Question

Solve for x and y: $\frac{a^2}{x}-\frac{b^2}{y}=0,\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b$

Answer 1

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

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

$a^{2}y-b^{2}x=0...\left(1\right)$

$\frac{a^{2}by+b^{2}ax}{xy}=a+b$

$a^{2}by+b^{2}ax=axy+bxy$

$a^{2}by+b^{2}ax-axy-bxy=0...\left(2\right)$

So equation (1) and (2)

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

$a^{2}by-b^{3}x=0$

$a^{2}by+b^{2}ax-axy-bxy=0$

________________________

$-b^{3}x-b^{2}ax+axy+bxy=0$

$-b^{2}x(b+a)+xy(a+b)=0$

$(a+b)(xy-b^{2}x)=0$

$a+b=0$

$xy-b^{2}x=0$

$xy=b^{2}x$

$y=b^2$

equation (1)

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

$a^2{b}^2-b^{2}x=0$

$a^2{b}^2=b^{2}x$

$x=a^2$