Question

Solve the pair of linear equations by using cross multiplication method
(iv) $2(ax-by)+a+4b=0,2(ax-ay)+b-4a=0$

Answer 1

$2ax-2by+(a+4b)=0$

$2ax-2ay+(b-4a)=0$

$\frac{x}{-8ab+2b^2-2a^2-8ab}=\frac{y}{-2a^2-8ab-8a^2+2ab}=\frac{1}{-4a^2+4ab}$

$\Rightarrow \frac{x}{2b^2-2a^2-16ab}=\frac{y}{-10a^2-6ab}=\frac{1}{-4a^2+4ab}$

$\Rightarrow \frac{x}{2(b^2-a^2-8ab)}=\frac{y}{-2a(5a+3b)}=\frac{1}{-4a(a-b)}$

$\Rightarrow \frac{x}{(b^2-a^2-8ab)}=\frac{y}{-a(5a+3b)}=\frac{1}{-2a(a-b)}$

$\Rightarrow \frac{x}{(b^2-a^2-8ab)}=\frac{1}{-2a(a-b)}$ and $\frac{y}{-a(5a+3b)}=\frac{1}{-2a(a-b)}$

$\Rightarrow x=\frac{b^2-a^2-8ab}{-2a(a-b)}$ and $y=\frac{-a(5a+3b)}{-2a(a-b)}$

$\Rightarrow x=\frac{b^2-a^2-8ab}{-2a(a-b)}$ and $y=\frac{(5a+3b)}{2(a-b)}$

$\therefore x=\frac{a^2-b^2+8ab}{2a(a-b)}$ and $y=\frac{5a+3b}{2(a-b)}$