Question

Solve for $x$ and $y$:

$\frac{y+7}{5}=\frac{2y-x}{4}+3x-5$, $\frac{7-5x}{2}+\frac{3-4y}{6}=5y-18$

• A. $x=3;y=12$
• B. $x=16;y=8$
• C. $x=1;y=7$
• D. $x=2;y=3$

Answer 1

The correct option is D $x=2;y=3$

Given equations are,

$\frac{y+7}{5}=\frac{2y-x}{4}+3x-5$

$\Rightarrow 4(y+7)=5(2y-x+12x-20)$

$\Rightarrow 4y+28=10y+55x-100$

$\Rightarrow 55x+6y=128$ ....(1)

and $\frac{7-5x}{2}+\frac{3-4y}{6}=5y-18$

$\Rightarrow 3(7-5x)+(3-4y)=6(5y-18)$

$\Rightarrow 21-15x+3-4y=30y-108$

$\Rightarrow 15x+34y=132$ ....(2)

Multiply (1) by 3 and (2) by 11, we get

$165x+18y=384$

$165x+374y=1452$

Subtracting above equations we get

$356y=1068⟹y=3$

Put value of y in (1)

we get $x=2$

$\therefore x=2,y=3$