## Absolute Value Equations Exercises

Absolute value equations practice problems with answers.

There are eleven (11) practice problems in this collection regarding absolute value equations. Whether you’re a beginner or seeking a challenge, there’s something here for you. As you tackle each problem, remember that in every attempt, right or wrong, fuels your mathematical growth. Good luck!

Problem 1: Solve the absolute value equation below.

[latex]\left| {x + 2} \right| = – 3[/latex]

Since the absolute value can’t be negative, therefore it has [latex]\color{red}\text{no solution}[/latex].

Problem 2: Solve the absolute value equation below.

[latex]\left| {2x \,- \,3} \right| = 7[/latex]

Solve the positive and negative cases.

Therefore, the solution set is [latex]\{ – 2,5\}[/latex].

Problem 3: Solve the absolute value equation below.

[latex] – 7\left| {x\, – \,2} \right| = – 21[/latex]

Isolate the absolute value expression on the left side by dividing both sides of the equation by [latex]-7[/latex]

Now, break the absolute value equation into two cases then solve.

Therefore, the solution set is [latex]{ – 1,5}[/latex].

Problem 4: Solve the absolute value equation below.

[latex]4\, – \,\left| {3x\,+ \,6} \right| = 1[/latex]

Isolate the absolute value expression by subtracting [latex]4[/latex] from both sides. Then, divide [latex]-1[/latex] from both sides.

Break the absolute value equation into two equations with positive and negative cases then solve.

Therefore, the solution set is [latex]\{ – 3,-1\}[/latex].

Problem 5: Solve the absolute value equation below.

[latex]\left| {x \,- \,{1 \over 2}} \right| + \left| {x \, – \,{1 \over 2}} \right| = 7[/latex]

The absolute value expressions on the left side are similar terms therefore we can combine them.

Finally, divide [latex]2[/latex] from sides of the equation to isolate the absolute value expression.

Now, we solve each case.

Therefore, the solution set is [latex]\{ – 3,4\}[/latex].

Problem 6: Solve the absolute value equation below.

[latex]\left| {3{x^2}\, – \,2x} \right| = 5[/latex]

Split the absolute value equation into two cases then solve.

Therefore, the solution set is [latex]\{ – 1,{\large{5 \over 3}}\}[/latex].

Problem 7: Solve the absolute value equation below.

[latex]{\Large\left| {{{x \,- \,2} \over {x\, + \,1}}} \right|} ={ \Large{1 \over 2}}[/latex]

Separate the absolute value equation into positive and negative cases then solve.

Therefore, the solution set is [latex]\{ 1,5\}[/latex].

Problem 8: Solve the absolute value equation below.

[latex]\left| {x\, – \,8} \right| = 2x\, -\, 1[/latex]

Consider the positive and negative cases then solve each equation.

Validate the answer [latex]x=-7[/latex] if it is indeed a solution by plugging it back into the original absolute value equation.

We arrived at a contradiction since [latex]15=-15[/latex] is false. That means [latex]x=-7[/latex] is not a solution.

Let’s also validate [latex]x=3[/latex].

It yields a true statement which means [latex]x=3[/latex] is a solution.

Therefore, the solution set is [latex]\{3\}[/latex].

Problem 9: Solve the absolute value equation below.

[latex]\left| {2x + 5} \right| = x + 7[/latex]

Break the absolute value equation into two cases of linear equations the solve.

Substitute [latex]x=2[/latex] back into the original absolute value equation to check if it’s true or not. If true, then it is a solution. If false, then it is not.

It checks! Therefore, [latex]x=2[/latex] is a solution.

We also validate [latex]x=-4[/latex] using the same procedure as above.

Since it gives a true statement that means [latex]x=-4[/latex] is also a solution.

Therefore, the solution set is [latex]\{ -4,2\}[/latex].

Problem 10: Solve the absolute value equation below.

[latex]\left| {2x + 20} \right| = \left| {3x – 5} \right|[/latex]

Set up the equations then solve.

Therefore, the solution set is [latex]\{ -3,25\}[/latex].

Problem 11: Solve the absolute value equation below.

[latex]\left| { – 9x + 10} \right| = \left| {11x} \right|[/latex]

Set up the absolute value equation into cases then solve.

Therefore, the solution set is [latex]\{ -5,1/2\}[/latex].

You may also be interested in these related math lessons or tutorials:

Solving Absolute Value Equations

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## 2.6: Solving Absolute Value Equations and Inequalities

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Learning Objectives

- Review the definition of absolute value.
- Solve absolute value equations.
- Solve absolute value inequalities.

## Absolute Value Equations

Recall that the absolute value 63 of a real number \(a\), denoted \(|a|\), is defined as the distance between zero (the origin) and the graph of that real number on the number line. For example, \(|−3|=3\) and \(|3|=3\).

In addition, the absolute value of a real number can be defined algebraically as a piecewise function.

\(| a | = \left\{ \begin{array} { l } { a \text { if } a \geq 0 } \\ { - a \text { if } a < 0 } \end{array} \right.\)

Given this definition, \(|3| = 3\) and \(|−3| = − (−3) = 3\).Therefore, the equation \(|x| = 3\) has two solutions for \(x\), namely \(\{±3\}\). In general, given any algebraic expression \(X\) and any positive number \(p\):

\(\text{If}\: | X | = p \text { then } X = - p \text { or } X = p\)

In other words, the argument of the absolute value 64 \(X\) can be either positive or negative \(p\). Use this theorem to solve absolute value equations algebraically.

Example \(\PageIndex{1}\):

Solve: \(|x+2|=3\).

In this case, the argument of the absolute value is \(x+2\) and must be equal to \(3\) or \(−3\).

Therefore, to solve this absolute value equation, set \(x+2\) equal to \(±3\) and solve each linear equation as usual.

\(\begin{array} { c } { | x + 2 | = 3 } \\ { x + 2 = - 3 \quad \quad\text { or } \quad\quad x + 2 = 3 } \\ { x = - 5 \quad\quad\quad\quad\quad\quad\quad x = 1 } \end{array}\)

The solutions are \(−5\) and \(1\).

To visualize these solutions, graph the functions on either side of the equal sign on the same set of coordinate axes. In this case, \(f (x) = |x + 2|\) is an absolute value function shifted two units horizontally to the left, and \(g (x) = 3\) is a constant function whose graph is a horizontal line. Determine the \(x\)-values where \(f (x) = g (x)\).

From the graph we can see that both functions coincide where \(x = −5\) and \(x = 1\). The solutions correspond to the points of intersection.

Example \(\PageIndex{2}\):

Solve: \(| 2 x + 3 | = 4\).

Here the argument of the absolute value is \(2x+3\) and can be equal to \(-4\) or \(4\).

\(\begin{array} { r l } { | 2 x + 3 | } & { = \quad 4 } \\ { 2 x + 3 = - 4 } & { \text { or }\quad 2 x + 3 = 4 } \\ { 2 x = - 7 } & \quad\quad\:\: { 2 x = 1 } \\ { x = - \frac { 7 } { 2 } } & \quad\quad\:\: { x = \frac { 1 } { 2 } } \end{array}\)

Check to see if these solutions satisfy the original equation.

