2(ax-by)+a+4b=0 2(bx+ay)+b-4a=0 Find the solution of …

The given system of equation may be written as
2(ax - by) + a + 4b = 0
So, 2ax - 2by + a + 4b ..............(i)
2(bx + ay) + b - 4a = 0
so, 2bx+2ay+b-4a=0................(ii)
compare (i) and (ii) with standard form, we get
a₁ = 2a, b₁ = -2b, c₁ = a + 4b
a₂ = 2b, b₂ = 2a, c₂ = b - 4a
By cross multiplication method
{tex} \frac{x}{{ - 2{b^2} + 8ab - 2{a^2} - 8ab}}{/tex} {tex}= \frac{{ - y}}{{2ab - 8{a^2} - 2ab - 8{b^2}}}{/tex} {tex} = \frac{1}{{4{a^2} + 4{b^2}}}{/tex}
{tex} \frac{x}{{ - 2{b^2} - 2{a^2}}} = \frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \frac{1}{{4{a^2} + 4{b^2}}}{/tex}
Now, {tex}\frac{x}{{ - 2{b^2} - 2{a^2}}} = \frac{1}{{4{a^2} + 4{b^2}}} {/tex}
{tex}⇒ x = \frac{{ - 1}}{2}{/tex}
And, {tex}\frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \frac{1}{{4{a^2} + 4{b^2}}} {/tex}
{tex}⇒ y = 2{/tex}
Therefore, the solution of the given pair of equations are {tex}\frac{{ - 1}}{2}{/tex} and 2 respectively.