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Chapter 2: Linear Equations

2.5 Absolute Value Equations

When solving equations with absolute values, there can be more than one possible answer. This is because the variable whose absolute value is being taken can be either negative or positive, and both possibilities must be accounted for when solving equations.

Example 2.5.1

When there are absolute values in a problem, it is important to first isolate the absolute value, then remove the absolute value by considering both the positive and negative solutions. Notice that, in the next two examples, all the numbers outside of the absolute value are moved to one side first before the absolute value bars are removed and both positive and negative solutions are considered.

Example 2.5.2

Example 2.5.3

Note: the objective in solving for absolute values is to isolate the absolute value to yield a solution.

Example 2.5.4

Often, the absolute value of more than just a variable is being taken. In these cases, it is necessary to solve the resulting equations before considering positive and negative possibilities. This is shown in the next example.

Example 2.5.5

Remember: the absolute value must be isolated first before solving.

Example 2.5.6

There exist two other possible results from solving an absolute value besides what has been shown in the above six examples.

Example 2.5.7

One final type of absolute value problem covered in this chapter is when two absolute values are equal to each other. There will still be both a positive and a negative result—the difference is that a negative must be distributed into the second absolute value for the negative possibility.

Example 2.5.8

For questions 1 to 24, solve each absolute value equation.

  • [latex]| x | = 8[/latex]
  • [latex]| n | = 7[/latex]
  • [latex]| b | = 1[/latex]
  • [latex]| x | = 2[/latex]
  • [latex]| 5 + 8a | = 53[/latex]
  • [latex]| 9n + 8 | = 46[/latex]
  • [latex]| 3k + 8 | = 2[/latex]
  • [latex]| 3 - x | = 6[/latex]
  • [latex]-7 | -3 - 3r | = -21[/latex]
  • [latex]| 2 + 2b | + 1 = 3[/latex]
  • [latex]7 | -7x - 3 | = 21[/latex]
  • [latex]| -4 - 3n | = 2[/latex]
  • [latex]8 | 5p + 8 | - 5 = 11[/latex]
  • [latex]3 - | 6n + 7 | = -40[/latex]
  • [latex]5 | 3 + 7m | + 1 = 51[/latex]
  • [latex]4 | r + 7 | + 3 = 59[/latex]
  • [latex]-7 + 8 | -7x - 3 | = 73[/latex]
  • [latex]8 | 3 - 3n | - 5 = 91[/latex]
  • [latex]| 5x + 3 | = | 2x - 1 |[/latex]
  • [latex]| 2 + 3x | = | 4 - 2x |[/latex]
  • [latex]| 3x - 4 | = | 2x + 3 |[/latex]
  • [latex]| 2x - 5 | = | 3x + 4 |[/latex]
  • [latex]| 4x - 2 | = | 6x + 3 |[/latex]
  • [latex]|3x+2|=|2x-3|[/latex]

Answer Key 2.5

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Solving Absolute Value Equations

Equations that have absolute value sign on one side the absolute value equation |ax + b| = c (c ≥ 0) can be solved by rewriting as two linear equations ax + b = c or ax + b = -c and then solving each equation separately. absolute values always give 2 equations example1: solve |x| = 2 solution: x = 2 or x = -2 example 2: solve |x + 1| = 2 solution: x + 1 = 2 or x + 1 = -2 x = 1 or x = -3 example 3 solve |3x - 4| = 5 solution: 3x - 4 = 5 or 3x - 4 = -5 3x = 9 or 3x = -1 x = 3 or x = -1/3 example 4 solve |4x + 7| = -3 solution: this equation has no solution, since an absolute value cannot be negative. example 5 : solve |2x - 6| = 0 solution: since positive and negative 0 mean the same thing, we only need one equation 2x - 6 = 0 2x = 6 x = 3 exercise 1: solve absolute value equations level 1 $$ \color{blue}{\left| {2x - 3} \right| = 5} $$ $ x = 4 , x = 1 $ $ x = - 4 , x = 1 $ $ x = 4 , x = - 1 $ $ x = -4 , x = - 1 $ level 2 $$ \color{blue}{\left| {3x + 7} \right| = 1} $$ $ x = - 2 , x = - \frac{8}{3} $ $ x = - 2 , x = \frac{8}{3} $ $ x = 2 , x = - \frac{8}{3} $ $ x = 2 , x = \frac{8}{3} $ equations that have absolute value signs on both sides.

