Question

Solve:

$a(x+y)+b(x-y)=a^2-ab+b^2$ and
$a(x+y)-b(x-y)=a^2+ab+b^2$.

Answer 1

Taking $x+y=u$ and $x-y=v$ the given system of equations becomes
$au+bu-(a^2-ab+b^2)=0$
$au-bv-(a^2+ab+b^2)=0$

By cross-multiplication, we have
$\Rightarrow \frac{u}{b\times -(a^2+ab+b^2)-(-b)\times -(a^2-ab+b^2)}=\frac{-v}{a\times -(a^2+ab+b^2)+a(a^2-ab+b^2)}=\frac{1}{a\times -b-a\times b}$

$\Rightarrow \frac{u}{-b(a^2+ab+b^2)--b(a^2-ab+b^2)}=\frac{-v}{-a(a^2+ab+b^2)+a(a^2-ab+b^2)}=\frac{1}{-ab-ab}$

$\Rightarrow \frac{u}{-b(a^2+ab+b^2+a^2-ab+b^2)}=\frac{-v}{-a(a^2+ab+b^2-a^2+ab-b^2)}=\frac{1}{-2ab}$

$\Rightarrow \frac{u}{-2b(a^2+b^2)}=\frac{-v}{-a\left(2ab\right)}=\frac{1}{-2ab}$

$\Rightarrow u=\frac{-2b(a^2+b^2)}{-2ab},v=\frac{2a^{2}b}{-2ab}\Rightarrow u=\frac{a^2+b^2}{a},v=-a$

Now, $u=\frac{a^2+b^2}{a}\Rightarrow x+y=\frac{a^2+b^2}{a}$ .(i)

and, $v=-a\Rightarrow x-y=-a$ ..(ii)

Adding equations (i) and (ii), we get

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

Substitutiing equation (ii) from equation (i), we get

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

Hence, the solution of the given system of equations is $x=\frac{b^2}{2a},y=\frac{2a^2+b^2}{2a}$.