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.
 A: 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.
A: Hint:
Multiply the first equation by $7$ and the second by $5$ than subtract and you find (as suggested in the comment of Benjamin  Moss) the equation
$$
18x^2-33xy+9y^2=0
$$
that is a homogeneous equation that can be solved with the substitution
$y=kx$
A: This is an intersection of two hyperbolae. The right hand side is the shift. Personally, I would first try to combine equations and substitute so that it simplifies as much as possible. For instance, the right hand side shifts in a particular direction, so if we rotate the coordinates by $45^\circ$:
$$x=u+v$$
$$y=u-v$$
we get the system of equations
$$3u^2+6uv-5v^2=10u$$
$$3u^2+12uv+5v^2=14u$$
If you add these equations and divide by $u$, you get:
$$6u+18v=24$$
$$u+3v=4$$
which is a straight line.
Now you can directly put this into any of the previous two equations and solve it as a quadratic equation.
For instance, expressing $u=4-3v$ and putting it into the first equation, you obtain
$$2v^2-9v+4=0$$
$$(2v-1)(v-4)=0$$
$$v=\{4,1/2\}$$
Now you just put this back into $u$ and then $x$ and $y$.
