Given ${x^2-x y+y^2 = 15, x y+x+y = 13}$ find the value of $x^2+6y$

Question

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.

Answer 1

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 $x+y$

Answer 2

Changing unknowns: $u=x+y, \ v=xy$ you get: $$ u^2-3v=15,\ \ \ u+v=13. $$ From here $$ u_1=6, \ \ v_1=7, $$ $$ u_2=-9, \ \ v_2=22 $$ etc.