Question

In the given equation $\frac{x+3}{5}=\frac{8-y}{4}=\frac{3(x+y)}{8}$ values of x and y will be

• A. $4$ and $-8$
• B. $10$ and $-12$
• C. $8$ and $-4$
• D. $12$ and $-4$

Answer 1

The correct option is D $12$ and $-4$

Suppose $\frac{x-3}{5}=\frac{8-y}{4}=\frac{3(x+y)}{8}=k$
$\therefore \frac{x+3}{5}=k\Rightarrow x=5k-3$ .......(i)
$\frac{8-y}{4}=k$
$\Rightarrow y=8-4k$ ....(ii)
and $\frac{3(x+y)}{8}=k$
$\Rightarrow 3(x+y)=8k$ ......(iii)
On substituting the value of x and y from equation (i) and (ii) in equation (iii),
$3(5k-3+8-4k)=8k$
$\Rightarrow 3k+15=8k\Rightarrow 5k=15\therefore k=\frac{15}{5}=3$
On substituting this value of k in equation (i) and (ii),
$x=5\times 3-3=12$
and $y=8-4\times 3=-4$.