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I need to solve the following system of equations. I did it by graphing because this seemed the most appropriate, but my student wanted to do it using Algebra...

I know the solution is the empty set, and I am very comfortable with solving and graphing each individual equation. I just wasn't sure if it was "legal" to split up the two instances of the absolute value equation and set each equal to a positive and negative version of the quadratic as as I would if the other side of the equation were a constant.

Any help is appreciated. Thanks in advance!

System with Abs Value

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  • $\begingroup$ It is indeed legal to split the domain depending on where the absolute value is positive or negative. $\endgroup$
    – Parcly Taxel
    Sep 24, 2017 at 5:52

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I assume you're trying to find solutions to $f(x)=g(x)$.

We split into two cases: when $x+1<0$ and $x+1\geq 0$, or in other words: $$\begin{equation} x<-1 \\x \geq -1 \end{equation}$$

Assume the first case, then $f(x)$ becomes: $$f(x)=-3\times (-x-1)=3x+3$$ so that we have a quadratic to solve for when we set $f(x)=g(x)$, looking for solutions in the interval $(-\infty, -1)$.

In the second case, $f(x)=-3\times(x+1)$ so that we have a (different) quadratic to solve for when we set $f(x)=g(x)$, looking for solutions in $[-1, \infty)$.

Note that we can keep $g(x)$ the same in both cases because in either case, $g(x)$ does not change.

This is illustrated when you draw these two graphs - which you've done with your student.

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    $\begingroup$ Thank you for your thoughtful response, Jihoon! $\endgroup$
    – J.E.M. Texas
    Sep 24, 2017 at 13:51

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