0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ It is indeed legal to split the domain depending on where the absolute value is positive or negative. $\endgroup$ – Parcly Taxel Sep 24 '17 at 5:52
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thank you for your thoughtful response, Jihoon! $\endgroup$ – J.E.M. Texas Sep 24 '17 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.