Find the real solutions for the system: $x^2+4xy-2y^2=5(x+y)$, $ 5x^2-xy-y^2=7(x+y)$ 
Find all real solutions for the system:
  $$\left\{
\begin{array}{l}
x^2+4xy-2y^2=5(x+y)\\ 
5x^2-xy-y^2=7(x+y)\\ 
\end{array}
\right. 
$$
  This is from a math olympiad training book. No answer provided. 

It is easy to spot that $(x,y)=(0,0)$ is a solution, by inspection. But I'm not being able to find other solutions, if they exist. I tried all tricks I know but was not getting to anything useful.
Hints and answers are appreciated. Sorry if this is a duplicate.

Note that $$\begin{align*}5\times\text{Eq.(2)}-7\times\text{Eq.(1)} \equiv 18x^2-33xy+9y^2=0 \\\equiv (y-3x)(3y-2x)=0\\\implies y=\frac{2x}{3} \text{ and } y=3x\end{align*}$$
Substituting $y=3x$ into $\text{Eq.(1)}$, we have, $$x^2+12x^2-2(9x^2)=5(4x) \implies x^2+4x=0 \\\implies \boxed{(x=0, y=0) \text{ and } (x=-4, y= -12)}$$
Similarly, $y=\frac{2x}{3}$ gives us, $$x^2+\frac{8x^2}{3}-\frac{8x^2}{9}=5(\frac{5x}{3}) \implies x^2-3x=0 \\\implies \boxed{(x=0,y=0) \text{ and } (x=3, y=2)}$$
Hope it helps.
