Question

Solve: $\frac{3}{x+y}+\frac{2}{x-y}and\frac{9}{x+y}-\frac{4}{x-y}=1$

Answer 1

$\frac{3}{x+y}+\frac{2}{x-y}=1$....(1)
$\frac{9}{x+y}-\frac{4}{x-y}=1$....(2)

$3(x-y)+2(x+y)=(x+y)(x-y)$
$3x-3y+2x+2y=x^2-y^2$
$5x-y=x^2-y^2$....(3)

$9(x-y)-4(x+y)=(x+y)(x-y)$
$9x-9y-4x-4y=x^2-y^2$
$5x-13y=x^2-y^2$...(4)

subtracting equation (4) and (3)
$5x-y=x^2-y^2$
$5x-13y=x^2-y^2$
we get
$12y=0$
$y=0$
put in equation (3)
$5x-0=x^2-0$
$5x=x^2$
$x=5$