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Absolute value equations

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The Absolute value equations exercise appears under the Algebra I Math Mission and Mathematics II Math Mission . This exercise practices solving absolute value and recognizing how these are different from but related to linear equations.

Types of Problems [ ]

There is one type of problem in this exercise:

Solve the absolute value equation

Strategies [ ]

Knowledge of the properties of equality and some experience with absolute value would help when performing this exercise.

• Once the absolute value is isolated, the absolute value can you be though of under it's distance definition to increase efficiency.
• Treat the absolute value as a variable, and isolate the absolute value first. Then the absolve value equation can be split into a "positive" equation and a "negative equation."
• If the absolute value expression is equal to a negative number there is no solution.

Real-life Applications [ ]

• Absolute value can be used in applications connecting algebra to geometry.
• The absolute value function is an example of a continuous but non-differentiable function (specifically at zero).
• Absolute value extends into the concept of modulus for complex numbers.
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1.2: Absolute Value Equations

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• Page ID 45028

• Darlene Diaz
• Santiago Canyon College via ASCCC Open Educational Resources Initiative

When solving equations with absolute value, the solution may result in more than one possible answer because, recall, absolute value is just distance from zero. Since the integer \(−4\) has distance \(4\) units from zero, and 4 has distance 4 units from zero, then there are two integers that have distance \(4\) from zero, \(−4,\: 4\). We extend this concept to algebraic absolute value equations. This is illustrated in the following example.

Example \(\PageIndex{1}\)

Solve for \(x\): \(|x|=7\)

\[\begin{array}{rl}|x|=7&\text{Expression in the absolute value can be positive or negative} \\ x=7\text{ or }x=-7&\text{Solution}\end{array}\nonumber\]

Let’s think about the solution set. The equation is asking for all numbers in which the distance from zero is \(7\). Well, there are two integers that have a distance \(7\) from zero, \(−7\) and \(7\). Hence, the solution set \(\{−7, 7\}\).

The first set of rules for working with negative numbers came from \(7^{\text{th}}\) century India. However, in 1758, more than a thousand years later, British mathematician Francis Maseres claimed that negatives “Darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple.”

Definition: Absolute Value

Absolute value for linear equations in one variable is given by \[\text{If }|x|=a,\text{ then }x=a\text{ or }x=-a\nonumber\] where \(a\) is a real number.

When we have an equation with absolute value, it is important to first isolate the absolute value, then remove the absolute value by applying the definition.

Example \(\PageIndex{2}\)

Solve for \(x\): \(5+|x|=8\)

\[\begin{array}{rl}5+|x|=8&\text{Isolate the absolute value by subtracting }5\text{ from each side} \\ |x|=3&\text{Rewrite as two linear equations} \\ x=3\text{ or }x=-3&\text{Solution}\end{array}\nonumber\]

Thus, the solution set is \(\{-3,3\}\).

Example \(\PageIndex{3}\)

Solve for \(x\): \(-4|x|=-20\)

\[\begin{array}{rl}-4|x|=-20&\text{Isolate the absolute value} \\ \frac{-4|x|}{-4}=\frac{-20}{-4}&\text{Divide each side by }-4 \\ |x|=5&\text{Rewrite as two linear equations} \\ x=5\text{ or }x=-5&\text{Solution}\end{array}\nonumber\]

Thus, the solution set is \(\{-5,5\}\).

Never combine the inside of the absolute value with factors or terms from outside the absolute value. We always have to isolate the absolute value first, then apply the definition to obtain two equations without the absolute value.

Example \(\PageIndex{4}\)

Solve for \(y\): \(5|y| − 4 = 26\)

\[\begin{array}{rl}5|y|-4=26&\text{Isolate the absolute value term by adding }4\text{ to each side} \\ 5|y|=30&\text{Divide each side by }5 \\ \\ |y|=6&\text{Rewrite as two linear equations} \\ y=6\quad\text{or}\quad y=-6&\text{Solution}\end{array}\nonumber\]

Thus, the solution set is \(\{-6,6\}\).

Absolute Value Equations with Different Solutions

Often, we will have linear arguments inside the absolute value which changes the solution. Previously, all solution sets have been opposite integers, but in these cases, the solution sets contain different sized numbers.

