Solve the following pair equations for x and y.
$$\dfrac{a^2}{x}-\dfrac{b^2}{y}=0\ ,\ \ \dfrac{a^2b}{y}+\dfrac{b^2a}{y}=a+b\ ;\ \ x\neq0, \ y\neq0$$
Given equations are,
$$\dfrac{a^2}{x}-\dfrac{b^2}{y}=0\quad...(i)$$
$$\dfrac{a^2b}{y}+\dfrac{b^2a}{y}=a+b\quad...(ii)$$
Multiplying equation $$(i)$$ by $$a$$ and adding to equation $$(ii)$$, we gwt
$$\dfrac{a^aa}{x}-\dfrac{b^2a}{y}=0 \\ \dfrac{a^2b}{x}+\dfrac{b^2a}{y}=a+b$$
$$\ \overline{\ \ \ \ \dfrac{a^2a}{x}-\dfrac{a^2b}{x}=a+b\ \ \ \ \ }$$
$$ \Rightarrow \dfrac{a^2}{x}(a+b)=a+b\\\Rightarrow x=\dfrac{a^2(a+b)}{a+b}=a^2$$
Putting the value of $$x$$ in equation $$(i)$$, we get
$$\dfrac{a^2}{a^2}-\dfrac{b^2}{y}=0\\ \Rightarrow 1-\dfrac{b^2}{y}=0 \\ \Rightarrow \dfrac{b^2}{y}=1\\ \Rightarrow y=b^2$$
Hence, $$x=a^2, y=b^2$$