Solving Absolute Value Equations and Inequalities

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As we saw earlier, an absolute value (designated by |  |) means take the positive value of whatever is between the two bars. The absolute value is always positive, so you can think of it as the distance from 0. So \(\left| 3 \right|=3\) and \(\left| {-3} \right|=3\). It’s as simple as that!

(Note that we also address absolute values here in the Piecewise Functions section, and here in the Rational Functions, Equations and Inequalities section.)

Solving Absolute Value Equations

Solving absolute equations isn’t too difficult; we just have to separate the equation into two different equations (once we isolate the absolute value), since we don’t if what’s inside the absolute value is positive or negative (we can do this with a number line if we want).

I like to then make the expression on the right hand side (without the variables) both positive and negative and split the equation that way.

The other thing we have to remember is that we must check our answers, since we may get extraneous solutions (solutions that don’t work).

There are a few cases with absolute value equations or inequalities that you may see where you don’t have to go any further. One is when we have isolated the absolute value, and it is set equal to a negative number, such as \(\left| {x-5} \right|=-4\), or \(\left| {x-5} \right|\le -4\), for example. Since an absolute value can never be negative, we have no solution for this case.

The other is when the absolute value is greater than a negative number, such as  \(\left| {x-5} \right|>-4\)  for example. In this case our answer is all real numbers, since an absolute value is always positive.

Here are more problems:



\(\begin{array}{c}\color{#800000}{{\left| x \right|-3=20}}\\\left| x \right|=23\\\swarrow \,\,\searrow \\x=23\,\,\,\,\,\,\,\,\,\,x=-23\\x=-23,\,\,23\end{array}\)

Note that we have to first simplify, then separate the absolute value problem into two separate equations, since we don’t know if what the value of \(x\) is positive or negative.


Try the answers in the original equation to make sure they work!

\(\displaystyle \begin{array}{c}\color{#800000}{{3\left| {x-2} \right|+5=14}}\\3\left| {x-2} \right|=9\\\left| {x-2} \right|=3\\\swarrow \,\,\,\,\,\,\,\,\,\,\searrow \\x-2\,=\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,x-2=-3\\\underline{{\,\,\,\,\,\,\,+2=+2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,+2=+2}}\\\,\,\,\,\,x\,\,\,=\,\,\,\,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,=-1\end{array}\)


Here’s one a little bit more complicated. Note that we still have to simplify first to separate the absolute value from the rest of the numbers.


Check the answers; the work!

\(\displaystyle \begin{array}{c}\color{#800000}{{\left| {{{x}^{2}}-2x} \right|+1=9}}\\\left| {{{x}^{2}}-2x} \right|=8\\\swarrow \,\,\,\,\,\,\,\,\,\,\,\,\searrow \\{{x}^{2}}-2x\,=\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}-2x\,\,=\,-8\\{{x}^{2}}-2x\,\,-8=0\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}-2x\,\,+8=0\\\left( {x-4} \right)\left( {x+2} \right)=0\,\,\,\,\,\,\,\,\,\,x=\,\,\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 1 \right)\left( 8 \right)}}}}{{2\left( 1 \right)}}\\x=\,\,4,\,\,-2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1\pm 1\sqrt{7}i\end{array}\)

This one’s a little trickier, since it’s not linear. After simplifying, separate the absolute value equations into two; 8 positive, and 8 negative.


For the second case, we’ll have to use the quadratic formula to get the roots: \(\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\). Note that we get some complex roots (since we had to take the square root of a negative number).


We’ll have to check all the answers;  they all work!

\(\begin{array}{c}\color{#800000}{{\frac{2}{{\left| {x-2} \right|}}=5}}\\\,\frac{2}{{\left| {x-2} \right|}}=\frac{5}{1}\\2\times 1=5\left| {x-2} \right|\\\,\left| {x-2} \right|=\frac{2}{5}\\\swarrow \,\,\,\,\searrow \\x-2=\frac{2}{5}\,\,\,\,\,\,\,\,\,\,x-2=-\frac{2}{5}\\x=\,2\frac{2}{5},\,\,\,1\frac{3}{5}\end{array}\)

Here’s another nonlinear equation where the absolute value is originally in the denominator (so it’s a rational function). We need to treat the absolute value like a variable, and get it out from the denominator by cross multiplying.


