How to solve simultaneous equations with both of the equations being quadratic equations.

I'm familiar with how to solve simultaneous equations with 1 quadratic equation but not 2. I've looked all of the internet for a thread that has covered this, but I can't seem to find one. $$\begin{cases} (x+y)^2 = 1\\ \\ (3x+2y)(x-y) = -5 \end{cases}$$ Could someone please describe the process?

• Neither of these are equations. – Peter Foreman Jul 12 '19 at 7:43
• My bad, accidentally forgot to add RHS. – Henry Page Jul 12 '19 at 7:45
• @HenryPage, Is it not $$3x+2y?$$ – lab bhattacharjee Jul 12 '19 at 7:48
• Equation $1$ implies that $y=-x\pm1$ plug that into equation $2$ and solve for four values of $x$ and hence eight values of $y$ by using $y=-x\pm1$. – Peter Foreman Jul 12 '19 at 7:49
• my bad, again. yes it is – Henry Page Jul 12 '19 at 7:49

What if $$x=0$$
Else set $$y=mx$$ to find $$\dfrac1{-5}=\dfrac{(1+m)^2}{(3+2m)(1-m)}$$ which is a quadratic equation in $$m$$
The first equation is two lines. Treat them separately. First do $$x+y=1$$, substitute into equation 2 for two solutions. Then do $$x+y=-1$$, substitute into equation 2 for two more solutions.