by Henry Page
Last Updated July 12, 2019 08:20 AM

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?

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.

Updated February 19, 2018 19:20 PM

Updated November 09, 2017 00:20 AM

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