To solve

x^2/y+y^2/x=35/3...(1)

x+y=10..(2).

From (2), we get y = 10-x. We put y = 10-x in (1):

x^2/(10-x) +(10-x)^2/x = 35/3.

3(x^3 +(10-x)^3 ) + 35(10-x)x = 350x-35x^2.

3{x^3+10^3-300x+30x^2-x^3 = 350x-35x^2.

3{1000-300x+30x^2} = 350x-35x^2.

3000-900x+90x^2-350x+35x^2 = 0.

125x^2-1250x+3000 = 0.

x^2-10x+24 = 0.

(x-6)(x-4) = 0.

So x= 6 or x= 4.

When x = 6, from (2), we get x+y = 10,. So y = 10-6 = 4.

When x=4, from (2), x+y = 10. So y = 10-4 = 6.

Therefore (x,y) = (6,4). Or (x,y) = (4,6).

We'll multiply the first equation by 3xy:

3(x^3 + y^3) = 35xy

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

3(x+y)(x^2 - xy + y^2) = 35xy

We'll substitute x+y by the second equation:

x+y=10

30(x^2 - xy + y^2) = 35xy

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

x^2 + y^2 = 100 - 2xy

30(100 - 3xy) = 35xy

3000 - 90xy = 35xy

125xy = 3000

xy = 24

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

x^2 - Sx + P = 0

x^2 - 10x + 24 = 0

x1 = [10+sqrt(100 - 96)]/2

x1 = (10+2)/2

x1 = 6

x2 = (10-2)/2

x2 = 4

**So, the solutions of the symmetric system are: {4 ; 6} and {6 ; 4}.**