Question

Solve the pair of linear equations by using cross multiplication method
(i)$\frac{x}{a}+\frac{y}{b}=a+b,\frac{x}{a^2}+\frac{y}{b^2}=2$

Answer 1

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

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

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

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

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

$\Rightarrow \frac{b^{2}x}{b-a}=\frac{a^{2}y}{b-a}=\frac{a^3{b}^3}{ab(a-b)}$

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

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

$\Rightarrow x=-a^2,y=-b^2$