Question
Solve for $x$ and $y$:
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=\frac{\text{a}+\text{b}}{\text{b}}$
$\frac{\text{x}}{\text{a}^2}+\frac{\text{y}}{\text{b}^2}=2$

Answer

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=\frac{\text{a}+\text{b}}{\text{b}}\dots(\text{i})$
$\frac{\text{x}}{\text{a}^2}+\frac{\text{y}}{\text{b}^2}=2\ \dots(\text{ii})$
Multiplying (i) by $b$ and (ii) by $b^2$ and subtract, we get
$\Rightarrow\frac{\text{bx}}{\text{a}}-\frac{\text{b}^2\text{x}}{\text{a}^2}=\text{ab}+\text{b}^2-\text{2b}^2$
$\Rightarrow\text{x}=\frac{\big(\text{a}\text{b}-\text{b}^2\big)\text{a}^2}{\big(\text{ab}-\text{b}^2\big)}$
$\Rightarrow\text{x}=\text{a}^2$
Substituting $x = a^2$ in (i), we get
$\text{a}+\frac{\text{y}}{\text{b}}=\frac{\text{a}+\text{b}}{\text{b}}$
$\Rightarrow\frac{\text{y}}{\text{b}}=\frac{\text{a}+\text{b}}{\text{b}}-\text{a}$
$\Rightarrow\text{y}=\text{b}^2$
So, $x = a^2$ and $y = b^2$

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