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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.
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 $y$
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.