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Multiply by three the second equation and add this to the first one, and you'll get (check this!)
Now take it from here.
Here is another way through, which simply depends on seeing the factor $x+y$ in both parts of what is given, reducing the cubic to a quadratic as follows (though the elimination of the constant term will always give a homogeneous cubic even without noticing this factor):
Multiply the first equation by $15$ and the second by $19$ to obtain:
Then we either have $x=-y$, which is easily eliminated, or divide through by $x+y\neq 0$ to give $$15x^2-34xy+15y^2=0$$ which factorises (quadratic formula) as $$(5x-3y)(3x-5y)=0$$
Either of these factors gives a linear relation which can be plugged into either of the original equations to give a solution.
Another method :
Suppose $x,y$ be the roots of $z^2+az+b=0$
$x+y=-a$ and $xy=b$
$a=-8$ and $b=120/8=15$
Solve of $z^2-8z+15=0$ for $x$ and $y$