Then we can continue to solve, and divide up the equations to get the two answers. (Note that if we ended up with “2” as one of the answers, we’d have to eliminate it, since we can’t have a denominator that equals 0).


Check the answers – they work!

{{\,\left| {x-2} \right|=-\left| {-2} \right|}}\,\\\,\left| {x-2} \right|=-2\\\emptyset ,\text{ or no solution}\end{array}
Since the absolute value can never be a negative number, we can stop right away, and not separate the equations.

Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it. In this case, we have to separate in four cases, just to be sure we cover all the possibilities. We then must check for extraneous solutions, possible solutions that don’t work.


Solving with Absolute Value

Here’s another way to approach the absolute value problem above, using number lines (sort of like sign charts):

Solving Absolute Values using Number LinesSolving Absolute Value Inequalities

When dealing with absolute values and inequalities (just like with absolute value equations), we have to separate the equation into two different ones, if there are any variables inside the absolute value bars.


We first have to get the absolute value all by itself on the left.

Now we have to separate the equations. We get the first equation by just taking away the absolute value sign away on the left. The easiest way to get the second equation is to take the absolute value sign away on the left, and do two things on the right:  reverse the inequality sign, and change the sign of everything on the right (even if we have variables over there).

We also have to think about whether or not to use “or” or “and” between the two new equations. The way I remember this is that with a \(>\,\text{or}\,\,\ge  \) sign, you can remember “gore”: greater than uses “or”.

With a  \(<\,\text{or}\,\,\le  \) sign, think “land”: less than uses “and”.

GORE: Greater Than ­­­- OR

LAND:  Less Than – AND

Note that statement with “or” is a disjunction, which means that it works if only one (or both) parts are true. A statement with “and” is a conjunction, which means it only works if both parts are true.

And again, if we get something like \(\left| {x+3} \right|<0\) (or a negative number), there is no solution, and something like \(\left| {x+3} \right|\ge 0\) (or a negative number), there are infinite solutions (all real numbers).


Here are some examples:

Here are more examples that have absolute value rational function inequalities:

Rational Absolute Value Problem


Let’s do a simple one first, where we can handle the absolute value just like a factor, but when we do the checking, we’ll take into account that it is an absolute value.


\(\displaystyle \,\frac{{\left| {x+4} \right|}}{{x-1}}<0\)


The problem calls for \(<0\), so we look for the minus signs, but we can’t include –4.

Even with the absolute value, we can set each factor to 0, so we get –4 and 1 as critical values. We use open circles since we have a \(<\) sign (we’d have to for the 1 anyway, since it’s on the bottom).


Then we check each interval with random points to see if the factored form of the quadratic is positive or negative, making sure we include the absolute value. It’s a little different, since we have 2 minuses in a row without a “bounce” in the graph.


We want \(<\) from the problem, so we look for the \(-\) signs, but can’t include the –4 since it has a circle on it. The interval is \(\displaystyle \left( {-\infty ,-4} \right)\cup \left( {-4,1} \right)\).

\(\displaystyle \,\left| {\frac{2}{{x+2}}} \right|\ge 4\)


\(\displaystyle \begin{align}\frac{2}{{x+2}}\ge 4\text{ }\,\,\,&\text{or }\,\,\,\frac{2}{{x+2}}\le -4\\\frac{2}{{x+2}}-4\ge 0\text{ }\,\,\,&\text{or }\,\,\,\frac{2}{{x+2}}+4\le 0\\\frac{2}{{x+2}}-\frac{{4\left( {x+2} \right)}}{{x+2}}\ge 0\text{ }\,\,\,&\text{or }\,\,\,\frac{2}{{x+2}}+\frac{{4\left( {x+2} \right)}}{{x+2}}\le 0\\\frac{{-6-4x}}{{x+2}}\ge 0\text{ }\,\,\,&\text{or }\,\,\,\frac{{4x+10}}{{x+2}}\le 0\end{align}\)



This one’s a little more complicated since we don’t have a 0 on the right.


Let’s first separate the absolute value into two separate inequalities. Then we need to get everything to the left side to have 0 on the right first. Simplify with a common denominator.


Our critical values are \(\displaystyle -\frac{5}{2}\), –2, and \(\displaystyle -\frac{3}{2}\). The sign chart is to the left; we can use the original inequality and put T where it works and F where it doesn’t. We see the solution is: \(\displaystyle \left[ {-\frac{5}{2},-2} \right)\cup \left( {-2,-\frac{3}{2}} \right]\).


