We have,

⇒ b²x + a²y = a³b + ab³ …(i)

x + y = 2ab …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (ii) by a², so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

b²x + a²y = a³b + ab³

x + y = 2ab [×a²]

⇒ b²x – a²x = - a³b + ab³

⇒ (b² – a²)x = ab(b² – a²)

⇒ x = ab

Substitute x = ab in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (ii), we get

(ab) + y = 2ab

⇒ y = ab

Hence, we have x = ab and y = ab.