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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?

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    $\begingroup$ Neither of these are equations. $\endgroup$ – Peter Foreman Jul 12 '19 at 7:43
  • $\begingroup$ My bad, accidentally forgot to add RHS. $\endgroup$ – Henry Page Jul 12 '19 at 7:45
  • $\begingroup$ @HenryPage, Is it not $$3x+2y?$$ $\endgroup$ – lab bhattacharjee Jul 12 '19 at 7:48
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    $\begingroup$ 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 $y=-x\pm1$. $\endgroup$ – Peter Foreman Jul 12 '19 at 7:49
  • $\begingroup$ my bad, again. yes it is $\endgroup$ – Henry Page Jul 12 '19 at 7:49
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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 $m$

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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.

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