Note that we can’t include –2 as a solution (soft bracket) since it would make the denominator 0.

Math Notes

\(\displaystyle \frac{{\left| {x-1} \right|}}{{\left| {3x+1} \right|}}>1\)


\(x-1\) is positive,

\(3x+1\) is positive:


\(\displaystyle \begin{array}{c}\frac{{x-1}}{{3x+1}}>1\\\frac{{x-1}}{{3x+1}}-1>0\\\frac{{x-1}}{{3x+1}}-\frac{{3x+1}}{{3x+1}}>0\\\frac{{-2x-2}}{{3x+1}}>0\end{array}\)

\(x-1\)  is positive,

\(3x+1\) is negative:


\(\displaystyle \begin{array}{c}\frac{{x-1}}{{-3x-1}}>1\\\frac{{x-1}}{{-3x-1}}-1>0\\\frac{{x-1}}{{-3x-1}}-\frac{{-3x-1}}{{-3x-1}}>0\\\frac{{4x}}{{-3x-1}}>0\end{array}\)

\(x-1\)  is negative,

\(3x+1\) is positive:


\(\displaystyle \begin{array}{c}\frac{{-x+1}}{{3x+1}}>1\\\frac{{-x+1}}{{3x+1}}-1>0\\\frac{{-x+1}}{{3x+1}}-\frac{{3x+1}}{{3x+1}}>0\\\frac{{-4x}}{{3x+1}}>0\end{array}\)

\(x-1\)  is negative,

\(3x+1\) is negative:


\(\displaystyle \begin{array}{c}\frac{{-x+1}}{{-3x-1}}>1\\\frac{{-x+1}}{{-3x-1}}-1>0\\\frac{{-x+1}}{{-3x-1}}-\frac{{-3x-1}}{{-3x-1}}>0\\\frac{{2x+2}}{{-3x-1}}>0\end{array}\)

We need to separate into four cases, since we don’t know whether \(x-1\) and \(3x+1\) are positive or negative, since they are absolute values.


We probably could have just used two cases, since the absolute values are on the top and bottom of same fraction, but this way is safer.


The problem calls for \(>0\), so we look for the plus sign intervals, and make sure they work in the original inequality.



The answer is \(\displaystyle \left( {-1,\,\,-\frac{1}{3}} \right)\cup \left( {-\frac{1}{3},0} \right)\). We have to “skip over” (asymptote) \(\displaystyle -\frac{1}{3}\) (so we don’t divide by 0), and use soft brackets, since the inequality is \(>\) and not \(\ge \).


Put the inequality in your graphing calculator to check your answer!

Rational Inequality with Absolute Value Notes

\(\displaystyle \frac{{\left| x \right|}}{{x-1}}>\frac{{x+1}}{{2x+1}}\)


\(x\) is positive:


\(\displaystyle \begin{array}{c}\frac{x}{{x-1}}>\frac{{x+1}}{{2x+1}}\\\frac{x}{{x-1}}-\frac{{x+1}}{{2x+1}}>0\\\frac{{x\left( {2x+1} \right)-\left( {x+1} \right)\left( {x-1} \right)}}{{\left( {x-1} \right)\left( {2x+1} \right)}}>0\\\frac{{{{x}^{2}}+x+1}}{{\left( {x-1} \right)\left( {2x+1} \right)}}>0\end{array}\)


\(x\) is negative:


\(\displaystyle \begin{array}{c}\frac{{-x}}{{x-1}}>\frac{{x+1}}{{2x+1}}\\\frac{{-x}}{{x-1}}-\frac{{x+1}}{{2x+1}}>0\\\frac{{-x\left( {2x+1} \right)-\left( {x+1} \right)\left( {x-1} \right)}}{{\left( {x-1} \right)\left( {2x+1} \right)}}>0\\\frac{{-3{{x}^{2}}-x+1}}{{\left( {x-1} \right)\left( {2x+1} \right)}}>0\end{array}\)


The problem calls for \(>0\), so we look for the plus sign intervals. But we have to throw away any intervals where the sign doesn’t agree with our conditions of \(x\) (positive or negative), such as the interval \(\left( {.434,1} \right)\) (\(x\) is supposed to be negative). Tricky!

