Question

Solve the following equations:
$\frac{x-a}{a^2}+\frac{y-b}{b^2}=\frac{1}{x-b}-\frac{1}{y-a}-\frac{1}{a-b}=0$.

Answer 1

$⟹\frac{x-a}{a^2}=\frac{b-y}{b^2}$
$⟹x=a+\frac{a^2}{b^2}(b-y)$
Putting in equation $\left(II\right)$
$⟹\frac{1}{a+\frac{a^2}{b^2}(b-y)-b}-\frac{1}{y-a}=\frac{1}{a-b}$
$⟹\frac{1}{a{b}^2+a^{2}b-a^{2}y-b^3}-\frac{1}{y-a}=\frac{1}{a-b}$
$⟹\frac{y-a-a{b}^2-a^{2}b+a^{2}y+b^3}{(y-a)(a{b}^2+a^{2}b-a^{2}y-b^3)}$
$⟹$ On solving we get:-
$x=\frac{a^2}{b},y=\frac{b^2}{a}$
$x=\frac{2ab-a^2}{b},y=\frac{b(2a-b)}{a}$