Question

On solving $\frac{25}{x+y}-\frac{3}{x-y}=1$, $\frac{40}{x+y}+\frac{2}{x-y}=5$, we get

• A. $x=8$, $y=6$
• B. $x=4$, $y=6$
• C. $x=6$, $y=4$
• D. None of these

Answer 1

Answer: (C)

$x=6$, $y=4$

$\frac{25}{x+y}-\frac{3}{x-y}=1$ ......(i)
$\frac{40}{x+y}+\frac{2}{x-y}=5$ ......(ii)
Multiply equation $\left(i\right)$ by $2$ and equation $\left(ii\right)$ by $3$
$\Rightarrow \frac{50}{x+y}-\frac{6}{x-y}=2$ and $\frac{120}{x+y}+\frac{6}{x-y}=15$

Adding the new obtained equations, we get

$\frac{170}{x+y}=17$
$\Rightarrow x+y=10$ ......(iii)
Substitute $x+y=10$ in equation $\left(i\right)$
$\frac{25}{10}-\frac{5}{x-y}=1$
$\Rightarrow \frac{3}{x-y}=\frac{25}{10}-1$
$\frac{3}{x-y}=\frac{15}{10}$
$x-y=2$ ........(iv)
Adding equation $\left(iii\right)$ and $\left(iv\right)$
$2x=12$ $\Rightarrow$ $x=6$
Put $x=6$ in equation $\left(iii\right)$
$6+y=10$ $\Rightarrow$ $y=4$
So, the solution is $x=6$ and $y=4$