We need to separate into two cases, since we don’t know whether \(x\) is positive or negative. After separating into two cases, we need to get all the variables to one side and set to 0, so we can find the boundary values or critical points using a sign chart).

We need to look at the negative part of the sign chart when \(x\) is negative, and the positive part of the sign chart when \(x\) is positive.

When \(x\) is positive, the numerator yields no “real” critical points. When \(x\) is negative, we need to use the Quadratic Formula to get the roots (critical points) of the numerator:

\(\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}=\frac{{1\pm \sqrt{{{{{\left( {-1} \right)}}^{2}}-4\left( {-3} \right)\left( 1 \right)}}}}{{-6}}\)

\(\displaystyle =\frac{{1\pm \sqrt{{13}}}}{{-6}}\approx -.768,\,\,\,.434\)

The answer is \(\displaystyle \left( {.768,-\frac{1}{2}} \right)\cup \left( {1,\infty } \right)\).

Put the inequality in your graphing calculator to check your answer!


Graphs of Absolute Value Functions

Note that you can put absolute values in your Graphing Calculator (and even graph them!) by hitting MATH, scroll right to NUM, and then hitting 1 (abs) or ENTER.

Absolute Value functions typically look like a V (upside down if the absolute value is negative), where the point at the V is called the vertex. For the absolute value parent function, the vertex is at \(\left( {0,0} \right)\).

We looked at absolute value parent functions and their transformations here in the Absolute Value Transformations section, and absolute value functions as piecewise equations here in the Piecewise Functions section.

Note that the general form for the absolute value function is  \(f\left( x \right)=a\left| {x-h} \right|+k\), where \(\left( {h,k} \right)\) is the vertex. If \(a\) is positive, the function points down (like a V); if \(a\) is negative, the function points up (like an upside down V).

Here’s a graph of the parent function, and also a transformation:

Absolute Value Graphs

Applications of Absolute Value Functions

Absolute Value Functions are in many applications, especially in those involving V-shaped paths and margin of errors, or tolerances.

Here are some examples absolute value “word” problems that you may see:

Problem Solution
Two students are bouncing-passing a ball between them. The first student bounces the ball from 6 feet high and it bounces 5 feet away from her. The second student is 4 feet away from where the ball bounced.


Create an absolute value equation to represent the situation.

How high the did the ball bounce for the second student to catch it?

Let’s first draw a picture on the coordinate system. We can arbitrarily put the first point on the \(y\)-axis:

Let’s get the equation of this absolute value function: since the vertex is at \(\left( {5,0} \right)\), we have \(f\left( x \right)=a\left| {x-5} \right|+0=a\left| {x-5} \right|\). To find \(a\) (which should be positive), we can use the non-vertex point \(\left( {0,6} \right)\) for \(\left( {x,f\left( x \right)} \right)\):        \(\displaystyle 6=a\left| {0-5} \right|;\,\,\,\,\,\,\,6=5a;\,\,\,\,\,\,a=\frac{6}{5}\).

So the equation of the path of the ball is \(\displaystyle f\left( x \right)=\frac{6}{5}\left| {x-5} \right|\). Since the second student is 4 feet away from the vertex, we need to use \(9\,\,\left( {4+5} \right)\) for \(x\) to get the height of the ball at that point: \(\displaystyle f\left( x \right)=\frac{6}{5}\left| {9-5} \right|=\frac{6}{5}\times 4=\frac{{24}}{5}=4\frac{4}{5}\) feet.

Suppose that a coordinate grid is placed over a putt-putt golf course, where Amy is playing golf. The golf ball starts at \(\left( {3,2} \right)\), the hole is at \(\left( {5,2} \right)\), and Amy wants to bank the ball (bounce the ball off of a wall) at \(\left( {4,6} \right)\). 


Write an equation for the path of the ball.  

Let’s draw a picture on a coordinate system:

Let’s get the equation of this absolute value function:  since the vertex is at \(\left( {4,6} \right)\), we have \(f\left( x \right)=a\left| {x-4} \right|+6\). To find \(a\) (which should be negative), we can either put one of the non-vertex points in for \(x\) and \(y\), or notice that the graph is upside down and has a vertical stretch of 4. Let’s go ahead and use point \(\left( {3,2} \right)\) in:    \(2=a\left| {3-4} \right|+6;\,\,\,\,\,\,\,2=a+6;\,\,\,\,\,\,a=-4\).