Example \(\PageIndex{5}\)

Solve for \(t\): \(|2t − 1| = 7\)

\[\begin{array}{rl} |2t-1|=7&\text{The absolute value term is isolated. Rewrite as two linear equations.} \\ 2t-1=7\quad\text{or}\quad 2t-1=-7&\text{Solve each equation.}\end{array}\nonumber\]

Notice we have two equations to solve where each equation results in a different solution. In any case, we solve as usual.

\[\begin{array}{lll} 2t-1=7&& 2t-1=-7 \\ 2t=8&\text{or}& 2t=-6 \\ t=4&&t=-3\end{array}\nonumber\]

Thus, the solution set is \(\{-3,4\}\).

Multiple-Step Absolute Value Equations

Example \(\pageindex{6}\).

Solve for \(x\): \(2 − 4|2x + 3| = −18\)

To isolate the absolute value, we first apply the addition rule for equations. Then apply the multiplication rule for equations.

\[\begin{array}{rl}2-4|2x+3|=-18&\text{Isolate the absolute value term by subtracting }2\text{ from each side} \\ -4|2x+3|=-20&\text{Divide each side by }-4 \\ |2x+3|=5&\text{Rewrite as two linear equations.} \\ 2x+3=5\quad\text{or}\quad 2x+3=-5\end{array}\nonumber\]

Solve each equation.

\[\begin{array}{lll}2x+3=5&&2x+3=-5 \\ 2x=2&\text{or}& 2x=-8 \\ x=1&&x=-4\end{array}\nonumber\]

We now have obtained two solutions, \(x = 1\) and \(x = −4\). Thus, the solution set is \(\{−4, 1\}\).

Equations with Two Absolute Values

In this case, we have an absolute value on each side of the equals sign. However, even though there are two absolute values, we apply the same process. Recall, methods never change, only problems.

Example \(\PageIndex{7}\)

Solve for \(m\): \(|2m-7|=|4m+6|\)

In order to apply the definition, we rewrite this equation as two linear equations, but with the left side as its positive and negative value: \[\begin{array}{rl}|2m-7|=|4m+6|&\text{Rewrite as two linear equations} \\ 2m-7=\color{blue}{4m+6}\color{black}{}\quad\text{or}\quad 2m-7=\color{blue}{-(4m+6)}\color{black}{}\end{array}\nonumber\]

Now, we can solve as usual. Be sure to distribute the negative for the equation on the right.

\[\begin{array}{lll} &&2m-7=\color{blue}{-}\color{black}{}(4m+6) \\ 2m-7=4m+6&&2m-7=\color{blue}{-}\color{black}{}4m\color{blue}{-}\color{black}{}6 \\ -13=2m&&6m-7=-6 \\ -\frac{13}{2}=m&\text{or}&6m=1 \\ &&m=\frac{1}{6}\end{array}\nonumber\]

Thus gives the solutions, \(m=-\frac{13}{2}\) or \(m=\frac{1}{6}\). Thus, the solution set is \(\left\{-\frac{13}{2},\frac{1}{6}\right\}\).

In Example \(\PageIndex{7}\) , because there are absolute value expressions on both sides of the equation, we could have easily applied the definition to the left side and obtained \[\color{blue}{2m-7}\color{black}{}=4m-6\quad\text{or}\quad\color{blue}{-(2m-7)}\color{black}{}=4m-6\nonumber\]

Then solved each linear equation as usual and obtained the same results.

Special Cases

As we are solving absolute value equations, it is important to be aware of special cases. Remember, the result after evaluating absolute value must always be non-negative.

Example \(\PageIndex{8}\)

Solve for \(x\): \(7 + |2x − 5| = 4\)

\[\begin{array}{rl}7+|2x-5|=4&\text{Isolate the absolute value term by subtracting }7\text{ from each side} \\ |2x-5|=-3&X\text{ False}\end{array}\nonumber\]

Careful! Observe the absolute value of \(2x − 5\) is a negative number. This is impossible with absolute value because the result after evaluating absolute value must always be non-negative . Thus, we say this equation has no solution .

Absolute Value Equations Homework

Exercise \(\pageindex{1}\).