Check \(x=-\frac{7}{2}\) | Check \(x=\frac{1}{2}\) |

\(\begin{aligned} | 2 x + 3 | & = 4 \\ \left| 2 \left( \color{Cerulean}{- \frac { 7 } { 2 }} \right) + 3 \right| & = 4 \\ | - 7 + 3 | & = 4 \\ | - 4 | & = 4 \\ 4 & = 4 \:\:\color{Cerulean}{✓} \end{aligned}\) | \(\begin{array} { r } { | 2 x + 3 | = 4 } \\ { \left| 2 \left( \color{Cerulean}{\frac { 1 } { 2 }} \right) + 3 \right| = 4 } \\ { | 1 + 3 | = 4 } \\ { | 4 | = 4 } \\ { 4 = 4 } \:\:\color{Cerulean}{✓} \end{array}\) |

The solutions are \(-\frac{7}{2}\) and \(\frac{1}{2}\).

To apply the theorem, the absolute value must be isolated. The general steps for solving absolute value equations are outlined in the following example.

Example \(\PageIndex{3}\):

Solve: \(2 |5x − 1| − 3 = 9\).

Step 1 : Isolate the absolute value to obtain the form \(|X| = p\).

\(\begin{aligned} 2 | 5 x - 1 | - 3 & = 9 \:\:\:\color{Cerulean} { Add\: 3\: to\: both\: sides. } \\ 2 | 5 x - 1 | & = 12 \:\:\color{Cerulean} { Divide\: both\: sides\: by\: 2 } \\ | 5 x - 1 | & = 6 \end{aligned}\)

Step 2 : Set the argument of the absolute value equal to \(±p\). Here the argument is \(5x − 1\) and \(p = 6\).

\(5 x - 1 = - 6 \text { or } 5 x - 1 = 6\)

Step 3 : Solve each of the resulting linear equations.

\(\begin{array} { r l } { 5 x - 1 = - 6 \quad\:\:\text { or } \quad\quad5 x - 1 } & { \:\:\:\:\:\:= 6 } \\ { 5 x = - 5 }\quad\:\quad\quad\quad\quad\quad\quad\: & { 5 x = 7 } \\ { x = - 1 } \quad\quad\quad\quad\quad\quad\quad\:\:& { x = \frac { 7 } { 5 } } \end{array}\)

Step 4 : Verify the solutions in the original equation.

Check \(x=-1\) | Check \(x=\frac{7}{5}\) |

\(\begin{aligned} 2 | 5 x - 1 | - 3 & = 9 \\ 2 | 5 ( \color{Cerulean}{- 1}\color{Black}{ )} - 1 | - 3 & = 9 \\ 2 | - 5 - 1 | - 3 & = 9 \\ 2 | - 6 | - 3 & = 9 \\ 12 - 3 & = 9 \\ 9 & = 9 \color{Cerulean}{✓}\end{aligned}\) | \(\begin{aligned} 2 | 5 x - 1 | - 3 & = 9 \\ 2 \left| 5 \left( \color{Cerulean}{\frac { 7 } { 5 }} \right) - 1 \right| - 3 & = 9 \\ 2 | 7 - 1 | - 3 & = 9 \\ 2 | 6 | - 3 & = 9 \\ 12 - 3 & = 9 \\ 9 & = 9 \color{Cerulean}{✓} \end{aligned}\) |

The solutions are \(-1\) and \(\frac{7}{5}\)

Exercise \(\PageIndex{1}\)

Solve: \(2 - 7 | x + 4 | = - 12\).

www.youtube.com/v/G0EjbqreYmU

Not all absolute value equations will have two solutions.

Example \(\PageIndex{4}\):

Solve: \(| 7 x - 6 | + 3 = 3\).

Begin by isolating the absolute value.

\(\begin{array} { l } { | 7 x - 6 | + 3 = 3 \:\:\:\color{Cerulean} { Subtract\: 3\: on\: both\: sides.} } \\ { \quad | 7 x - 6 | = 0 } \end{array}\)

Only zero has the absolute value of zero, \(|0| = 0\). In other words, \(|X| = 0\) has one solution, namely \(X = 0\). Therefore, set the argument \(7x − 6\) equal to zero and then solve for \(x\).

\(\begin{aligned} 7 x - 6 & = 0 \\ 7 x & = 6 \\ x & = \frac { 6 } { 7 } \end{aligned}\)

Geometrically, one solution corresponds to one point of intersection.

The solution is \(\frac{6}{7}\).

Example \(\PageIndex{5}\):

Solve: \(|x+7|+5=4\).

\(\begin{aligned} | x + 7 | + 5 & = 4 \:\:\color{Cerulean} { Subtract \: 5\: on\: both\: sides.} \\ | x + 7 | & = - 1 \end{aligned}\)

In this case, we can see that the isolated absolute value is equal to a negative number. Recall that the absolute value will always be positive. Therefore, we conclude that there is no solution. Geometrically, there is no point of intersection.

There is no solution, \(Ø\).

If given an equation with two absolute values of the form \(| a | = | b |\), then \(b\) must be the same as \(a\) or opposite. For example, if \(a=5\), then \(b = \pm 5\) and we have:

\(| 5 | = | - 5 | \text { or } | 5 | = | + 5 |\)

In general, given algebraic expressions \(X\) and \(Y\):

\(\text{If} | X | = | Y | \text { then } X = - Y \text { or } X = Y\).

In other words, if two absolute value expressions are equal, then the arguments can be the same or opposite.

Example \(\PageIndex{6}\):

Solve: \(| 2 x - 5 | = | x - 4 |\).

Set \(2x-5\) equal to \(\pm ( x - 4 )\) and then solve each linear equation.

\(\begin{array} { c } { | 2 x - 5 | = | x - 4 | } \\ { 2 x - 5 = - ( x - 4 ) \:\: \text { or }\:\: 2 x - 5 = + ( x - 4 ) } \\ { 2 x - 5 = - x + 4 }\quad\quad\quad 2x-5=x-4 \\ { 3 x = 9 }\quad\quad\quad\quad\quad\quad \quad\quad x=1 \\ { x = 3 \quad\quad\quad\quad\quad\quad\quad\quad\quad\:\:\:\:} \end{array}\)

To check, we substitute these values into the original equation.

Check \(x=1\) | Check \(x=3\) |

\(\begin{aligned} | 2 x - 5 | & = | x - 4 | \\ | 2 ( \color{Cerulean}{1}\color{Black}{ )} - 5 | & = | ( \color{Cerulean}{1}\color{Black}{ )} - 4 | \\ | - 3 | & = | - 3 | \\ 3 & = 3 \color{Cerulean}{ ✓}\end{aligned}\) | \(\begin{aligned} | 2 x - 5 | & = | x - 4 | \\ | 2 ( \color{Cerulean}{3}\color{Black}{ )} - 5 | & = | ( \color{Cerulean}{3}\color{Black}{ )} - 4 | \\ | 1 | & = | - 1 | \\ 1 & = 1 \color{Cerulean}{✓}\end{aligned}\) |

As an exercise, use a graphing utility to graph both \(f(x)= |2x-5|\) and \(g(x)=|x-4|\) on the same set of axes. Verify that the graphs intersect where \(x\) is equal to \(1\) and \(3\).

The solutions are \(1\) and \(3\).

Exercise \(\PageIndex{2}\)

Solve: \(| x + 10 | = | 3 x - 2 |\).

www.youtube.com/v/CskWmsQCBMU

## Absolute Value Inequalities

We begin by examining the solutions to the following inequality:

\(| x | \leq 3\)

The absolute value of a number represents the distance from the origin. Therefore, this equation describes all numbers whose distance from zero is less than or equal to \(3\). We can graph this solution set by shading all such numbers.

