Question

Solve:

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

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

Answer 1

The given system of equations may be written as

$\frac{1}{a}\cdot x+\frac{1}{b}\cdot y-(a+b)=0$

$\frac{1}{a^2}\cdot x+\frac{1}{b^2}\cdot y-2=0$

By cross-multiplication, we have

$\Rightarrow \frac{x}{\frac{1}{b}\times (-2)-\frac{1}{b^2}\times -(a+b)}=\frac{-y}{\frac{1}{a}\times -2-\frac{1}{a^2}\times -(a+b)}=\frac{1}{\frac{1}{a}\times \frac{1}{b^2}-\frac{1}{a^2}\times \frac{1}{b}}$

$\Rightarrow \frac{x}{-\frac{2}{b}+\frac{a}{b^2}+\frac{1}{b}}=\frac{-y}{-\frac{2}{a}+\frac{1}{a}+\frac{b}{a^2}}=\frac{1}{\frac{1}{a{b}^2}-\frac{1}{a^{2}b}}$

$\Rightarrow \frac{x}{\frac{a}{b^2}-\frac{1}{b}}=\frac{-y}{-\frac{1}{a}+\frac{b}{a^2}}=\frac{1}{\frac{1}{a{b}^2}-\frac{1}{a^{2}b}}$

$\Rightarrow \frac{x}{\frac{a-b}{b^2}}=\frac{y}{\frac{a-b}{a^2}}=\frac{1}{\frac{a-b}{a^2{b}^2}}$

$\Rightarrow x=\frac{a-b}{b^2}\times \frac{1}{\frac{a-b}{a^2{b}^2}}=a^2$

$y=\frac{a-b}{a^2}\times \frac{1}{\frac{a-b}{a^2{b}^2}}=b^2$

Hence, $x=a^2,y=b^2$ is the solution of the given system of equations.