\(|x| = 8\)

Exercise \(\PageIndex{2}\)

\(|b| = 1\)

Exercise \(\PageIndex{3}\)

\(|5 + 8a| = 53\)

Exercise \(\PageIndex{4}\)

\(|3k + 8| = 2\)

Exercise \(\PageIndex{5}\)

\(|9 + 7x| = 30\)

Exercise \(\PageIndex{6}\)

\(|8 + 6m| = 50\)

Exercise \(\PageIndex{7}\)

\(|6 − 2x| = 24\)

Exercise \(\PageIndex{8}\)

\(−7| − 3 − 3r| = −21\)

Exercise \(\PageIndex{9}\)

\(7| − 7x − 3| = 21\)

Exercise \(\PageIndex{10}\)

\(\frac{|−4b − 10|}{8} = 3\)

Exercise \(\PageIndex{11}\)

\(8|x + 7| − 3 = 5\)

Exercise \(\PageIndex{12}\)

\(5|3 + 7m| + 1 = 51\)

Exercise \(\PageIndex{13}\)

\(3 + 5|8 − 2x| = 63\)

Exercise \(\PageIndex{14}\)

\(|6b − 2| + 10 = 44\)

Exercise \(\PageIndex{15}\)

\(−7 + 8| − 7x − 3| = 73\)

Exercise \(\PageIndex{16}\)

\(|5x + 3| = |2x − 1|\)

Exercise \(\PageIndex{17}\)

\(|3x − 4| = |2x + 3|\)

Exercise \(\PageIndex{18}\)

\(\left|\frac{4x-2}{5}\right|=\left|\frac{6x+3}{2}\right|\)

Exercise \(\PageIndex{19}\)

\(|n| = 7\)

Exercise \(\PageIndex{20}\)

\(|x| = 2\)

Exercise \(\PageIndex{21}\)

\(|9n + 8| = 46\)

Exercise \(\PageIndex{22}\)

\(|3 − x| = 6\)

Exercise \(\PageIndex{23}\)

\(|5n + 7| = 23\)

Exercise \(\PageIndex{24}\)

\(|9p + 6| = 3\)

Exercise \(\PageIndex{25}\)

\(|3n − 2| = 7\)

Exercise \(\PageIndex{26}\)

\(|2 + 2b| + 1 = 3\)

Exercise \(\PageIndex{27}\)

\(\frac{|-4-3n|}{4}=2\)

Exercise \(\PageIndex{28}\)

\(8|5p + 8| − 5 = 11\)

Exercise \(\PageIndex{29}\)

\(3 − |6n + 7| = −40\)

Exercise \(\PageIndex{30}\)

\(4|r + 7| + 3 = 59\)

Exercise \(\PageIndex{31}\)

\(5 + 8| − 10n − 2| = 101\)

Exercise \(\PageIndex{32}\)

\(7|10v − 2| − 9 = 5\)

Exercise \(\PageIndex{33}\)

\(8|3 − 3n| − 5 = 91\)

Exercise \(\PageIndex{34}\)

\(|2 + 3x| = |4 − 2x|\)

Exercise \(\PageIndex{35}\)

\(\left|\frac{2x-5}{3}\right|=\left|\frac{3x+4}{2}\right|\)

Exercise \(\PageIndex{36}\)

\(\frac{|-n+6|}{6}=0\)

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Algebra 1 : How to solve absolute value equations

Study concepts, example questions & explanations for algebra 1, all algebra 1 resources, example questions, example question #1 : how to solve absolute value equations.

Solve the absolute value equation:

Solve for x.

No solution

So therefore, the solution is x = –7, 1.

Example Question #3 : How To Solve Absolute Value Equations

Find the solution to x for |x – 3| = 2.

|x – 3| = 2 means that it can be separated into x – 3 = 2 and x – 3 = –2.

So both x = 5 and x = 1 work.

x – 3 = 2 Add 3 to both sides to get x = 5

x – 3 = –2 Add 3 to both sides to get x = 1

Example Question #4 : How To Solve Absolute Value Equations

Solve for x:

Because of the absolute value signs,

Subtract 2 from both sides of both equations:

Example Question #5 : How To Solve Absolute Value Equations

There are two answers to this problem:

Example Question #6 : How To Solve Absolute Value Equations

Example Question #7 : How To Solve Absolute Value Equations

Example Question #8 : How To Solve Absolute Value Equations

What is the solution set of this equation?

