Solve this system of equations for real numbers Solve this system of equations for real numbers:
$$x^2+xy=3, $$
$$4y^2+3xy=22. $$
I think it's pretty obvious what to do to start, I think we should sum the two equations and get that:
$$x^2+4xy+4y^2=25$$
$$(x+2y)^2=5^2$$
$$\Rightarrow x+2y=5,-5$$
I'm not sure where to go from here, I've tried factorizing the equations and substituting what we've got but I've not gotten anywhere. Hints are appreciated.

$$x^2+xy=3$$
$$4y^2+3xy=22$$
Summing these two terms we get
$$x^2+4xy+4y^2=25$$
$$(x+2y)^2=25$$
$$=>x+2y=5,-5$$
First option is $x+2y=5$
$$x+2y=5$$
$$x=5-2y$$
If we put this into the first equation we get
$$(5-2y)^2+(5-2y)\cdot y=3$$
$$25-15y+2y^2=3$$
$$-2y^2+15y-22=0$$
$$-2y^2+4y+11y-22=0$$
$$(-2y+11)(y-2)=0$$
Since their product is zero one of the factors must be zero, so we have two options
$$-2y+11=0$$
$$y=\frac{11}{2}$$
or
$$y-2=0$$
$$y=2$$
We know that $x=5-2y$, so
$$y=\frac{11}{2}$$
$$x=-6$$
or
$$y=2$$
$$x=1$$
These same steps work for $x=2y=-5$ and we end up with the answer being
$$(x,y)\in{(-6,\frac{11}{2}),(1,2),(-1,-2),(6,\frac{-11}{2})}$$
