I was helping out in a learning support class today and we were working through some absolute value problems when something like $|x + 4| - 5 = 10$ came up and both students I was working with split the problem up into

$|x + 4| - 5 = 10$ and $|x + 4| - 5 = -10$

when I tried to explain that this was not correct, I couldn't seem to find a way to put it concisely. I ended up just telling them, "make sure that the absolute value is by itself" which I feel like doesn't really explain the reasoning and the students didn't seem to be able to connect it with what we had been doing previously with absolute values and inequalities. Any suggestions on a better way to explain this?

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    $\begingroup$ Welcome to the site! Did you consider graphing the equation? $\endgroup$ Commented Oct 31, 2014 at 2:17
  • $\begingroup$ Ah great idea! we were graphing the inequalities and they were really getting it, i will try that next time i see them ! $\endgroup$
    – celeriko
    Commented Oct 31, 2014 at 2:22
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    $\begingroup$ A classical approach would be to scaffold with a sequence of problems, and to try and figure out (along the way) where the student's confusion begins. A simple sequence of problems would be to solve for $x$ in the following: A. $x - 5 = 10$. B. $|x| = 10$. C. $|x| - 5 = 10$. D. $|x+4| - 5 = 10$. The fundamental suggestion I have, more than the particular A-D above, is to ask the students to justify their reasoning as they solve the problems. Not just: "Then I did this." Instead: "Then I did this because ______." $\endgroup$ Commented Oct 31, 2014 at 21:35
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    $\begingroup$ This isn't enough to clear up the problem, but to convince them that there is a problem have them try their current technique on the equations $|x| = 1$, and then on $|x| - 1 = 0$. Point out that these are equivalent equations, and so should have the same solution. This should be enough to get them bothered, and interested in resolving the discrepancy. $\endgroup$
    – NiloCK
    Commented Dec 4, 2014 at 20:07
  • $\begingroup$ Related: All solutions of $|2x|=x−1$ are extraneous $\endgroup$
    – ryang
    Commented Mar 29, 2021 at 10:38

5 Answers 5


One of the reasons your students are putting the $\pm$ on the 10 is probably because someone told them that when you remove a pair of absolute value signs that the $\pm$ goes on the number after the $=$. They are simply doing what they thought they were told to do.

So I try to avoid telling them this sort of thing. I like to tell them is that $|x|$ is equal to $+x$ or $-x$ depending on the value of $x$. In short, $|x| = \pm x$.

This may seem like it's not any different, but the key point I stress is that the $|x|$ itself and all by itself is equal to $\pm x$ regardless of whether it's in an equation or not.

So this basically means that when you remove the absolute value signs, the plus or minus attaches itself to whatever had the absolute value signs on it before. So the solution is this:

$$ |x+4| - 5 = 10\\ \pm(x+4) - 5 = 10\\ +(x+4) - 5 = 10 \quad \text{or} \quad -(x+4) - 5 = 10\\ x+4 - 5 = 10 \quad \text{or} \quad -x - 4 - 5 = 10\\ x - 1 = 10 \quad \text{or} \quad -x - 9 = 10\\ x = 11 \quad \text{or} \quad -x = 19\\ x = 11 \quad \text{or} \quad x = -19 $$

Note that I do this even for the simple absolute value equations like $|x| = 3$, beginning with $\pm x = 3$, even though it introduces an extra step. I do this because I know in advance that it will prevent problems later, and also it reinforces that it is all to do with the meaning of the absolute value signs, not the equation per se.

This approach works quite well when there are two sets of absolute values in an equation such as $|x+2| + |x-2| = 5$, because you get $\pm(x+2) \pm (x-2) = 5$ so you can much more easily see that there are in fact four options.

Something else it is worth stressing is that they should always check their solution to see if it works in the original equation. Because $|x| = \pm x$ depending on the value of $x$ and we haven't been careful to specify which values of $x$ make it $+$ and which make it $-$ in our solution, sometimes we get extra solutions that don't exist, because the value of $x$ that we find doesn't match with when it's the $+$ or the $-$.

Graphing the function $y = |x+4| - 5$ and finding where it hits the horizontal line $y=10$ is also a good activity to help them see visually what the answer ought to be, but I find that while it helps them see that their answer is wrong, it doesn't help them see how to go about solving the equation on paper.

Note: of course the above equation-solving approach is not what I would actually use, because I always tell students they should try to deal with the things that are furthest from the $x$ first. So from their experience solving ordinary equations, they should deal with the $-5$ first like this:

$$ |x + 4| - 5 = 10\\ |x+4| = 15\\ \pm(x+4) = 15\\ +(x+4) = 15 \quad \text{or} \quad -(x+4) = 15\\ x + 4 = 15 \quad \text{or} \quad x + 4 = -15\\ x = 11 \quad \text{or} \quad x = -19 $$ However, I still put the $\pm$ on the thing that had the absolute value signs though, because this transfers to other situations.

