Question

$\frac{a}{x}-\frac{b}{y}=0$
$\frac{a{b}^2}{2}+\frac{a^{2}b}{y}=a^2+b^2$

Answer 1

$\frac{a}{x}-\frac{b}{y}=0$ -----(1)
$\frac{a{b}^2}{2}+\frac{a^{2}b}{y}=a^2+b^2$ ----(2)
Equation (1) $\times a^2$ , $\frac{a^3}{x}-\frac{a^{2}b}{y}=0$ ---(3)
equn(2)+equn (3)
$\frac{a{b}^2}{x}+\frac{a^2\text{/}b}{y}+\frac{a^3}{x}-\frac{a^2\text{/}b}{y}=a^2+b^2$
$\therefore \frac{a{b}^2}{x}+\frac{a^3}{x}=a^2+b^2$
$\therefore \frac{a(\text{/}{a}^2+b^2}{x}=\left(\text{/}{a}^2+b^3\right)$
$\therefore x=a$
From (1) $\frac{a}{x}-\frac{b}{y}=0$
$\therefore \frac{a}{x}-\frac{b}{y}$
$\therefore \frac{a}{a}-\frac{b}{y}$
$\therefore y=b$
$\therefore x=a\&y=b$