Example Question #9 : How To Solve Absolute Value Equations

Example Question #10 : How To Solve Absolute Value Equations

Solve each separately to obtain the solution set:

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How to Solve Absolute Value Equations

When Do You Flip the Inequality Sign?

Absolute value equations can be a little intimidating at first, but if you keep at it you'll soon be solving them easily. When you're trying to solve absolute value equations, it helps to keep the meaning of absolute value in mind.

The ​ absolute value ​ of a number ​ x ​, written | ​ x ​ |, is its distance from zero on a number line. For instance, −3 is 3 units away from zero, so the absolute value of −3 is 3. We write it like this: | −3 | = 3.

Another way to think about it is that ​ absolute value ​ is the positive "version" of a number. So the absolute value of −3 is 3, while the absolute value of 9, which is already positive, is 9.

Algebraically, we can write a ​ formula for absolute value ​ that looks like this:

Take an example where ​ x ​ = 3. Since 3 ≥ 0, the absolute value of 3 is 3 (in absolute value notation, that's: | 3 | = 3).

Now what if ​ x ​ = −3? It's less than zero, so | −3 | = −( −3). The opposite, or "negative," of −3 is 3, so | −3 | = 3.

Now for some absolute value equations. The general steps for solving an absolute value equation are:

Isolate the absolute value expression.

Solve the positive "version" of the equation.

Solve the negative "version" of the equation by multiplying the quantity on the other side of the equals sign by −1.

Take a look at the problem below for a concrete example of the steps.

Example: Solve the equation for ​ x ​:

You'll need to get | 3 + ​ x ​ | by itself on the left side of the equals sign. To do this, add 5 to both sides:

Solve for ​ x ​ as if the absolute value sign weren't there!

That's easy: Just subtract 3 from both sides.

So one solution to the equation is that ​ x ​ = 6.

Start again at | 3 + ​ x ​ | = 9. The algebra in the previous step showed that ​ x ​ could be 6. But since this is an absolute value equation, there's another possibility to consider. In the equation above, the absolute value of "something" (3 + ​ x ​) equals 9. Sure, the absolute value of positive 9 equals 9, but there's another option here too! The absolute value of −9 also equals 9. So the unknown "something" could also equal −9.

In other words:

The quick way to arrive at this second version is to multiply the quantity on the other side of the equals from the absolute value expression (9, in this case) by −1, then solve the equation from there.

Subtract 3 from both sides to get:

So the two solutions are: ​ x ​ = 6 or ​ x ​ = −12.

And there you have it! These kinds of equations take practice, so don't worry if you're struggling at first. Keep at it and it will get easier!

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• Khan Academy: Intro to absolute value equations and graphs
• Lamar University: Absolute Value Equations
• Wolfram MathWorld: Absolute Value

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How to solve absolute value equations

|x + 5| = 3.

Worksheet on Abs Val Equations Abs Val Eqn Solver

The General Steps to solve an absolute value equation are:

• Rewrite the absolute value equation as two separate equations, one positive and the other negative
• Solve each equation separately
• After solving, substitute your answers back into original equation to verify that you solutions are valid
• Write out the final solution or graph it as needed

It's always easiest to understand a math concept by looking at some examples so, check outthe many examples and practice problems below.

You can always check your work with our Absolute value equations solver too

Practice Problems

Example equation.

Solve the equation: | X + 5| = 3

Click here to practice more problems like this one , questions that involve variables on 1 side of the equation.

Some absolute value equations have variables both sides of the equation. However, that will not change the steps we're going to follow to solve the problem as the example below shows:

Solve the equation: |3 X | = X − 21

Solve the following absolute value equation: | 5X +20| = 80

Solve the following absolute value equation: | X | + 3 = 2 X

This first set of problems involves absolute values with x on just 1 side of the equation (like problem 2 ).

Solve the following absolute value equation: |3 X −6 | = 21

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About absolute value equations.

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