Certainly we can see that there are infinitely many solutions to \(|x|≤3\) bounded by \(−3\) and \(3\). Express this solution set using set notation or interval notation as follows:

\(\begin{array} { c } { \{ x | - 3 \leq x \leq 3 \} \color{Cerulean} { Set\: Notation } } \\ { [ - 3,3 ] \quad \color{Cerulean}{ Interval \:Notation } } \end{array}\)

In this text, we will choose to express solutions in interval notation. In general, given any algebraic expression \(X\) and any positive number \(p\):

\(\text{If} | X | \leq p \text { then } - p \leq X \leq p\).

This theorem holds true for strict inequalities as well. In other words, we can convert any absolute value inequality involving " less than " into a compound inequality which can be solved as usual.

Example \(\PageIndex{7}\):

Solve and graph the solution set: \(|x+2|<3\).

Bound the argument \(x+2\) by \(−3\) and \(3\) and solve.

\(\begin{array} { c } { | x + 2 | < 3 } \\ { - 3 < x + 2 < 3 } \\ { - 3 \color{Cerulean}{- 2}\color{Black}{ <} x + 2 \color{Cerulean}{- 2}\color{Black}{ <} 3 \color{Cerulean}{- 2} } \\ { - 5 < x < 1 } \end{array}\)

Here we use open dots to indicate strict inequalities on the graph as follows.

Using interval notation, \((−5,1)\).

The solution to \(| x + 2 | < 3\) can be interpreted graphically if we let \(f ( x ) = | x + 2 |\) and \(g(x)=3\) and then determine where \(f ( x ) < g ( x )\) by graphing both \(f\) and \(g\) on the same set of axes.

The solution consists of all \(x\)-values where the graph of \(f\) is below the graph of \(g\). In this case, we can see that \(|x + 2| < 3\) where the \(x\)-values are between \(−5\) and \(1\). To apply the theorem, we must first isolate the absolute value.

Example \(\PageIndex{8}\):

Solve: \(4 |x + 3| − 7 ≤ 5\).

\(\begin{array} { c } { 4 | x + 3 | - 7 \leq 5 } \\ { 4 | x + 3 | \leq 12 } \\ { | x + 3 | \leq 3 } \end{array}\)

Next, apply the theorem and rewrite the absolute value inequality as a compound inequality.

\(\begin{array} { c } { | x + 3 | \leq 3 } \\ { - 3 \leq x + 3 \leq 3 } \end{array}\)

\(\begin{aligned} - 3 \leq x + 3 \leq & 3 \\ - 3 \color{Cerulean}{- 3} \color{Black}{ \leq} x + 3 \color{Cerulean}{- 3} & \color{Black}{ \leq} 3 \color{Cerulean}{- 3} \\ - 6 \leq x \leq 0 \end{aligned}\)

Shade the solutions on a number line and present the answer in interval notation. Here we use closed dots to indicate inclusive inequalities on the graph as follows:

Using interval notation, \([−6,0]\)

Exercise \(\PageIndex{3}\)

Solve and graph the solution set: \(3 + | 4 x - 5 | < 8\).

Interval notation: \((0, \frac{5}{2})\)

www.youtube.com/v/sX6ppL2Fbq0

Next, we examine the solutions to an inequality that involves " greater than ," as in the following example:

\(| x | \geq 3\)

This inequality describes all numbers whose distance from the origin is greater than or equal to \(3\). On a graph, we can shade all such numbers.

There are infinitely many solutions that can be expressed using set notation and interval notation as follows:

\(\begin{array} { l } { \{ x | x \leq - 3 \text { or } x \geq 3 \} \:\:\color{Cerulean} { Set\: Notation } } \\ { ( - \infty , - 3 ] \cup [ 3 , \infty ) \:\:\color{Cerulean} { Interval\: Notation } } \end{array}\)

In general, given any algebraic expression \(X\) and any positive number \(p\):

\(\text{If} | X | \geq p \text { then } X \leq - p \text { or } X \geq p\).

The theorem holds true for strict inequalities as well. In other words, we can convert any absolute value inequality involving “ greater than ” into a compound inequality that describes two intervals.

Example \(\PageIndex{9}\):

Solve and graph the solution set: \(|x+2|>3\).

The argument \(x+2\) must be less than \(−3\) or greater than \(3\).

\(\begin{array} { c } { | x + 2 | > 3 } \\ { x + 2 < - 3 \quad \text { or } \quad x + 2 > 3 } \\ { x < - 5 }\quad\quad\quad\quad\quad\: x>1 \end{array}\)

Using interval notation, \((−∞,−5)∪(1,∞)\).

The solution to \(|x + 2| > 3\) can be interpreted graphically if we let \(f (x) = |x + 2|\) and \(g (x) = 3\) and then determine where \(f(x) > g (x)\) by graphing both \(f\) and \(g\) on the same set of axes.

The solution consists of all \(x\)-values where the graph of \(f\) is above the graph of \(g\). In this case, we can see that \(|x + 2| > 3\) where the \(x\)-values are less than \(−5\) or are greater than \(1\). To apply the theorem we must first isolate the absolute value.

Example \(\PageIndex{10}\):

Solve: \(3 + 2 |4x − 7| ≥ 13\).

\(\begin{array} { r } { 3 + 2 | 4 x - 7 | \geq 13 } \\ { 2 | 4 x - 7 | \geq 10 } \\ { | 4 x - 7 | \geq 5 } \end{array}\)

\(\begin{array} &\quad\quad\quad\quad\:\:\:|4x-7|\geq 5 \\ 4 x - 7 \leq - 5 \quad \text { or } \quad 4 x - 7 \geq 5 \end{array}\)

\(\begin{array} { l } { 4 x - 7 \leq - 5 \text { or } 4 x - 7 \geq 5 } \\ \quad\:\:\:\:{ 4 x \leq 2 } \quad\quad\quad\:\:\: 4x\geq 12\\ \quad\:\:\:\:{ x \leq \frac { 2 } { 4 } } \quad\quad\quad\quad x\geq 3 \\ \quad\quad{ x \leq \frac { 1 } { 2 } } \end{array}\)

Shade the solutions on a number line and present the answer using interval notation.

Using interval notation, \((−∞,\frac { 1 } { 2 }]∪[3,∞)\)

Exercise \(\PageIndex{4}\)

Solve and graph: \(3 | 6 x + 5 | - 2 > 13\).

Using interval notation, \(\left( - \infty , - \frac { 5 } { 3 } \right) \cup ( 0 , \infty )\)

www.youtube.com/v/P6HjRz6W4F4

Up to this point, the solution sets of linear absolute value inequalities have consisted of a single bounded interval or two unbounded intervals. This is not always the case.

Example \(\PageIndex{11}\):

Solve and graph: \(|2x−1|+5>2\).

\(\begin{array} { c } { | 2 x - 1 | + 5 > 2 } \\ { | 2 x - 1 | > - 3 } \end{array}\)

Notice that we have an absolute value greater than a negative number. For any real number x the absolute value of the argument will always be positive. Hence, any real number will solve this inequality.

Geometrically, we can see that \(f(x)=|2x−1|+5\) is always greater than \(g(x)=2\).

All real numbers, \(ℝ\).

Example \(\PageIndex{12}\):

Solve and graph: \(|x+1|+4≤3\).

\(\begin{array} { l } { | x + 1 | + 4 \leq 3 } \\ { | x + 1 | \leq - 1 } \end{array}\)

In this case, we can see that the isolated absolute value is to be less than or equal to a negative number. Again, the absolute value will always be positive; hence, we can conclude that there is no solution.

