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I don’t mean to bother the community with something so easy, but for the life of me, I can’t remember how to do these. I even forgot what they were called, and I’m referring to the use of the “$ | $” symbol in math. I think I may have been studying too much all day. This is my last section, and if I can’t even remember the name, then I can’t look it up.

Can some please tell me what these types of math problems are called and how to solve them? I just need to see someone do it. I remember it was easy, but it’s been sometime since I’ve last seen these. I really do appreciate everyone helping me study today. I couldn’t of done it without ya! ^_^

Problem. Solve for $ |5 x + 20| = 5 $ for $ x $.

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    $\begingroup$ | means "absolute value" in this case. It essentially means "if its negative, make it positive". |-1| = 1, |2| = 2 etc. $\endgroup$
    – Loocid
    Commented May 4, 2015 at 5:25

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$|\cdot|$ denotes the absolute value here. To solve for $x$, you can break the problem into two cases to get rid of the absolute value.

Consider the cases $5x+20<0$ and $5x+20\geq0$ separately and solve for $x$.

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    $\begingroup$ Or simply $5x+20=\pm5$ $\endgroup$
    – Alexey Burdin
    Commented May 4, 2015 at 5:29
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    $\begingroup$ @AlexeyBurdin Sure, but in general for problems with absolute values breaking the problem into cases like I propose is the method to apply. For example to solve for $\frac{x}{x+7-|2x-1|}\leq1$ one would consider the cases $x<1/2$ and $x\geq1/2$. $\endgroup$
    – Guest
    Commented May 4, 2015 at 5:34
  • $\begingroup$ @Guest for you example, all you have to do is isolate the absolute value and then solve. You do not need to consider those cases. You want to make it easier on yourself. $\endgroup$
    – Julian Rachman
    Commented May 4, 2015 at 5:38
  • $\begingroup$ @JulianRachman How would you do that? $\endgroup$
    – Guest
    Commented May 4, 2015 at 5:45
  • $\begingroup$ Cases are better for inequalities, sure. @Guest $\endgroup$
    – Alexey Burdin
    Commented May 4, 2015 at 5:47
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As an alternative using $|a|=\sqrt{a^2}$ $$ \begin{align} |5x+20|&=5\\ \sqrt{(5x+20)^2}&=5 \\(5x+20)^2&=25 \\ 25x^2+200x+400&=25\\ x^2+8x+15&=0 \\(x+3)(x+5)&=0 \\ x =-3,-5\end{align}$$

Absolute value is sometimes called Modulus

I thought I'd upload a picture whilst I was using Geogebra. The intersections are shown. enter image description here

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The equation $$\lvert a+b\rvert=c$$ may be read as "the distance from the number $a$ to the number $-b$ is $c$."

With this in mind, the equation $$ \lvert5x+20 \rvert=5 $$ is the collection of all $x$ so that the distance from the number $5x$ to the number $-20$ is $5$. Can you find all such $x$?

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$$|5x+20|=5$$

We want to consider both $5$ and $-5$ for this absolute value function. So we solve for both $5$ and $-5$.

$$5x+20=5$$ $$5x=-15$$ $$\boxed{x=-3}$$ $$5x+20=-5$$ $$5x=-25$$ $$\boxed{x=-5}$$

If we plug $-3$ and $-5$ back into the original equation, we should get an equivalent statement. So the final answer is $\boxed{x\in \{-5,-3\}}$.

If you would like more information on solving absolute value equations and inequalities, please check this out.

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  • $\begingroup$ Good answer @JulianRachman $\endgroup$
    – user210387
    Commented May 4, 2015 at 6:06
  • $\begingroup$ @Rememberme thanks! $\endgroup$
    – Julian Rachman
    Commented May 4, 2015 at 6:08

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