Given ${x^2-x y+y^2 = 15, x y+x+y = 13}$ find the value of $x^2+6y$ Both x and y are real numbers and x > y .
Given ${x^2-x y+y^2 = 15, x y+x+y = 13}$ find the value of $x^2+6y$ .
I tried solving the second equation to get $y=(13-x)/(x+1)$ and substituted that in equation one to get a quartic equation in x : $x^4+3 x^3-25 x^2-69 x+154 = 0$ .
I tried a few  rational root test guesses to no avail. Then I checked wolfram alpha to find out that it has no rational roots. So I think there must be a better way to compute $x^2+6y$ if x > y.

HINT:
We have , $$x^2-xy+y^2+3(xy+x+y)=15+3\cdot13$$
$$\implies (x+y)^2+3(x+y)-54=0$$
Solve for $x+y$
