Question

$\frac{10}{x+y}+\frac{2}{x-y}=4;\frac{15}{x+y}-\frac{5}{x-y}=-2$

Answer 1

$\frac{10}{x+y}+\frac{2}{x-y}=4$

$\frac{15}{x+y}-\frac{5}{x-y}=-2$

Let $u=\frac{1}{x+y}$ and $v=\frac{1}{x-y}$

$\Rightarrow 10u+2v=4$

or $5u+v=2$ .......$\left(1\right)$

$\Rightarrow 15u-5v=-2$ .......$\left(2\right)$

$3\times \left(1\right)-\left(2\right)$

$\Rightarrow 15u+3v-15u+5v=6+2$

$\Rightarrow 8v=8$

$\therefore v=1$

Put $v=1$ in $\left(1\right)$ we get

$5u+1=2$

$\Rightarrow 5u=2-1=1$

$\therefore u=\frac{1}{5}$

$\therefore x+y=\frac{1}{5}$ and $x-y=1$

$\Rightarrow x+y+x-y=\frac{1}{5}+1=\frac{6}{5}$

$\Rightarrow 2x=\frac{6}{5}$

$\Rightarrow x=\frac{3}{5}$

Put $x=\frac{3}{5}$ in $x+y=\frac{1}{5}$

$\Rightarrow y=\frac{1}{5}-x=\frac{1}{5}-\frac{3}{5}=\frac{1-3}{5}=\frac{-2}{5}$

$\therefore x=\frac{3}{5}y=\frac{-2}{5}$