Geometrically, we can see that \(f(x)=|x+1|+4\) is never less than \(g(x)=3\).

Answer : \(Ø\)

In summary, there are three cases for absolute value equations and inequalities. The relations \(=, <, \leq, > \) and \(≥\) determine which theorem to apply.

## Case 1: An absolute value equation:

\(\begin{array} { c } { \text { If } | X | = p } \\ { \text { then } X = - p \text { or } X = p } \end{array}\) |

## Case 2: An absolute value inequality involving " less than ."

\(\begin{array} { c } { \text { If } | X | \leq p } \\ { \text { then } - p \leq X \leq p } \end{array}\) |

## Case 3: An absolute value inequality involving " greater than ."

\(\begin{array} { c } { \text { If } | X | \geq p } \\ { \text { then } X \leq - p \text { or } X \geq p } \end{array}\) |

## Key Takeaways

- To solve an absolute value equation, such as \(|X| = p\), replace it with the two equations \(X = −p\) and \(X = p\) and then solve each as usual. Absolute value equations can have up to two solutions.
- To solve an absolute value inequality involving “less than,” such as \(|X| ≤ p\), replace it with the compound inequality \(−p ≤ X ≤ p\) and then solve as usual.
- To solve an absolute value inequality involving “greater than,” such as \(|X| ≥ p\), replace it with the compound inequality \(X ≤ −p\) or \(X ≥ p\) and then solve as usual.
- Remember to isolate the absolute value before applying these theorems.

Exercise \(\PageIndex{5}\)

- \(|x| = 9\)
- \(|x| = 1\)
- \(|x − 7| = 3\)
- \(|x − 2| = 5\)
- \(|x + 12| = 0\)
- \(|x + 8| = 0\)
- \(|x + 6| = −1\)
- \(|x − 2| = −5\)
- \(|2y − 1| = 13\)
- \(|3y − 5| = 16\)
- \(|−5t + 1| = 6\)
- \(|−6t + 2| = 8\)
- \(\left| \frac { 1 } { 2 } x - \frac { 2 } { 3 } \right| = \frac { 1 } { 6 }\)
- \(\left| \frac { 2 } { 3 } x + \frac { 1 } { 4 } \right| = \frac { 5 } { 12 }\)
- \(|0.2x + 1.6| = 3.6\)
- \(|0.3x − 1.2| = 2.7\)
- \(| 5 (y − 4) + 5| = 15\)
- \(| 2 (y − 1) − 3y| = 4\)
- \(|5x − 7| + 3 = 10\)
- \(|3x − 8| − 2 = 6\)
- \(9 + |7x + 1| = 9\)
- \(4 − |2x − 3| = 4\)
- \(3 |x − 8| + 4 = 25\)
- \(2 |x + 6| − 3 = 17\)
- \(9 + 5 |x − 1| = 4\)
- \(11 + 6 |x − 4| = 5\)
- \(8 − 2 |x + 1| = 4\)
- \(12 − 5 |x − 2| = 2\)
- \(\frac{1}{2} |x − 5| − \frac{2}{3} = −\frac{1}{6}\)
- \(\frac { 1 } { 3 } \left| x + \frac { 1 } { 2 } \right| + 1 = \frac { 3 } { 2 }\)
- \(−2 |7x + 1| − 4 = 2\)
- \(−3 |5x − 3| + 2 = 5\)
- \(1.2 |t − 2.8| − 4.8 = 1.2\)
- \(3.6 | t + 1.8| − 2.6 = 8.2\)
- \(\frac{1}{2} |2 (3x − 1) − 3| + 1 = 4\)
- \(\frac{2}{3} |4 (3x + 1) − 1| − 5 = 3\)
- \(|5x − 7| = |4x − 2|\)
- \(|8x − 3| = |7x − 12|\)
- \(|5y + 8| = |2y + 3|\)
- \(|7y + 2| = |5y − 2|\)
- \(|5 (x − 2)| = |3x|\)
- \(|3 (x + 1)| = |7x|\)
- \(\left| \frac { 2 } { 3 } x + \frac { 1 } { 2 } \right| = \left| \frac { 3 } { 2 } x - \frac { 1 } { 3 } \right|\)
- \(\left| \frac { 3 } { 5 } x - \frac { 5 } { 2 } \right| = \left| \frac { 1 } { 2 } x + \frac { 2 } { 5 } \right|\)
- \(|1.5t − 3.5| = |2.5t + 0.5|\)
- \(|3.2t − 1.4| = |1.8t + 2.8|\)
- \(|5 − 3 (2x + 1)| = |5x + 2|\)
- \(|3 − 2 (3x − 2)| = |4x − 1|\)

1. \(−9, 9\)

3. \(4, 10\)

5. \(−12\)

7. \(Ø\)

9. \(−6, 7\)

11. \(−1, \frac{7}{5}\)

13. \(1, \frac{5}{3}\)

15. \(−26, 10\)

17. \(0, 6\)

19. \(0, \frac{14}{5}\)

21. \(−\frac{1}{7}\)

23. \(1, 15\)

25. \(Ø\)

27. \(−3, 1\)

29. \(4, 6\)

31. \(Ø\)

33. \(−2.2, 7.8\)

35. \(−\frac{1}{6}, \frac{11}{6}\)

37. \(1, 5\)

39. \(−\frac{5}{3}, −\frac{11}{7}\)

41. \(\frac{5}{4} , 5\)

43. \(−\frac{1}{13} , 1\)

45. \(−4, 0.75\)

47. \(0, 4\)

Exercise \(\PageIndex{6}\)

Solve and graph the solution set. In addition, give the solution set in interval notation.

- Solve for \(x: p |ax + b| − q = 0\)
- Solve for \(x: |ax + b| = |p + q|\)

1. \(x = \frac { - b q \pm q } { a p }\)

Exercise \(\PageIndex{7}\)

- \(|x| < 5\)
- \(|x| ≤ 2\)
- \(|x + 3| ≤ 1\)
- \(|x − 7| < 8\)
- \(|x − 5| < 0\)
- \(|x + 8| < −7\)
- \(|2x − 3| ≤ 5\)
- \(|3x − 9| < 27\)
- \(|5x − 3| ≤ 0\)
- \(|10x + 5| < 25\)
- \(\left| \frac { 1 } { 3 } x - \frac { 2 } { 3 } \right| \leq 1\)
- \(\left| \frac { 1 } { 12 } x - \frac { 1 } { 2 } \right| \leq \frac { 3 } { 2 }\)
- \(|x| ≥ 5\)
- \(|x| > 1\)
- \(|x + 2| > 8\)
- \(|x − 7| ≥ 11\)
- \(|x + 5| ≥ 0\)
- \(|x − 12| > −4\)
- \(|2x − 5| ≥ 9\)
- \(|2x + 3| ≥ 15\)
- \(|4x − 3| > 9\)
- \(|3x − 7| ≥ 2\)
- \(\left| \frac { 1 } { 7 } x - \frac { 3 } { 14 } \right| > \frac { 1 } { 2 }\)
- \(\left| \frac { 1 } { 2 } x + \frac { 5 } { 4 } \right| > \frac { 3 } { 4 }\)

1. \(( - 5,5 )\);

3. \([ - 4 , - 2 ]\);

5. \(\emptyset\);

7. \([ - 1,4 ]\);

9. \(\left\{ \frac { 3 } { 5 } \right\}\);

11. \([ - 1,5 ]\);

13. \(( - \infty , - 5 ] \cup [ 5 , \infty )\);

15. \(( - \infty , - 10 ) \cup ( 6 , \infty )\);

17. \(\mathbb { R }\);

19. \(( - \infty , - 2 ] \cup [ 7 , \infty )\);

21. \(\left( - \infty , - \frac { 3 } { 2 } \right) \cup ( 3 , \infty )\);

23. \(( - \infty , - 2 ) \cup ( 5 , \infty )\);

Exercise \(\PageIndex{8}\)

Solve and graph the solution set.

