Question

Solve the equations
$\frac{10}{x+y}+\frac{2}{x-y}=4,\frac{15}{x+y}-\frac{5}{x-y}=-2$

Answer 1

$10(x-y)+2(x+y)=4(x+y)\times (x-y)$

$12x-8y=4(x^2-y^2)$

$3x-2y=x^2-y^2$ ----------(1)

$15(x-y)-5(x+y)=-2(x+y)\times (x-y)$

$10x-16y=-2(x^2-y^2)$

$-5x+8y=x^2-y^2$ ------------(2)

From eqn. (1) and (2) we get,

$3x-2y=-5x+8y$

$8x=10y$

$x=\frac{5}{4}y$

On putting value of $x=\frac{5}{4}y$ in eqn. (1) we get,

or, $3\times \frac{5}{4}y-2y={\left(\frac{5}{4}y\right)}^2-y^2$

or, $\frac{7}{4}y=\frac{25}{16}{y}^2-y^2$

or, $\frac{7}{4}y=\frac{9}{16}{y}^2$

or, $28y=9y^2$

or, $y(9y-28)=0$

or, $y=0$ or $y=\frac{28}{9}$

$y$ cannot be zero in this case so, $y=\frac{28}{9}$

So, $x=\frac{5}{4}\times \frac{28}{9}$

So, $x=\frac{35}{9}$