(a-b)x+(a+b)y=2a2-2b2 (a+b)(x+y)=4ab Solve this by cross …

The given system of equations are:
(a - b)x + (a + b)y = 2a² - 2b²
So, (a - b)x + (a + b)y - 2a² - 2b² = 0
 (a - b)x + (a + b)y - 2(a² - b²) = 0 ........(i)
And (a + b)(x + y) = 4ab
So, (a +b)x + (a + b)y - 4ab = 0 ..........(ii)
The given system of equation is in the form of
a₁x + b₁y - c₁ = 0
and a₂x + b₂y - c₂ = 0
Compare (i) and (ii) , we get
a₁ = a - b, b₁ = a + b, c₁ = -2(a² + b²)
a₂ = a + b, b₂ = a + b, c₂ = -4ab
By cross-multiplication method
$\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}}$ $= \frac{{ - y}}{{2(a - b)({a^2} + {b^2})}}$ $= \frac{1}{{ - 2b(a + b)}}$
Now, $\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}} = \frac{1}{{ - 2b(a + b)}}$ 

$⇒ x = \frac{{2ab - {a^2} + {b^2}}}{b}$
And, $\frac{{ - y}}{{2(a - b)({a^2} + {b^2})}} = \frac{1}{{ - 2b(a + b)}}$ 
$⇒ y = \frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}$
The solution of the system of equations are $\frac{{2ab - {a^2} + {b^2}}}{b}$ and $\frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}$ respectively.