Question

Solve the following systems of equations by the method of cross-multiplication:
$a^{2}x+b^{2}y=c^2$
$b^{2}x+a^{2}y=d^2$

Answer 1

$a^{2}x+b^{2}y=c^2{b}^{2}x+a^{2}y=d^2$

On simplification$,$

$a^{2}x+b^{2}y-c^2=0b^{2}x+a^{2}y-d^2=0$

On cross multiplication$,$

$\frac{x}{-b^2{d}^2+a^2{c}^2}=\frac{-y}{-a^2{d}^2+b^2{c}^2}=\frac{1}{a^4-b^4}\therefore x=\frac{a^2{c}^2-b^2{d}^2}{a^4-b^4}y=\frac{a^2{d}^2-b^2{c}^2}{a^4-b^4}$