Question

Solve for x and y:

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

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

Answer 1

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

Multiply by ( $\frac{1}{a}$) both sides, we get

$\frac{x}{a^2}+\frac{y}{ab}=1+\frac{b}{a}$ → (1)

Given other linear equation as

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

Subtract (2) from (1), we get
$\frac{x}{a^2}+\frac{y}{ab}=1+\frac{b}{a}$
$\frac{x}{a^2}+\frac{y}{b^2}=2$
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$\left(\frac{y}{ab}\right)-\left(\frac{y}{b^2}\right)=\left(\frac{b}{a}\right)-1$
⇒ $y\left[\frac{(b-a)}{a{b}^2}\right]=\frac{(b-a)}{a}$
∴ $y=b^2$
Put y = b^2 in $\frac{x}{a^2}+\frac{y}{b^2}=2$

⇒ $\frac{x}{a^2}+\frac{b^2}{b^2}=2$

⇒ $\frac{x}{a^2}+1=2$
⇒ $\frac{x}{a^2}=1$
∴ $x=a^2$