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

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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)**

This is a symmetrical system and we'll solve it using the sum and the product.

x + y = S (sum)

x*y = P (product)

We'll multiply the 2nd equation by xy both sides:

x^3 + y^3 = 9xy/2

We'll re-write the equations of the system:

x + y = 3

x^3 + y^3 = 9xy/2

We'll write the sum of cubes from the second equation:

x^3 +y^3 = (x+y)(x^2 + xy + y^2)

Now, we'll substitute all by S and P.

S = 3

3(S^2 - 2P) = 9P/2

S^2 - 2P = 3P/2

9 - 2P = 3P/2

18 - 6P = 3P

9P = 18

P = 2

We'll form the quadratic equation knowing the sum and the product:

x^2 - 3x + 2 = 0

x = 1 and y = 2

x = 2 and y = 1

**The solutions of the system are (1 ; 2) and (2 ; 1).**