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## Solving absolute value equations and inequalities

- Absolute equations I
- Absolute equations II
- Absolute equations III

The absolute number of a number a is written as

$$\left | a \right |$$

And represents the distance between a and 0 on a number line.

An absolute value equation is an equation that contains an absolute value expression. The equation

$$\left | x \right |=a$$

Has two solutions x = a and x = -a because both numbers are at the distance a from 0.

To solve an absolute value equation as

$$\left | x+7 \right |=14$$

You begin by making it into two separate equations and then solving them separately.

$$x+7 =14$$

$$x+7\, {\color{green} {-\, 7}}\, =14\, {\color{green} {-\, 7}}$$

$$x+7 =-14$$

$$x+7\, {\color{green} {-\, 7}}\, =-14\, {\color{green} {-\, 7}}$$

An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative.

The inequality

$$\left | x \right |<2$$

Represents the distance between x and 0 that is less than 2

Whereas the inequality

$$\left | x \right |>2$$

Represents the distance between x and 0 that is greater than 2

You can write an absolute value inequality as a compound inequality.

$$\left | x \right |<2\: or

$$-2<x<2$$

This holds true for all absolute value inequalities.

$$\left | ax+b \right |<c,\: where\: c>0$$

$$=-c<ax+b<c$$

$$\left | ax+b \right |>c,\: where\: c>0$$

$$=ax+b<-c\: or\: ax+b>c$$

You can replace > above with ≥ and < with ≤.

When solving an absolute value inequality it's necessary to first isolate the absolute value expression on one side of the inequality before solving the inequality.

Solve the absolute value inequality

$$2\left |3x+9 \right |<36$$

$$\frac{2\left |3x+9 \right |}{2}<\frac{36}{2}$$

$$\left | 3x+9 \right |<18$$

$$-18<3x+9<18$$

$$-18\, {\color{green} {-\, 9}}<3x+9\, {\color{green} {-\, 9}}<18\, {\color{green} {-\, 9}}$$

$$-27<3x<9$$

$$\frac{-27}{{\color{green} 3}}<\frac{3x}{{\color{green} 3}}<\frac{9}{{\color{green} 3}}$$

$$-9<x<3$$

## Video lesson

Solve the absolute value equation

$$4 \left |2x -1 \right | -2 = 10$$

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## Solving Absolute Value Equations and Inequalities

As we saw earlier in the Negative Numbers and Absolute Value section, 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 . As an example, $ \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; 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 . Then, make the expression on the right-hand side (without the variables) both positive and negative and solve each equation; typically, we will get two answers . We must check our answer s , since we may get extraneous solutions (solutions that don’t work). .

There are a few cases with absolute value equations or inequalities where we don’t have to even solve! 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.

Note that we can always solve absolute value equations and inequalities graphically , as shown below.

Here are some problems:

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.

For example, when the expression $ 3x-2$ is negative, the absolute value of that expression is the negation of it, or $ -3x+2$, to make it positive in the equation. Play around with some numbers and you’ll see this!

When we get all the possible answers, check for extraneous solutions , since we’re dealing with absolute value. We found two answers that worked: $ \displaystyle x=\frac{3}{2}$ and $ x=-1$. You can also put the equation in your graphing calculator to check your answers!

Here’s another way to approach the absolute value problem above, using number lines :

Now draw number lines for each absolute value, and then for the whole equation above . We see for the last number line that for $ <-2$, we’ll use $ 2-3x$ and $ -x-2$, between $ -2$ and $ \displaystyle \frac{2}{3}$, we’ll use $ 2-3x$ and $ x+2$ , and $ \displaystyle >\frac{2}{3}$, we’ll use $ 3x-2$ and $ x+2$.

After solving for $ x$ in the original equation, we have to check to make sure each value we get for $ x$ falls into the correct interval of the number line. For example, $ \displaystyle -\frac{3}{2}$ isn’t $ <-2$, so we have to “throw it away”. $ -1$ is between $ -2$ and $ \displaystyle \frac{2}{3}$, so it works, and $ \displaystyle \frac{3}{2}$ is $ \displaystyle >\frac{2}{3}$, so it works.

