How to solve the system x+y=3 and x^2/y+y^2/x=9/2?

We have to solve the system x + y = 3 and x^2/y + y^2/x =
9/2

x + y = 3

=> x = 3
- y

Substitute this in x^2/y + y^2/x =
9/2

=> (3 - y)^2 / y + y^2 / (3 - y) =
9/2

=> (3 - y)(3 - y)^2 + y* y^2 / (3 - y) =
(9/2)*y*(3 - y)

=> (3 - y)^3 + y^3 = (9/2)(3y -
y^2)

=> 2(3^3 - 3*9*y + 3*3*y^2 - y^3 +
y^3) = 27y - 9y^2

=> 54 - 81y + 27y^2 =
0

=> y^2 - 3y + 2 =
0

=>y^2 - 2y - y + 2 =
0

=>y( y - 2) - 1( y - 2) =
0

=> (y - 1)(y - 2) =
0

=> y = 1 and y
=2

=> x = 3 - y = 2 and
1.

The values of x and y are (2, 1) and (1,
2)