- \(|3 (2x − 1)| > 15\)
- \(|3 (x − 3)| ≤ 21\)
- \(−5 |x − 4| > −15\)
- \(−3 |x + 8| ≤ −18\)
- \(6 − 3 |x − 4| < 3\)
- \(5 − 2 |x + 4| ≤ −7\)
- \(6 − |2x + 5| < −5\)
- \(25 − |3x − 7| ≥ 18\)
- \(|2x + 25| − 4 ≥ 9\)
- \(|3 (x − 3)| − 8 < −2\)
- \(2 |9x + 5| + 8 > 6\)
- \(3 |4x − 9| + 4 < −1\)
- \(5 |4 − 3x| − 10 ≤ 0\)
- \(6 |1 − 4x| − 24 ≥ 0\)
- \(3 − 2 |x + 7| > −7\)
- \(9 − 7 |x − 4| < −12\)
- \(|5 (x − 4) + 5| > 15\)
- \(|3 (x − 9) + 6| ≤ 3\)
- \(\left| \frac { 1 } { 3 } ( x + 2 ) - \frac { 7 } { 6 } \right| - \frac { 2 } { 3 } \leq - \frac { 1 } { 6 }\)
- \(\left| \frac { 1 } { 10 } ( x + 3 ) - \frac { 1 } { 2 } \right| + \frac { 3 } { 20 } > \frac { 1 } { 4 }\)
- \(12 + 4 |2x − 1| ≤ 12\)
- \(3 − 6 |3x − 2| ≥ 3\)
- \(\frac{1}{2} |2x − 1| + 3 < 4\)
- 2 |\frac{1}{2} x + \frac{2}{3} | − 3 ≤ −1\)
- \(7 − |−4 + 2 (3 − 4x)| > 5\)
- \(9 − |6 + 3 (2x − 1)| ≥ 8\)
- \(\frac { 3 } { 2 } - \left| 2 - \frac { 1 } { 3 } x \right| < \frac { 1 } { 2 }\)
- \(\frac { 5 } { 4 } - \left| \frac { 1 } { 2 } - \frac { 1 } { 4 } x \right| < \frac { 3 } { 8 }\)

1. \(( - \infty , - 2 ) \cup ( 3 , \infty )\);

3. \(( 1,7 )\);

5. \(( - \infty , 3 ) \cup ( 5 , \infty )\);

7. \(( - \infty , - 8 ) \cup ( 3 , \infty )\);

9. \(( - \infty , - 19 ] \cup [ - 6 , \infty )\);

11. \(\mathbb { R }\);

13. \(\left[ \frac { 2 } { 3 } , 2 \right]\);

15. \(( - 12 , - 2 )\);

17. \(( - \infty , 0 ) \cup ( 6 , \infty )\);

19. \([ 0,3 ]\);

21. \(\frac { 1 } { 2 }\);

23. \(\left( - \frac { 1 } { 2 } , \frac { 3 } { 2 } \right)\);

25. \(\left( 0 , \frac { 1 } { 2 } \right)\);

27. \(( - \infty , 3 ) \cup ( 9 , \infty )\);

Exercise \(\PageIndex{9}\)

Assume all variables in the denominator are nonzero.

- Solve for \(x\) where \(a, p > 0: p |ax + b| − q ≤ 0\)
- Solve for \(x\) where \(a, p > 0: p |ax + b| − q ≥ 0\)

1. \(\frac { - q - b p } { a p } \leq x \leq \frac { q - b p } { a p }\)

Exercise \(\PageIndex{10}\)

Given the graph of \(f\) and \(g\), determine the \(x\)-values where:

(a) \(f ( x ) = g ( x )\)

(b) \(f ( x ) > g ( x )\)

(c) \(f ( x ) < g ( x )\)

1. (a) \(−6, 0\); (b) \((−∞, −6) ∪ (0, ∞)\); (c) \((−6, 0)\)

3. (a) \(Ø\); (b) \(ℝ\); (c) \(Ø\)

Exercise \(\PageIndex{11}\)

- Make three note cards, one for each of the three cases described in this section. On one side write the theorem, and on the other write a complete solution to a representative example. Share your strategy for identifying and solving absolute value equations and inequalities on the discussion board.
- Make your own examples of absolute value equations and inequalities that have no solution, at least one for each case described in this section. Illustrate your examples with a graph.

1. Answer may vary

63 The distance from the graph of a number \(a\) to zero on a number line, denoted \(|a|\).

64 The number or expression inside the absolute value.

- Inspiration

## Leveled Practice Cards Activity for Absolute Value Equations

By: Author Sarah Carter

Posted on Published: June 10, 2021 - Last updated: March 16, 2023

Categories Absolute Value Equations , Absolute Value , Dry Erase Activities , Review Games

This blog post contains Amazon affiliate links. As an Amazon Associate, I earn a small commission from qualifying purchases.

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I created these leveled practice cards for solving absolute value equations when I taught Algebra 1. There are three levels of problems so that students can start with problems of an appropriate difficulty level.

Level 1 Questions – Absolute value bars are already isolated.

Level 2 Questions – Students have one step to do in order to isolate the absolute value bars.

Level 3 Questions – Students have multiple steps to do in order to isolate the absolute value bars.

When I typed these up, I originally thought I would cut the cards apart. But, I actually ended up leaving them intact and using them as a review game.

I put all three levels of questions at the front of the room with a stack of dry erase pockets . Each group of students sent someone up to the front of the room to grab a page of problems to slide in their dry erase pocket . Students could choose to start with Level 1, 2, or 3.

MATH = LOVE RECOMMENDS…

I cannot imagine teaching math without my dry erase pockets! They instantly make any activity more engaging and save me countless hours at the copy machine since I can use the same class sets of copies year after year.

Here are my current go-to recommendations:

- 8.5 x 11 Inch Dry Erase Pockets from Puroma
- 11 x 17 Inch Dry Erase Pockets from C-Line Products
- Expo Dry Erase Markers

If you don’t have a classroom set of dry erase pockets , you could also use heavy duty sheet protectors . But, I highly recommend investing in a classroom set of the pockets since they are so much more durable.

I encouraged students to switch levels when the problems became too easy.

Groups used their jumbo dry erase boards so that they could each have a working space. Some groups decided to work the same problem and compare their answers.

Other groups divvied up the questions and each solved their own individual questions. I was okay with this because they were still helping one another and learning from one another.

We used Julie Morgan’s 1 to 100 game to make this into a super easy review game. All you need is a 100 chart . I displayed the 100 chart on my SmartBoard. When students finished a problem, they raised their hand. I checked their work.

If it was correct, they got to go to the 100 chart and mark off a number with their group’s color/symbol. If it was incorrect, they got to rework the problem with their group to see where they had made a mistake.

I made color-coded answer keys to make it super easy to check students’ work.

Groups marked off the problems their group had solved on the dry erase pocket .

At the end of class, I used a random number generator to pick a number between 1 and 100. Whichever group had that number claimed on the hundred chart won a small prize. Usually I just gave them a piece of candy.