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    $\begingroup$ WOW! i REALLY like this strategy! i've never even thought of it this way but it makes a lot more sense and makes the learning of absolute values much more extensible. I will be sure to teach it this way next time it comes up, thank you! $\endgroup$
    – celeriko
    Commented Nov 4, 2014 at 1:57

I find it helpful to teach the concept that $|x - a|$ is exactly the distance of $x$ from $a$. This also generalizes easily to higher dimensions, where $|x - a|$ will be the norm of the vector $x - a$ (no matter which norm/metric you use). While teaching this, explain it using the number line, with a circle centred at $a$. Of course, since the only possible points are on the number line, you have to consider only the points where the circle intersects the line, which are at equal distance from $a$ and on either side of it. That is, if $|x - a| = d$, then the points (limits of $x$) will be at distance $d$ from $a$, and these will clearly be $a + d$ and $a - d$.

Once this concept is understood, $|x + 4| - 5 = 10$ becomes: "The distance of $x$ from $-4$, less $5$, is $10$, so the distance of $x$ from $-4$ is $15$, thus $x$ is either $-4 + 15 = 11$, or $-4 - 15 = -19$.

One other advantage of this approach is that it is also just as easy to solve inequalities involving $|\cdot|$. There is no need to remember complicated [for students] rules about changing the inequality with the sign.

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    $\begingroup$ this geometric approach to solving $|x -(-4)| = 15$ which is interpreted as the distance between the points represented by $x$ and $-4$ is $15$. to find $x$ you go $15$ units to the left and right of $-4$ arrive at $x=11$ and at $x = -19.$ $\endgroup$
    – abel
    Commented Dec 7, 2014 at 23:45

You should get the students to graph $y = |x + 4| -5$ and $y = 10$.

It's obvious where the solutions are then from that.

The first graph can be done by shifting the graph of y=$|x|$ to the left by 4 units then down by 4 units.


You can then relate +(x+4) and -(x+4) to the different branches.


One of the new foci in the Common Core State Standards is on the role of translations and other form of transformations and how they affect points in a plane. For example, see the Congruence standards in the high school standards: http://www.corestandards.org/Math/Content/HSG/CO/

|x+4|−5=10 and |x+4|−5=−10

When I taught algebra 2, I'd have students experiment with these first without using absolute value in order to get familiar with moving equations to the left, to the right, and up and down. You can do this with lines, but i think that it works a lot nicer with parabolas. Using graphing calculators at this point is fairly nice, as students are convinced a lot easier of things when they can see them instantly :-)

Try doing this with x^2 and then x^2 -5. the -5 moves the parabola down 5 points on the y axis. Then (x+4)^2 - 5 moves the parabola - okay, this part is a bit tricky, it moves it left. My students were always totally baffled by this. I would have them do a table: x x^2 x+4 (x+4)^2 -6 36 -2 4 -5 25 -1 1 -4 16 0 0 -3 9 1 1 -2 4 2 4 -1 1 3 9 0 0 4 16

So, interestingly enough, when you add 4 before squaring, it means that -4 becomes 0, -3 becomes 1, -2 becomes 2, and -1 becomes 3. Hence, the parabola moves left, not right!

Your first one, |x+4| - 5 = 10 - we can think of as intersecting y= |x+4| -5 and y=10. Start by graphing y = |x|. Now move it down 5 points on the y axis. Then move it - again, tricky! - 4 spaces left on the x axis.

Then graph y= 10 and find where it intersects.

I think that showing this with other functions like parabolas helps make this a lot clearer. As in, this is a type of operation you can do on functions, not just on absolute value equations.


I like the answer by @DavidButlerUofA above, but here's how I'd be prone to do this (in synch with similar issues like solving by factoring and completing the square). I'd throw up the initial theorem: "For all real $a > 0$, $|u| = a$ means $u = a$ or $u = -a$." (Or using the plus-or-minus symbol for shorthand.) I'd point out a procedure: to solve such an equation we first need to isolate the absolute-value expression. And then I'd run a few warm-up recognition exercises of the ilk: "Is the absolute-value expression isolated? Can we use this transformation right now?".

And I have recently evolved to require checking solutions all the time, in an attempt to make sure students have that as a tool to try to catch these errors. In the specific case of your supported students, I'd totally let them go through with the faulty transformation, then check at the end, and seeing that it doesn't check, then thinking through where they broke the process.


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