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

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?

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#### Answers 2

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 $$m$$

lab bhattacharjee
July 12, 2019 07:49 AM

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.

Empy2
July 12, 2019 08:17 AM

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