We have $ \displaystyle x=\frac{3}{2}$ and $ x=-1$. √

## Solving Absolute Value Inequalities

Note that we learned about Linear Inequalities here .

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

First, get the absolute value all by itself on the left (remember to reverse the inequality sign when multiplying or dividing by a negative number). Now, 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 uses OR LAND: Less Than uses 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).

Also, remember to use open brackets for inequalities that aren’t inclusive ($ <$ and $ <$) and closed brackets for inequalities that are inclusive and include the boundary point ($ \le $ and $ \ge $).

Here are some examples:

There are examples of rational functions with absolute values here in the Rational Functions, Equations, and Inequalities section .

## 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 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:

Without using a t-chart, we can see that the vertex is at $ \left( {-2,1} \right)$ and the graph is upside-down because of the negative sign. It’s also stretched vertically by a factor of 3 and horizontal by a factor of $ \displaystyle \frac{1}{2}$ (or stretched vertically by a factor of 6 ); thus, other points down can be drawn by going back and forth 1 and down 6 .

You can solve absolute value equations and equalities with graphing ; here are some examples of solving inequalities:

## 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:

Here are examples that are absolute value inequality applications . Use this rule of thumb : the absolute value of a difference is usually on the left-hand side, the amount that differs or varies is usually on the right-hand side, with a $ <$ or $ \le $ sign in between.

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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You can also go to the Mathway site here , where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

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## Unit 1: Linear equations and inequalities

About this unit.

Linear equations and inequalities are the foundation of many advanced math topics, such as functions, systems, matrices, and calculus. Learn how to master them and unlock new possibilities for your future studies and careers in engineering, finance, computer science, and more.

## Solving equations with one unknown

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## 1-5 Absolute Value Equations and Inequalities

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## 1.4: Absolute value inequalities

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Using the notation from the previous section, we now solve inequalities involving the absolute value. These inequalities may be solved in three steps:

- Step 1: Solve the corresponding equality . The solution of the equality divides the real number line into several subintervals.
- Step 2: Using step 1, check the inequality for a number in each of the subintervals. This check determines the intervals of the solution set.
- Step 3: Check the endpoints of the intervals.

Here are some examples for the above solution method.

## Example \(\PageIndex{1}\)

Solve for \(x\) :

- \(|x+7|<2\)
- \(|3x-5|\geq 11\)
- \(|12-5x|\leq 1\)
- We follow the three steps described above. In step 1, we solve the corresponding equality, \(|x+7|=2\) . \(x+7=2\) & \(x+7=-2\) \[\begin{array}{l|l} x+7=2 & x+7=-2 \\ \Longrightarrow x=-5 & \Longrightarrow x=-9 \end{array} \nonumber \] The solutions \(x=-5\) and \(x=-9\) divide the number line into three subintervals:

Now, in step 2, we check the inequality for one number in each of these subintervals.

\[ \begin{array}{c|c|c} \text { Check: } \quad x=-10 & \text { Check: } \quad x=-7 & \text { Check: } \quad x=0 \\ |(-10)+7| \stackrel{?}{<} 2 & |(-7)+7| \stackrel{?}{<} 2 & |0+7| \stackrel{?}{<} 2 \\ |-3| \stackrel{?}{<} 2 & |0| \stackrel{?}{<} 2 & |7| \stackrel{?}{<} 2 \\ 3 \stackrel{?}{<} 2 & 0 \stackrel{?}{<} 2 & 7 \stackrel{?}{<} 2 \\ \text { false } & \text { true } & \text { false } \end{array} \nonumber \]

Since \(x=-7\) in the subinterval given by \(-9<x<-5\) solves the inequality \(|x+7|<2\) , it follows that all numbers in the subinterval given by \(-9<x<-5\) solve the inequality. Similarly, since \(x=-10\) and \(x=0\) do not solve the inequality, no number in these subintervals will solve the inequality. For step 3, we note that the numbers \(x=-9\) and \(x=-5\) are not included as solutions since the inequality is strict (that is we have \(<\) instead of \(\leq\) ).The solution set is therefore the interval \(S=(-9,-5)\) . The solution on the number line is:

- We follow the steps as before. First, in step 1, we solve \(|3x-5|=11\) . \[\begin{array}{l|l} 3 x-5=11 & 3 x-5=-11 \\ \Longrightarrow 3 x=16 & \Longrightarrow 3 x=-6 \\ \Longrightarrow x=\dfrac{16}{3} & \Longrightarrow x=-2 \end{array} \nonumber \]

The two solutions \(x=-2\) and \(x=\dfrac{16}{3}=5\dfrac {1}{3}\) divide the number line into the subintervals displayed below. \[x<-2 \hspace{1in} -2<x<5\frac 1 3 \hspace{1in} 5\frac 1 3<x \nonumber\]

For step 2, we check a number in each subinterval. This gives:

\[\begin{array}{c|c|c|c} \text { Check: } x=-3 & \text { Check: } \quad x=1 & \text { Check: } \quad x=6 \\ |3 \cdot(-3)-5| \stackrel{?}{\geq} 11 & |3 \cdot 1-5| \stackrel{?}{\geq} 11 & |3 \cdot 6-5| \stackrel{?}{\geq} 11 \\ |-9-5| \stackrel{?}{\geq} 11 & |3-5| \stackrel{?}{\geq} 11 & |18-5| \stackrel{?}{\geq} 11 \\ |-14| \stackrel{?}{\geq} 11 & |-2| \stackrel{?}{\geq} 11 & |13| \stackrel{?}{\geq} 11 \\ 14 \stackrel{?}{\geq} 11 & 2 \stackrel{?}{\geq} 11 & 13 \stackrel{?}{\geq} 11 \\ \text { true } & \text { false } & \text { true } \end{array} \nonumber \]

For step 3, note that we include \(-2\) and \(5\dfrac {1}{3}\) in the solution set since the inequality is “greater than or equal to” (that is \(\geq\) , as opposed to \(>\) ). Furthermore, the numbers \(-\infty\) and \(\infty\) are not included, since \(\pm\infty\) are not real numbers.

The solution set is therefore the union of the two intervals: \[S=\Big(-\infty,-2\Big]\cup \Big[5\dfrac {1}{3}, \infty\Big) \nonumber \]

- To solve \(|12-5x|\leq 1\) , we first solve the equality \(|12-5x|=1\) . \[\begin{array}{l|l} 12-5 x=1 & 12-5 x=-1 \\ \Longrightarrow-5 x=-11 & \Longrightarrow-5 x=-13 \\ \Longrightarrow x=\frac{-11}{-5}=2.2 & \Longrightarrow x=\frac{-13}{-5}=2.6 \end{array} \nonumber \]

This divides the number line into three subintervals, and we check the original inequality \(|12-5x|\leq 1\) for a number in each of these subintervals.

\[\begin{array}{c|c|c|c} \text {Interval: } \quad x<2.2 & \text {Interval: } \quad 2.2<x<2.6 & \text {Interval: } \quad 2.6<x\\ \text {Check: } \quad x=1 & \text {Check: } \quad x=2.4 & \text {Check: } \quad x=3 \\ |12-5 \cdot 1| \stackrel{?}{\leq} 1 & |12-5 \cdot 1| \stackrel{?}{\leq} 1 & |12-5 \cdot 3| \stackrel{?}{\leq} 1 \\ |12-5| \stackrel{?}{\leq} 1 & |12-12| \stackrel{?}{\leq} 1 & |12-15| \stackrel{?}{\leq} 1 \\ |7| \stackrel{?}{\leq} 1 & |0| \stackrel{?}{\leq} 1 & |-3| \stackrel{?}{\leq} 1 \\ 7 \stackrel{?}{\leq} 1 & 0 \stackrel{?}{\leq} 1 & 3 \stackrel{?}{\leq} 1 \\ \text { false } & \text { true } & \text { false } \end{array} \nonumber \]

The solution set is the interval \(S=[2.2,2.6]\) , where we included \(x=2.2\) and \(x=2.6\) since the original inequality “less than or equal to” ( \(\leq\) ) includes the equality.

Alternatively, whenever you have an absolute value inequality you can turn it into two inequalities.

Here are a couple of examples.

