Question

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

Answer 1

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

$3(x-y)+2(x+y)=2(x^2-y^2)$ $9(x-y)-4(x+y)=x^2-y^2$

$9x-3y+2x+2y=2(x^2-y^2)$ $9x-9y-4x-4y=x^2-y^2$

$5x-y=2(x^2-y^2)$ ………$\left(1\right)$ $5x-13y=x^2-y^2$

$10x-26y=2(x^2-y^2)$ ………..$\left(2\right)$

Subtract $\left(1\right)$ from $\left(2\right)$

$10x-26y-(5x-y)=0$

$5x-25y=0$

$5x=25y$

$x=5y\to$ put in $\left(1\right)$

$5\left(5y\right)-y=2\left(\right(5y{)}^2-y^2)$

$25y-y=48y^2$

$48y^2-24y=0$

$24y(2y-1)=0$

$y=0$ & $y=\frac{1}{2}$ $\to$ put in $\left(1\right)$

For $y=0$ For $y=\frac{1}{2}$

$5x-0=2(x^2-0)$ $5x-\frac{1}{2}=2(x^2-{\left(\frac{1}{2}\right)}^2)$

$2x^2-5x=0$ $5x-\frac{1}{2}=2x^2-\frac{1}{2}$

$x(2x-5)=0$ $2x^2-5x=0$

$x=0$ & $x=\frac{5}{2}$ $x(2x-5)=0$

$x=0,$ & $x=\frac{5}{2}$.