If we have absolute value signs on both sides of the equation, we can play the same game with two choices as follows.

Solve |3x + 4| = | 2x - 3|

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

3x + 4 = 2x - 3 or 3x + 4 = -2x + 3

3x = 2x - 7 or 3x = -2x - 1

x = -7 or 5x = -1

x = -7 or x = -1/5

Exercise 2: Solve absolute value equations

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Module 10: Linear Equations

Solving equations containing absolute values, learning outcomes.

  • Solve equations containing absolute values

Solving One-Step Equations Containing Absolute Values with Addition

The absolute value of a number or expression describes its distance from [latex]0[/latex] on a number line. Since the absolute value expresses only the distance, not the direction of the number on a number line, it is always expressed as a positive number or [latex]0[/latex].

For example, [latex]−4[/latex] and  [latex]4[/latex] both have an absolute value of  [latex]4[/latex] because they are each [latex]4[/latex] units from [latex]0[/latex] on a number line—though they are located in opposite directions from [latex]0[/latex] on the number line.

When solving absolute value equations and inequalities , you have to consider both the behavior of absolute value and the properties of equality and inequality.

Because both positive and negative values have a positive absolute value, solving absolute value equations means finding the solution for both the positive and the negative values.

Let’s first look at a very basic example.

[latex] \displaystyle \left| x \right|=5[/latex]

This equation is read “the absolute value of  [latex]x[/latex] is equal to five.” The solution is the value(s) that are five units away from [latex]0[/latex] on a number line.

You might think of [latex]5[/latex] right away; that is one solution to the equation. Notice that [latex]−5[/latex] is also a solution because [latex]−5[/latex] is [latex]5[/latex] units away from  [latex]0[/latex] in the opposite direction. So, the solution to this equation [latex] \displaystyle \left| x \right|=5[/latex] is [latex]x = −5[/latex] or [latex]x = 5[/latex].

Solving Equations of the Form [latex]|x|=a[/latex]

For any positive number [latex]a[/latex], the solution of [latex]\left|x\right|=a[/latex] is

[latex]x=a[/latex]  or  [latex]x=−a[/latex]

[latex]x[/latex] can be a single variable or any algebraic expression.

You can solve a more complex absolute value problem in a similar fashion.

Solve [latex] \displaystyle \left| x+5\right|=15[/latex].

This equation asks you to find what number plus [latex]5[/latex] has an absolute value of [latex]15[/latex]. Since 15 and [latex]−15[/latex] both have an absolute value of [latex]15[/latex], the absolute value equation is true when the quantity [latex]x + 5[/latex] is [latex]15[/latex] or  [latex]x + 5[/latex] is [latex]−15[/latex], since [latex]|15|=15[/latex] and [latex]|−15|=15[/latex]. So, you need to find out what value for  [latex]x[/latex] will make this expression equal to [latex]15[/latex], as well as what value for [latex]x[/latex] will make the expression equal to [latex]−15[/latex]. Solving the two equations you get

[latex] \displaystyle \begin{array}{l}x+5=15\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,{x+5=-15}\\\underline{\,\,\,\,\,-5\,\,\,\,-5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-5\,\,\,\,\,\,\,\,\,-5}\\x\,\,\,\,\,\,\,\,\,=\,\,10\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,=-20\end{array}[/latex]

You can check these two solutions in the absolute value equation to see if [latex]x=10[/latex] and [latex]x = −20[/latex] are correct.

Solving One-Step Equations Containing Absolute Values With Multiplication

In the last section, we saw examples of solving equations with absolute values where the only operation was addition or subtraction. Now we will see how to solve equations with absolute value that include multiplication.

Remember that absolute value refers to the distance from zero. You can use the same technique of first isolating the absolute value, then setting up and solving two equations to solve an absolute value equation involving multiplication.

Solve [latex] \displaystyle \left| 2x\right|=6[/latex].

This equation asks you to find what number times [latex]2[/latex] has an absolute value of [latex]6[/latex].

Since [latex]6[/latex] and [latex]−6[/latex] both have an absolute value of [latex]6[/latex], the absolute value equation is true when the quantity [latex]2x[/latex] is [latex]6[/latex] or  [latex]2x[/latex] is [latex]−6[/latex], since [latex]|6|=6[/latex] and [latex]|−6|=6[/latex].

