# 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.

• Good start. If $x=5-2y$, say, then $(5-2y)^2+(5-2y)\times y=3$. Can you finish from here?
– lulu
Dec 15, 2021 at 20:44
• I'll add the solution ASAP. Dec 15, 2021 at 21:04
• @Cookie Great! You should post self answer to this question. :) Dec 15, 2021 at 21:06
• Got it. And there's one other solution (along the same lines) as well.
– lulu
Dec 15, 2021 at 21:07
• Note: as squares and square roots are involved you might have picked up some spurious solutions. You should check each candidate just to confirm that they all work. In principle you could just check algebraically that the calculation doesn't pick up any fake solutions but in practice I always think it's best to simply check. Good way to verify that no algebraic errors occurred.
– lulu
Dec 15, 2021 at 21:13

$$x^2+xy=3$$ $$4y^2+3xy=22$$
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 $$(x,y)\in{(-6,\frac{11}{2}),(1,2),(-1,-2),(6,\frac{-11}{2})}$$
As $$xy\ne0$$ and both the equations are of the same degree, WLOG set $$y=mx$$ so that $$3=x^2(1+m)\ \ \ \ (1)$$
$$\dfrac{22}3=\dfrac{(4m^2+3m)x^2}{(1+m)x^2}$$
We can solve for $$m$$ and use the values in $$(1)$$