## Example \(\PageIndex{2}\)

Solve for \(x\) : \(|12-5x|\leq 1\)

Note that \(|12-5x|\leq 1\) implies that

\[-1\leq 12-5x\leq1 \nonumber \]

\[-13\leq -5x\leq -11 \nonumber \]

and by dividing by \(-5\) (remembering to switch the direction of the inequalities when multiplying or dividing by a negative number) we see that

\[\frac{13}{5}\geq x\geq \frac{11}{5} \nonumber \]

or in interval notation, we have the solution set

\[S=\left[\frac{11}{5},\frac{13}{5}\right] \nonumber\]

## Example \(\PageIndex{3}\)

If \(|x+6|>2\) then either \(x+6>2\) or \(x+6<-2\) so that either \(x>-4\) or \(x<-8\) so that in interval notation the solution is \(S=(-\infty,-8)\cup(-4,\infty)\).

There is a geometric interpretation of the absolute value on the number line as the distance between two numbers:

distance between \(a\) and \(b\) is \(|b-a|\) which is also equal to \(|a-b|\)

This interpretation can also be used to solve absolute value equations and inequalities.

## Example \(\PageIndex{4}\)

- \(|x-6|=4\)
- \(|x-6|\leq 4\)
- \(|x-6|\geq 4\)
- Consider the distance between \(x\) and \(6\) to be \(4\) on a number line:

There are two solutions, \(x=2\) or \(x=10\) . That is, the distance between \(2\) and \(6\) is \(4\) and the distance between \(10\) and \(6\) is \(4\) .

- Numbers inside the braces above have distance \(4\) or less. The solution is given on the number line as:

In interval notation, the solution set is the interval \(S=[2,10]\) . One can also write that the solution set consists of all \(x\) such that \(2\leq x\leq 10\) .

- Numbers outside the braces above have distance \(4\) or more. The solution is given on the number line as:

In interval notation, the solution set is the interval \((-\infty,2]\) and \([10,\infty)\) , or in short it is the union of the two intervals:

\[S= (\infty,2]\cup [10,\infty) \nonumber \]

One can also write that the solution set consists of all \(x\) such that \(x\leq 2\) or \(x\geq 10\) .

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1.2 I can identify increasing, decreasing, and the average rate of change of a given or table of values. 1.2 I can find a linear regression line and use it to predict values. 1.3 I can graph absolute value equations, identifying transformations. 1.4 I can identify min, max, vertex, end behavior, and compare absolute value graphs and tables. 1.5 ...

Step 2: Set the argument of the absolute value equal to ± p. Here the argument is 5x − 1 and p = 6. 5x − 1 = − 6 or 5x − 1 = 6. Step 3: Solve each of the resulting linear equations. 5x − 1 = − 6 or 5x − 1 = 6 5x = − 5 5x = 7 x = − 1 x = 7 5. Step 4: Verify the solutions in the original equation. Check x = − 1.

About this unit. This topic covers: Solving absolute value equations. Graphing absolute value functions. Solving absolute value inequalities.

Quiz: solving absolute value equations and inequalities 1.4.3. 10 terms. demiann04. Preview. Math Chapter 4 vocabulary. 15 terms. peyton_vance76. Preview. Solving Absolute Value Equations. 10 terms. Carliana2. ... Can be anything except for the number that makes the absolute value expression equal zero. (-∞,-4)U(-4,∞) IxI≥0 , IxI>-3, or ...

D. x = 5/2 or x = 1. Solve the inequality. Graph the solution. 8|x + 3/4| < 2. C. - 1 < x < -1/2. A furniture maker uses the specification 19.88 ≤ w ≤ 20.12 for the width w in inches of a desk drawer. Write the specification as an absolute value inequality. B. |w - 20| ≤ 0.12. Study with Quizlet and memorize flashcards containing terms ...

This algebra video tutorial shows you how to solve absolute value equations with inequalities and how to plot the solution on a number line and write the ans...

Review how to solve equations and inequalities involving absolute value with these interactive flashcards.

The absolute number of a number a is written as $$\left | a \right |$$ And represents the distance between a and 0 on a number line. An absolute value equation is an equation that contains an absolute value expression. The equation $$\left | x \right |=a$$ Has two solutions x = a and x = -a because both numbers are at the distance a from 0.

