Question

Solve the following system of linear equations:
$3(bx+ay)=a-6b,3(ax+by)=-(6a+b)$

Answer 1

$3bx+3ay=a-6b⟶\left(i\right)3ax-3by=-6a-b⟶\left(ii\right)$

Now multiply equation $\left(i\right)\times a$ and equation$\left(ii\right)\times b$

$3abx+3a^{2}y=a^2-6ba⟶\left(iii\right)3abx-3b^{2}y=-6ab-b^2⟶\left(iv\right)$

On substraction $\left(iii\right)$ and $\left(iv\right)$ $,$

$\Rightarrow 3y(a^2+b^2)=a^2+b^2\Rightarrow y=\frac{1}{3}$

Add $y$ in equation $\left(iii\right),$

$x=-2$

$\therefore x=-2$ and $y=\frac{1}{3}$