I like this review game structure because it is super easy to implement. I used problems that I had typed up. You can do this with a regular worksheet! And, students really are motivated to do as many problems as possible since claiming more numbers on the board increases their chance of winning.

At the same time, I like that because of the way probability/randomness works, the underdog team can still often win!

## Free Download of Leveled Practice Cards Activity for Solving Absolute Value Equations

Leveled Practice Cards for Solving Absolute Value Equations (PDF) (2103 downloads )

Leveled Practice Cards for Solving Absolute Value Equations (Editable Publisher File ZIP) (1099 downloads )

## More Activities for Teaching Absolute Value Equations and Inequalities

Friday 24th of September 2021

Thank you for sharing. Love your ideas and resources!

Sarah Carter

Tuesday 28th of September 2021

Thanks Linda!

## Absolute Value | Overview & Practice Problems

Anita Dunn graduated from Saint Mary's College with a Bachelor's of Science in Mathematics, and graduated from Purdue University with a Master's of Science in Mathematics. She has been certified as a Developmental Education Specialist through the Kellogg Institute. She has more than 10 years of experience as a college professor.

- Instructor Zach Pino

## How do you solve absolute value problems?

First, isolate the absolute value portion of the equation. Then look at what the absolute value portion is equal to. If it is equal to a negative number, then there is no solution. If it is equal to zero, there is one solution. If it is equal to a positive number, such as a, there are two solutions. Set the inside of the absolute value to a and solve, and set it equal to -a and solve.

## What is an example of an absolute value equation?

An absolute value equation is an equation with an absolute value. One example could be: |x+6|+5 = 10. Absolute value equations can have 2, 1 or no solutions.

## Table of Contents

Absolute value overview, how to solve equations with absolute values, absolute value practice problems, lesson summary.

The absolute value of a number is defined as the distance of that number from 0 on the number line. The notation of the absolute value of a real number x is given by {eq}|x| {/eq}.

Note that -5 is 5 spaces from 0 on the number line, and so {eq}|-5|=5 {/eq}.

Likewise, 5 is 5 spaces from 0 on the number line, so {eq}|5|=5 {/eq}.

As zero is zero spaces from 0 on the number line, {eq}|0|=0 {/eq}.

Continuing in this way, note that the absolute value basically makes negative numbers into positive numbers, and positive numbers stay positive. It is also important to note that the absolute value of a number is the distance of that number from zero, and as distances are always positive or equal to zero, all absolute values are equal to either a positive number or zero.

Examples of how to solve absolute values :

Example 1: {eq}|-7|=7 {/eq}.

Example 2: {eq}|3.5|=3.5 {/eq}.

An absolute value equation is an equation with an absolute value. Often, these absolute value equations have one variable, such as x. So, how does one solve an equation in one variable with an absolute value? Recall that to solve a linear equation in one variable, first simplify each side of the equation as much as possible. Then, get all of the constant values by themselves on one side of the equation, and all of the variable values on the other side of the equation. Then, divide the equation by the coefficient of the variable. This will result in the solution of the equation. The same idea holds for absolute value equations, except instead of trying to get the variable by itself, try to get the absolute value by itself. In a way, this is like peeling an onion. Start by adding or subtracting any constants that are not inside the absolute value, and then divide by any number that is in front of the absolute value. Now, the absolute value is by itself on one side of the equation, and the equation can be solved following the steps outlined in the following section.

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- 0:10 Review of Absolute Values
- 1:06 Using Onions to Solve…
- 3:22 Onions and Absolute Values
- 4:39 Onions and Complex…
- 6:30 Lesson Summary

Remember, when solving an equation, one is trying to find every possible value that could be substituted in for the variable and the equation ends up being a true statement. With that in mind, consider the following situations:

- Solve {eq}|x| = 5 {/eq}

As mentioned previously, {eq}|-5|=5 {/eq} and {eq}|5|=5 {/eq}. Considering the process of finding absolute values, both 5 and -5 can be substituted in for x and the equation holds true, both 5 and -5 are solutions of the equation. Thus, {eq}x = -5, 5 {/eq}.

- Solve {eq}|x|=0 {/eq}.

Note that the only number whose absolute value is equal to 0 is the number 0. Thus, {eq}x=0 {/eq}.

- Solve {eq}|x|=-6 {/eq}.

As mentioned previously, the absolute value of any number is either a positive number or 0. This is because absolute values are defined as a distance, and distances are never negative. Thus, this equation has no solution.

Keeping in mind what has been learned from the three example equations, the following are the steps to solve an absolute value equation.

- First, isolate the absolute value on one side of the equation.
- If the absolute value is equal to a negative number, then there is no solution. If it is equal to a positive number or zero, continue to the next step.
- If the absolute value is equal to zero, simply remove the absolute value bars and solve the resulting equation.
- If the absolute value is equal to a positive number, say a , then rewrite the equation as two different equations. One equation will be the inside of the absolute value equal to a , and the other will be the inside of the absolute value equal to -a .

When completing step one, think of peeling an onion. When peeling an onion, one will start at the outside and work their way in. The same idea applies here. Start with the numbers farthest away from the absolute value and continue until the absolute value is alone on one side of the equation.

For example, when isolating the absolute value of the equation {eq}2|x| - 4 = 6 {/eq}, first, add four to both sides of the equation: {eq}2|x| =10 {/eq}. Next, divide both sides by 2: {eq}|x|=5 {/eq}. Note that this equation has been solved previously, so further work will not be completed at this time.

Find the solution or solutions for each of the following equations, if possible.

Example 1 : {eq}|3x -5| +5 = 10 {/eq}

Thinking in terms of the onion analogy, first subtract 5 from each side to isolate the absolute value. {eq}|3x-5|=5 {/eq}

Noting that the absolute value is now isolated and that it is equal to a positive number, now write two equations: {eq}3x-5=5 {/eq} and {eq}3x-5=-5 {/eq} .

Solving each of the equations:

{eq}3x-5=5 \\ 3x=10 \\ x=\frac{10}{3} {/eq}

{eq}3x-5=-5 \\ 3x = 0 \\ x=0 {/eq}

Thus, the solutions are {eq}x = 0, \frac{10}{3} {/eq}

Example 2: Solve {eq}2|3x-6|-10=-4 {/eq}.

Again thinking of the onion analogy, start by adding 10 to both sides, and after that divide by 2. Note, even though the equation is currently equal to a negative number, the absolute value has not yet been isolated. Therefore, no conclusion about whether or not there will be a solution can yet be reached.

{eq}2|3x-6|-10= -4 \\ 2|3x -6| = 6 \\ |3x-6| = 3 {/eq}

Note that the absolute value is now isolated and it is equal to a positive number, so two equations need to be created: {eq}3x-6=3 {/eq} and {eq}3x-6=-3 {/eq}. Solving both of these:

{eq}3x-6=3 \\ 3x = 9 \\ x = 3 {/eq}

{eq}3x-6=-3 \\ 3x = 3 \\ x=1 {/eq}

Thus, the solutions are {eq}x = 1, 3 {/eq}.

Example 3: Solve {eq}|x+6| + 10 =4 {/eq}

Subtracting 10 from both sides in order to isolate the absolute value: {eq}|x+6| =-6 {/eq}. Note that the absolute value is now isolated, and it is equal to a negative number. As an absolute value is a measure of distance, it can not be equal to a negative number. Therefore, there is no solution to this equation.

Example 4 : Solve {eq}3|x+4|+6 = 6 {/eq}

Using the onion analogy once again, start by subtracting the six on both sides, and then by dividing each side by 3.

