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
x2(a+b)(a2−b2+2ab) =−y2(a−b)(a2+b2) =1−2b(a+b)
Now, x2(a+b)(a2−b2+2ab)=1−2b(a+b) 

⇒x=2ab−a2+b2b

And, −y2(a−b)(a2+b2)=1−2b(a+b) 
⇒y=(a−b)(a2−b2)b(a+b)
The solution of the system of equations are 2ab−a2+b2b and (a−b)(a2−b2)b(a+b) respectively.

Given that,

(a – b) x + (a + b) y = 2a² – 2b²

(a + b) (x + y) = 4ab

(a – b) x + (a + b) y – 2(a² – b²) = 0

(a + b)x +  (a + b)y – 4ab = 0

On comparing both the equation with the general form we get

 a₁ = a – b, b₁ = a + b, c₁ = -2,

a₂ = a + b, b₂ = a + b, c₂ = -4ab

Now by using cross multiplication we get

x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)

⇒ x/(-(a + b)4ab + 2(a + b) (a² – b²)) = y/(− 2(a² − b²)(a + b) + 4ab(a – b)) 

= 1/((a − b)(a + b) − (a + b)(a + b))

⇒ x/(2(a + b)(a² – b² + 2ab)) = 1/-2b(a + b)

x = (2ab – a² + b²)/b

and,

= -y/(2(a – b) (a² + b²) -2b (a + b)) = 1/ -2b(a + b)

y = (a – b)(a² + b²)/ b(a + b)

Hence, x = (2ab – a² + b²)/b and y = (a – b)(a² + b²)/ b(a + b)