This is the situation shown in Figure 4.4.1 4.4. 1 (a). The graph of y = |x| is therefore never below the graph of y = −5. Thus, the inequality |x| < −5 has no solution. An alternate approach is to consider the fact that the absolute value of x is always nonnegative and can never be less than −5.

The absolute value of a number is its distance from zero on the number line. We started with the inequality | x | ≤ 5. We saw that the numbers whose distance is less than or equal to five from zero on the number line were − 5 and 5 and all the numbers between − 5 and 5 (Figure 1.7.4 ). Figure 1.7.4.

Separate into four cases, since we don't know whether $ 3x-2$ and $ x+2$ are positive or negative. Since they are absolute values in the equations, they could be either, but still come out positive. For example, when the expression $ 3x-2$ is negative, the absolute value of that expression is the negation of it, or $ -3x+2$, to make it positive in the equation. Play around with some numbers ...

Section 3.1 Absolute Value Inequalities. A2.5.4 Solve equations and inequalities involving absolute values of linear expressions; Need a tutor? Click this link and get your first session free! Packet. a2_3.1_packet.pdf: File Size: 599 kb: File Type: pdf: Download File.

Objective: In this lesson, you will solve absolute value linear equations and inequalities and use graphs to verify and represent solutions. ... Unit 4; Lesson 1; Solving Absolute Value Equations and Inequalities. Flashcards. Learn. Test. Match. absolute value. Click the card to flip 👆 ...

This Equations and Inequalities Unit Bundle includes guided notes, homework assignments, two quizzes, a study guide and a unit test that cover the following topics: • Simplifying Radicals. • Classifying Numbers (the Real Number System) • Order of Operations (Includes absolute value and square roots) • Evaluating Expressions (Includes ...

Absolute value graphs make a V shape, but why do they do that? ... Solving equations & inequalities. Unit 3. Working with units. Unit 4. Linear equations & graphs. Unit 5. Forms of linear equations. ... Unit 10: Absolute value & piecewise functions. 600 possible mastery points. Mastered. Proficient. Familiar. Attempted.

Linear equations and inequalities: Unit test; About this unit. Linear equations and inequalities are the foundation of many advanced math topics, such as functions, systems, matrices, and calculus. ... Absolute value equations Get 3 of 4 questions to level up! Quiz 2. Level up on the above skills and collect up to 320 Mastery points Start quiz ...

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Each lesson has notes/class examples and homework examples. There is plenty of room for the student to show their work and write their answers in the space provided. An answer key is included. Download my FREE sample lessons: Algebra 2 - Unit 1 - Lesson 7 - Solving Absolute Value Equations & Inequalities Algebra 2 - Unit 10 - Lesson 1 ...

Products. $428.00 $675.44 Save $247.44. View Bundle. Equations and Inequalities MEGA Bundle (Algebra 2 - Unit 1) This is a MEGA Bundle of GUIDED NOTES WITH VIDEO LESSONS for your distance learning needs, homework, daily warm-up, content quizzes, mid-unit and end-unit assessments, review assignments, and cooperative activities for Algebra 2 ...

Math 3 Unit 1 Absolute Value Equations and Inequalities. 5.0 (1 review) Absolute value. Click the card to flip 👆. the distance from zero on a number line. Click the card to flip 👆. 1 / 22.

Here is your free content for this lesson! Absolute Value Equations and Inequalities - Word Docs & PowerPoints. To gain access to our editable content Join the Algebra 2 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards.

Step 1: Solve the corresponding equality. The solution of the equality divides the real number line into several subintervals. Step 2: Using step 1, check the inequality for a number in each of the subintervals. This check determines the intervals of the solution set. Step 3: Check the endpoints of the intervals.

Finding and Graphing Solution Sets of Inequalities. Home Link 4-10 English Español Selected Answers. ... 4-13. Absolute Value. Home Link 4-13 English Español Selected Answers. ... 4-15. Unit 4 Progress Check. Home Link 4-15 English Español Everyday Mathematics for Parents: What You Need to Know to Help Your Child Succeed.

Homework is also turned in through Remind. MATH 1215X COURSE DESCRIPTION: This 1-credit-hour course includes the first third of an Intermediate Algebra course, including problems in ratio and proportion, unit conversions, solving linear equations, and problems modeled by these, finding equations for lines and