So, you need to find out what value for [latex]x[/latex] will make this expression equal to [latex]6[/latex], as well as what value for [latex]x[/latex] will make the expression equal to [latex]−6[/latex].

Solving the two equations you get

[latex]2x=6\text { or } 2x=-6[/latex]

[latex]\frac{2x}{2}=\frac{6}{2}\text { or } \frac{2x}{2}=\frac{-6}{2}[/latex]

[latex]x=3\text { or } x=-3[/latex]

You can check these two solutions in the absolute value equation to see if [latex]x=3[/latex] and [latex]x=−3[/latex] are correct.

Solve [latex] \displaystyle\frac{1}{3}\left|k\right|=12[/latex].

Notice how this example is different from the last; [latex] \displaystyle\frac{1}{3}[/latex] is outside the absolute value grouping symbols. This means we need to isolate the absolute value first, then apply the definition of absolute value.

First, isolate the absolute value term by multiplying by the inverse of [latex] \displaystyle\frac{1}{3}[/latex]:

[latex]\begin{array}{r}\frac{1}{3}\left|k\right|=12\,\,\,\,\,\,\,\,\\\left(3\right)\frac{1}{3}\left|k\right|=\left(3\right)12\\\left|k\right|=36\,\,\,\,\,\,\,\,\end{array}[/latex]

Apply the definition of absolute value:

[latex] \displaystyle{k }=36\text { or } {k }=-36[/latex]

You can check these two solutions in the absolute value equation to see if [latex]k=36[/latex] and [latex]k=−36[/latex] are correct.

In the following video, you will see two examples of how to solve an absolute value equation, one with integers and one with fractions.

Solving Multi-Step Equations With Absolute Value

We can apply the same techniques we used for solving a one-step equation which contains absolute value to an equation that will take more than one step to solve.  Let’s start with an example where the first step is to write two equations: one equal to positive  [latex]26[/latex] and one equal to negative [latex]26[/latex].

Solve for [latex]p[/latex]. [latex]\left|2p–4\right|=26[/latex]

Write the two equations that will give an absolute value of [latex]26[/latex].

[latex] \displaystyle 2p-4=26\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,2p-4=\,-26[/latex]

Solve each equation for [latex]p[/latex] by isolating the variable .

[latex] \displaystyle \begin{array}{r}2p-4=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2p-4=\,-26\\\underline{\,\,\,\,\,\,+4\,\,\,\,+4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,+4\,\,\,\,\,\,\,+4}\\\underline{2p}\,\,\,\,\,\,=\underline{30}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{2p}\,\,\,\,\,=\,\underline{-22}\\2\,\,\,\,\,\,\,=\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,2\\\,\,\,\,\,\,\,\,\,p=15\,\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p=\,-11\end{array}[/latex]

Check the solutions in the original equation.

[latex] \displaystyle \begin{array}{r}\,\,\,\,\,\left| 2p-4 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 2p-4 \right|=26\\\left| 2(15)-4 \right|=26\,\,\,\,\,\,\,\left| 2(-11)-4 \right|=26\\\,\,\,\,\,\left| 30-4 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -22-4 \right|=26\\\,\,\,\,\,\,\,\,\,\,\,\,\left| 26 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -26 \right|=26\end{array}[/latex]

Both solutions check!

[latex]p=15[/latex] or [latex]p=-11[/latex]

In the next video, we show more examples of solving a simple absolute value equation.

Now let’s look at an example where you need to do an algebraic step or two before you can write your two equations. The goal here is to get the absolute value on one side of the equation by itself. Then we can proceed as we did in the previous example.

Solve for [latex]w[/latex]. [latex]3\left|4w–1\right|–5=10[/latex]

Isolate the term with the absolute value by adding  [latex]5[/latex] to both sides.

[latex]\begin{array}{r}3\left|4w-1\right|-5=10\\\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,+5\,\,\,+5}\\ 3\left|4w-1\right|=15\end{array}[/latex]

Divide both sides by [latex]3[/latex]. Now the absolute value is isolated.

[latex]\begin{array}{r} \underline{3\left|4w-1\right|}=\underline{15}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\\\left|4w-1\right|=\,\,5\end{array}[/latex]

Write the two equations that will give an absolute value of  [latex]5[/latex] and solve them.