{eq}3|x+4|+6 = 6 \\ 3|x+4| = 0 \\ |x+4| =0 {/eq}

As the absolute value is equal to zero, only one equation will be created and solved:

{eq}x+4 = 0 x = -4 {/eq}.

Thus, there is one solution, and it is {eq}x=-4 {/eq}.

The absolute value of a number is the distance between that number and zero on the number line. As distances are never negative values, neither are absolute values. To find the absolute value of a number, make the number a positive number. An absolute value equation is an equation with an absolute value. To solve an absolute value equation, first, isolate the absolute value on one side of the equation. To do this, consider the analogy of an onion. First, add or subtract any values that are being added outside the absolute value, and then divide any number which is being multiplied by the absolute value. Once that absolute value is isolated, look at what value it is equal to. If it is equal to a negative number, then no solution can be found. If it is equal to zero, simply remove the absolute value bars and solve the equation. If it is equal to a positive number, say a, then two equations need to be created. The first equation will be the inside of the absolute value equal to a, and the second equation will be the inside equal to -a. Then, both equations will be solved.

## Video Transcript

Review of absolute values.

When it comes to solving absolute value equations, the one most important thing to remember is that they always give you two answers. We get those two answers by splitting the equation up, but how and when to split it up is where it starts to get a little more complicated and also where the mistakes start happening. In order to avoid those mistakes, it really helps to have a conceptual understanding of what 'solving an equation' means.

So let's first ask ourselves a question that you probably don't hear every day; what does a mathematical equation have in common with an onion? Now, I suppose your answer could be 'they both make you cry,' but if that was your answer, I hope this video can change your mind. Well, just like an onion has a very center and then a bunch of layers on the outside, an equation has a very center (the number) and then a bunch of operations being done to that number.

## Using Onions to Solve Equations

So for example, an equation might tell us that we start with some number (lets call it x ), we multiply it by 5, then we subtract 1, then we divide all of that by 2, and after we've done that, we apparently end up with 7. This can be expressed as a normal mathematical equation, or using this onion analogy where the very middle of the onion is the number we start with and each thing we do to that number is a new layer on the outside. So when we're asked to 'solve' this equation, it's simply our job to figure out what was that number in the very beginning that made it that after we do all these things we end up with 7. It's our job to peel off the layers of the onion one at a time to work our way all the way back to the very center - the x .

Well the first layer I peel off is this divided by 2. After I divide it by 2, apparently I ended up with 7, which means that before I divided by 2, I must have had 14 because 14 divided by 2 is 7. So the way I can undo division is by multiplication because those are inverse operations. So by putting the little times 2 by this divide by 2, they undo each other - they cancel each other out - and over on the other side of the equation, I get 7 times 2 and I get 14. So that last step has been undone and now I have an equation with one less step.

The next layer I need to undo is the minus 1. We can undo subtraction with addition and by adding 1 to both sides of my equation, I can end up with 5 times x equals 15. Now the last layer I need to undo is multiplication by 5; I can undo multiplication with division. Those things undo and I'm left with 15 divided by 5, which is 3. This means that if we would have substituted 3 back into the very beginning and done all those different things to it, we would have ended up with 7.

## Onions and Absolute Values

Absolute value equations are done the exact same way. Le'ts take a look at the example |2 x +1| = 9. If we draw our onion analogy, we start with a number ( x ), we multiply that by 2, then we add 1, and after we've done those things, we take the absolute value and we end up with 9.

Because we're being asked to solve the equation, we have to figure out what was that number x that we plugged in way back in the beginning, and since the absolute value was the last thing to be done, it's the first thing we have to undo. I undo an absolute value by splitting it up into two separate equations; one that equals 9 and one that equals -9. Now it's just a matter of solving these two separate equations one step at a time; the first thing I have to undo is addition by 1. I undo addition with subtraction because those are inverse operations. I take one away from 9, one away from -9 and I end up with 8 or -10. Then I have to undo multiplication by 2; I undo multiplication with division and I end up with either x is equal to 4 or x is equal to -5.

## Onions and Complex Absolute Values

The last example will show us the final way we can complicate absolute value equations; having operations on the inside and the outside of the absolute value.

Take -2| x -4|+3=1. Looking at the onion picture for this problem shows that we start with x , we subtract 4, we take the absolute value, then we multiply by -2 and then we add 3. So the difference between this problem and the previous one is that the absolute value is not the very last thing being done to my variable, which means it's not the first thing we have to undo. A lot of people, when they see this problem, get so concentrated on the fact that there's an absolute value in it that they just think, 'split it in two equations right away!' That's not our job until we get to that layer.

First, we have to undo addition with subtraction so I minus 3 from both sides and I get -2 times the absolute value of x minus 4 is equal to -2 . Now I have to undo multiplication; I undo multiplication with division and I divide both sides by -2. Now I'm simply left with the absolute value of x minus 4 equals 1. Now the absolute value is the outermost layer of our onion, so it's time to undo it by splitting the equation into two. One where x minus 4 equals 1 and another where x minus 4 equals -1. After I've undone the absolute value, the last step to undo is the subtraction by 4; again, I undo subtraction with addition and I get my two answers that x is equal to 5 or 3.

To review, an equation is just a list of steps being done to an unknown value and the answer you get after those things are done.

You can undo an absolute value by breaking the equation into two separate ones - one equal to the positive answer and one equal to the negative.

If there are no steps on the outside of the absolute value, it's the first thing you can undo by splitting it into two equations. But if there are steps on the outside of the absolute value, you must to undo those first before you split the equation up.

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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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## SOLVING ABSOLUTE VALUE EQUATIONS PRACTICE PROBLEMS

Solve for x :

(1) 10 = 4 + |2y + 1|

(2) -1 = -|5x + 1|

(3) -2|3b - 7| - 9 = -9

(4) -3|5x + 1| + 4 = 4

(5) -2|x + 3| = 5

(6) -3|x - 5| = 7

(7) 0 = |6x – 9|

(8) 7 = |4k – 6| + 7

(9) |-1/5 – 1/2k| = 9/5

(10) |-1/6 – 2/9h| = 1/2

(11) -3|2 – 6x| + 5 = -10

(12) 5|1 – 2x| - 7 = 3

(13) |3x – 5| = |2x + 1|

(14) |8x + 9| = |8x - 1|

(15) |[(2x)/(x + 1)]| = 5

(16) |[(2x - 1)/(x + 3)]| = 2

(17) |[(x + 4)/(1 – 2x)]| = 3/4

1. Answer :

10 = 4 + |2y + 1|

Subtract 4 on both sides, we get

10 - 4 = 4 + |2y + 1| - 4

6 = |2y + 1|

|2y + 1| = 6

Now, we have an absolute value equation form.

2y + 1 = 6 (or) 2y + 1 = -6

2y = 5 (or) 2y = -7

y = 5/2 (or) y = -7/2

So, the solution of y is 5/2 or -7/2

2. Answer :

-1 = -|5x + 1|

Dividing by -1 on both sides, we get

1 = |5x + 1|

|5x + 1| = 1

5x + 1 = 1 (or) 5x + 1 = -1

5x = 0 (or) 5x = -2

x = 0 (or) x = -2/5

So, the solution of x is 0 or -2/5

3. Answer :

-2|3b - 7| - 9 = -9

Add 9 on both sides, we get

-2|3b - 7| - 9 + 9 = -9 + 9

-2|3b - 7| = 0

Dividing by -2 on both sides, we get

|3b - 7| = 0

Solving for b

3b - 7 = 0

3b = 7

b = 7/3

So, the solution of b is 7/3

4. Answer :

-3|5x + 1| + 4 = 4

-3|5x + 1| + 4 - 4 = 4 - 4

-3|5x + 1| = 0

Dividing by -3 on both sides, we get

|5x + 1| = 0

Solving for x

5x + 1 = 0

5x = -1

x = -1/5

So, the solution of x is -1/5.