So the equation of the path of the ball is \(f\left( x \right)=-4\left| {x-4} \right|+6\).

Here are examples that are absolute value inequality applications:

Problem Solution
\(x\) differs from 2.5 by less than .8


Express this situation as an absolute value inequality and solve for \(x\).


Since \(x\) differs from 2.5 by less than an amount, we could add or subtract 2.5 from \(x\) and that amount would have to be less than .8. By doing this, we can see that \(x\) would have to be between \(1.7\,\,\,(2.5-.8)\) and \(3.3\,\,\,(2.5+.8)\). We can put this into an absolute value equation since we don’t care if the difference of \(x\) and 2.5 is positive or negative: it still needs to be \(<\) .8:

\(\begin{array}{c}\left| {\,x-2.5} \right|<.8\\\,x-2.5<.8\text{ or }x-2.5>-.8\\\,x<3.3\text{ or }x>1.7\\\,\left( {1.7,\,\,3.3} \right)\end{array}\)

So \(x\) would need to be between 1.7 and 3.3. This makes sense since anything outside of these values would be more than .8 units from 2.5.

The thermostat at Lindsay’s house is set at 72°, but the temperature may vary by as much as


Write an absolute value inequality to model this situation, and solve for the range of possible temperatures, \(t\).

Let’s think about this with real numbers before we set up the absolute value inequality. Since the temperature can vary at most either way from 72°, Lindsay’s house could range anywhere from \(72-3\) degrees to \(72+3\) degrees, or from 69° to 75°, so we have \(72\pm 3\).

Now let’s try to write this using an absolute value. Since we have an “at most” , we will have something that should be \(\le \) 3. And we also know the difference of the temperature and 72 has to be in this range. But this difference could be positive or negative \((69-72=-3;\,\,\,\,\,\,75-72=3)\). Therefore, we can write this temperature range as an absolute value and solve:

.\(\displaystyle \begin{array}{c}\left| {\,t-72} \right|\le 3\\t-72\le 3\,\,\,\,\,\text{and}\,\,\,\,\,\,\,t-72\ge -3\,\,\,\\\,t\le 75\,\,\,\,\,\,\,\,\,\,\,\,t\ge 69\,\\69\le t\le 75\end{array}\)

This is tricky, but the way to do these problems is to always make an absolute value of the difference of the variable and the starting point (such as 72°) and make this  \(\le \)  the amount that varies (the ).

A bird is approaching Erin, a photographer, and she films it. She starts her video when the bird is 100 feet horizontally from her, and continues filming until the bird is at least 50 feet past her. The bird is flying at a rate of 30 feet per second.  


Write and solve an equation to find the times after Erin starts filming that the bird is 50 feet (horizontally) from her.

This one’s a little tricky since we have to start measuring the distance (and thus the time) from point 0 feet, which is 100 feet away from Erin. But then we need to capture the distance (which is rate times time, or 30t) and subtract it from 100, and add it to 100, and this needs to be within 50 feet of Erin.

So the absolute value inequality will be \(\left| {100-30t} \right|\,\,\le \,\,50\), or \(\left| {30t-100} \right|\,\,\le \,\,50\). Rule of thumb: The absolute value of the variable and starting point  \(\le \)  the part that varies.

At one restaurant, fresh lobsters are rejected if they weigh less than 1 pound, or more than 2.2 pounds.


Write an absolute value inequality to represent when lobsters are kept (not rejected) in the restaurant.

Here’s the trick for these types of problems: if we start out with the middle (average) of 1 and 2.2, we can go lower or higher by a certain amount and this would represent when lobsters are kept. Then we can turn the problem into an absolute value problem.


The average of the lowest and highest values is \(\displaystyle \frac{{1+2.2}}{2}=1.6\). Therefore, if we start out with 1.6, we can either go down to 1 (by subtracting .6) or up to 2.2 (but adding .6) to get lobsters that aren’t rejected.

Then can use the absolute value inequality \(\left| {x-1.6} \right|<.6\) to describe the weight of lobsters that aren’t rejected. Try it; it works!

Learn these rules, and practice, practice, practice!


Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

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On to Imaginary (Non-Real) and Complex Numbers – you’re ready!