Question

Solve the following system of equations.
$\frac{x}{a}+\frac{y}{b}=a+b;$

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

Answer 1

Given,

$\frac{x}{a}+\frac{y}{b}=a+b$

or, $x=a^2+ab-\frac{ay}{b}$......(1).

Now substituting the value of $x$ in (2) we get,

$\frac{x}{a^2}+\frac{y}{b^2}=2$.......(2)

or, $1+\frac{b}{a}-\frac{y}{ab}+\frac{y}{b^2}=2$

or, $1-\frac{b}{a}=\frac{y}{b^2}(1-\frac{b}{a})$

or, $y=b^2$ [ Since $a\ne b$]

Now putting the value of $y$ in (1) we get $x=a^2$.

So, $x=a^2$ and $y=b^2$.