5. Answer :

-2|x + 3| = 5

|x + 3| = -5/2

Now, we have an absolute value equation that is less than 0.

So, there is no solution for x.

6. Answer :

-3|x - 5| = 7

|x - 5| = -7/3

Now, we have an absolute value equation that is less than 0

7. Answer :

0 = |6x – 9|

|6x - 9| = 0

6x – 9 = 0

6x = 9

x = 9/6

x = 3/2

So, the solution of x is 3/2.

8. Answer :

7 = |4k – 6| + 7

Subtract 7 on both sides, we get

7 - 7 = |4k – 6| + 7 – 7

0 = |4k – 6|

|4k - 6| = 0

Solving for k

4k – 6 = 0

4k = 6

k = 6/4

k = 3/2

So, the solution of k is 3/2.

9. Answer :

|-1/5 – 1/2k| = 9/5

-1/2k – 1/5 = 9/5 (or) -1/2k – 1/5 = -9/5

-1/2k = 9/5 + 1/5 (or) -1/2k = -9/5 + 1/5

-1/2k = 2 (or) -1/2k = -8/5

k = -4 (or) k = 16/5

So, the solution of k is {-4 or 16/5}.

10. Answer :

|-1/6 – 2/9h| = 1/2

|-2/9h – 1/6| = 1/2

Solving for h

-2/9h – 1/6 = 1/2 (or) -2/9h – 1/6 = -1/2

-2/9h = 1/2 + 1/6 (or) -2/9h = -1/2 + 1/6

Taking least common multiple,

-2/9h = 4/6 (or) -2/9h = -2/6

-2/9h = 2/3 (or) -2/9h = -1/3

h = -[(2/3)(9/2)] (or) h = [(1/3)(9/2)]

h = -3 (or) h = 3/2

So, the solution of h is {-3 or 3/2}.

11. Answer :

-3|2 – 6x| + 5 = -10

-3|-6x + 2| + 5 = -10

Subtract 5 on both sides, we get

-3|-6x + 2| + 5 - 5 = -10 - 5

-3|-6x + 2| = -15

|-6x + 2| = 5

-6x + 2 = 5 (or) -6x + 2 = -5

-6x = 3 (or) -6x = -7

x = -3/6 (or) x = 7/6

x = -1/2 (or) x = 7/6

So, the solution of x is {-1/2 or 7/6}

12. Answer :

5|1 – 2x| - 7 = 3

5|-2x + 1| - 7 = 3

Add 7 on both sides, we get

5|-2x + 1| - 7 + 7 = 3 + 7

5|-2x + 1| = 10

Dividing by 5 on both sides, we get

|-2x + 1| = 2

-2x + 1 = 2 (or) -2x + 1 = -2

-2x = 1 (or) -2x = -3

x = -1/2 (or) x = 3/2

So, the solution of x is {-1/2 or 3/2}.

13. Answer :

|3x – 5| = |2x + 1|

By using absolute value equation property,

3x – 5 = 2x + 1 (or) 3x – 5 = -(2x + 1)

3x – 2x = 1 + 5 (or) 3x + 2x = -1 + 5

x = 6 (or) 5x = 4

x = 6 (or) x = 4/5

So, the solution of x is {6 or 4/5}.

14. Answer :

|8x + 9| = |8x - 1|

Solving for x,

8x + 9 = 8x - 1 (or) 8x + 9 = -(8x – 1)

8x – 8x = -1 - 9 (or) 8x + 8x = 1 - 9

0 = -10 (or) 16x = -8

x = -1/2

So, the solution of x is -1/2.

15. Answer :

|[(2x)/(x + 1)]| = 5

[(2x)/(x + 1)] = 5 (or) [(2x)/(x + 1)] = -5

2x = 5(x + 1) (or) 2x = -5(x + 1)

2x = 5x + 5 (or) 2x = -5x – 5

2x – 5x = 5 (or) 2x + 5x = -5

-3x = 5 (or) 7x = -5

x = -5/3 (or) x = -5/7

So, the solution of x is {-5/3 or -5/7}.

16. Answer :

|[(2x - 1)/(x + 3)]| = 2

[(2x - 1)/(x + 3)] = 2 (or) [(2x - 1)/(x + 3)] = -2

2x - 1 = 2(x + 3) (or) 2x - 1 = -2(x + 3)

2x = 2x + 3 + 1 (or) 2x = -2x – 6 + 1

By combining like terms,

2x – 2x = 4 (or) 2x + 2x = -5

0 = 4 (or) 4x = -5

x = -5/4

So, the solution of x is -5/4.

17. Answer :

|[(x + 4)/(1 – 2x)]| = 3/4

[(x + 4)/(1 – 2x)] = 3/4 (or) [(x + 4)/(1 – 2x)] = -3/4

4(x + 4) = 3(1 – 2x) (or) 4(x + 4) = -3(1 – 2x)

4x + 16 = 3 – 6x (or) 4x + 16 = -3 + 6x

4x + 6x = 3 - 16 (or) 4x - 6x = -3 - 16

10x = -13 (or) -2x = -19

x = -13/10 (or) x = 19/2

So, the solution of x is {-13/10 or 19/2}.

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There are eleven (11) practice problems in this collection regarding absolute value equations. Whether you're a beginner or seeking a challenge, there's something here for you. As you tackle each problem, remember that in every attempt, right or wrong, fuels your mathematical growth.

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This topic covers: - Solving absolute value equations - Graphing absolute value functions - Solving absolute value inequalities

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Success Criteria. Write and solve equations involving absolute value. • I can write the two linear equations related to a given absolute value equation. • I can solve equations involving one or two absolute values. • I can identify special solutions of absolute value equations. Solving an Absolute Value Equation. EXPLORE IT.

I created these leveled practice cards for solving absolute value equations when I taught Algebra 1. There are three levels of problems so that students can start with problems of an appropriate difficulty level.

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Now, we have an absolute value equation form. 5x + 1 = 1 (or) 5x + 1 = -1. 5x = 0 (or) 5x = -2. x = 0 (or) x = -2/5. So, the solution of x is 0 or -2/5. 3. Answer : -2|3b - 7| - 9 = -9. Add 9 on both sides, we get.

Free math problem solver answers your algebra homework questions with step-by-step explanations.

To solve absolute value problems with one solution, identify expression in absolute value bars. Recall absolute value is zero only if expression is zero. Make expression equal right-hand side. Use algebra to find x values that satisfy equation. Check solutions by plugging in. Graph solutions on number line. Mark points.

The Algebra Calculator is a versatile online tool designed to simplify algebraic problem-solving for users of all levels. Here's how to make the most of it: Begin by typing your algebraic expression into the above input field, or scanning the problem with your camera. After entering the equation, click the 'Go' button to generate instant solutions.

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The Algebra 1 course, often taught in the 9th grade, covers Linear equations, inequalities, functions, and graphs; Systems of equations and inequalities; Extension of the concept of a function; Exponential models; and Quadratic equations, functions, and graphs. Khan Academy's Algebra 1 course is built to deliver a comprehensive, illuminating, engaging, and Common Core aligned experience!

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