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.
Best Answer
$$x^2+xy=3$$ $$4y^2+3xy=22$$
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 $$(x,y)\in{(-6,\frac{11}{2}),(1,2),(-1,-2),(6,\frac{-11}{2})}$$