Solve the sistem x^2/y+y^2/x=35/3 x+y=10

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).