6$

Given, $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$

By taking first two. we get,

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

$\Rightarrow 3x+3y-24=2x+4y-28$
$\Rightarrow x-y=-4$ ......(i)

By taking last two. we get,

$\frac{x+2y-14}{3}=\frac{3x-y}{4}$

$\Rightarrow 4x+8y-56=9x-3y$
$\Rightarrow 5x-11y=-56$ .....(ii)

On multiplying equation (i) by $5$. we get,
$5x-5y=-20$ ....(iii)

On subtracting (ii) from (iii). we get,
$5x-5y-(5x-11y)=-20-(-56)$
$\therefore 6y=36$
$\therefore y=6$

Substitute $y=6$ in (i). we get,
$x-6=-4$
$\therefore x=2$

$\therefore x=2,y=6$