Question

Solve the following systems of equations by the method of cross-multiplication:
$ax+by=a^2$
$bx+ay=b^2$

Answer 1

Given,

$ax+by-a^2=0$

$bx+ay-b^2=0$

$a_1=a,b_1=b,c_1=-a^2$

$a_2=b,b_2=a,c_2=-b^2$

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

$\frac{x}{a^3-b^3}=\frac{1}{a^2-b^2}$

solving the above equation, we get,

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

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

solving the above equation, we get,

$y=-\frac{ab}{a+b}$