Question

Solve by reducing the equation method.
$\frac{30}{x-y}+\frac{44}{x+y}=10$
$\frac{40}{x-y}+\frac{55}{x+y}=13$

Answer 1

Let $\frac{1}{x-y}=A$ and $\frac{1}{x+y}=B$

so, equation will be $30A+44B=10$

and $40A+55B=13$

on solving, we get $B=\frac{1}{11}$ and $A=\frac{1}{5}$

so, $A=\frac{1}{5}=\frac{1}{x-y}$

and $B=\frac{1}{11}=\frac{1}{x+y}$

so, $x-y=5$

and $x+y=11$

on solving, we get $x=8$ and $y=3$

Hence, the answer is $8,3.$