[latex] \displaystyle \begin{array}{r}4w-1=5\,\,\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,\,\,4w-1=-5\\\underline{\,\,\,\,\,\,\,+1\,\,+1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,\,\,+1\,\,\,\,\,+1}\\\,\,\,\,\,\underline{4w}=\underline{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{4w}\,\,\,\,\,\,\,=\underline{-4}\\4\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\\\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\text{or}\,\,\,\,\,-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

[latex] \displaystyle \begin{array}{r}\,\,\,\,\,3\left| 4w-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4w-1\, \right|-5=10\\\\3\left| 4\left( \frac{3}{2} \right)-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4w-1\, \right|-5=10\\\\\,\,\,\,\,\,3\left| \frac{12}{2}-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4(-1)-1\, \right|-5=10\\\\\,\,\,\,\,\,\,\,3\left| 6-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| -4-1\, \right|-5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left(5\right)-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| -5 \right|-5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15-5=10\\10=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10=10\end{array}[/latex]

Both solutions check

[latex]w=-1\,\,\,\,\text{or}\,\,\,\,w=\frac{3}{2}[/latex]

In the two videos that follow, we show examples of how to solve an absolute value equation that requires you to isolate the absolute value first using mathematical operations.

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What is Absolute Value?

Practice questions, solving absolute value equations – methods & examples.

Solving equations containing an absolute value is as simple as working with regular linear equations . Before we can embark on solving absolute value equations, let’s take a review of what the word absolute value means.

In mathematics, the absolute value of a number refers to the distance of a number from zero, regardless of direction. The absolute value of a number x is generally represented as | x | = a, which implies that, x = + a and -a.

We say that the absolute value of a given number is the positive version of that number . For example, the absolute value of negative 5 is positive 5, and this can be written as: | − 5 | = 5.

Other examples of absolute values of numbers include:  |− 9| = 9, |0| = 0, − |−12| = −12 etc. From these examples of absolute values, we simply define absolute value equations as equations containing expressions with absolute value functions.

How to Solve Absolute Value Equations?

The following are the general steps for solving equations containing absolute value functions:

  • Isolate the expression containing the absolute value function.
  • Get rid of the absolute value notation by setting up the two equations so that in the first equation, the quantity inside absolute notation is positive. In the second equation, it is negative. You will remove the absolute notation and write the quantity with its suitable sign.
  • Calculate the unknown value for the positive version of the equation.
  • Solve for the negative version of the equation, in which you will first multiply the value on the other side of the equal sign by -1, and then solve.

In addition to the above steps, there are other important rules you should keep in mind when solving absolute value equations.

  • The ∣x∣is always positive: ∣x∣ → +x.
  • In | x| = a, if the  a on the right is a positive number or zero, then there is a solution.

9

Solve the equation for x: |3 + x| − 5 = 4.

  • Isolate the absolute value expression by applying the Law of equations. This means, we add 5 to both sides of the equation to obtain;

| 3 + x | − 5 + 5 = 4 + 5

| 3 + x |= 9

  • Calculate for the positive version of the equation. Solve the equation by assuming the absolute value symbols.

| 3 +  x  | = 9 → 3 +  x  = 9

Subtract 3 from both sides of the equation.

3 – 3 + x = 9 -3

  • Now calculate for the negative version of the equation by multiplying 9 by -1.

3 +  x  | = 9 → 3 +  x  = 9 × ( −1)

Also subtract 3 from both side to isolate x.

3 -3 + x = – 9 -3

Therefore 6 and -12 are the solutions.

Solve for all real values of x such that | 3x – 4 | – 2 = 3.

  • Isolate the equation with absolute function by add 2 to both sides.

= | 3x – 4 | – 2 + 2 = 3 + 2

= | 3x – 4 |= 5

Assume the absolute signs and solve for the positive version of the equation.

| 3x – 4 |= 5→3x – 4 = 5

Add 4 to both sides of the equation.

3x – 4 + 4 = 5 + 4

Divide: 3x/3 =9/3

Now solve for the negative version by multiplying 5 by -1.

3x – 4 = 5→3x – 4 = -1(5)

3x – 4 = -5

3x – 4 + 4 = – 5 + 4

Divide by 3 on both sides.

Therefore, 3 and 1/3 are the solutions.

Solve for all real values of x: Solve | 2 x  – 3 | – 4 = 3

Add 4 to both sides.

| 2 x  – 3 | -4 = 3 →| 2 x  – 3 | = 7

Assume the absolute symbols and solve for the positive version of x.

2 x  – 3 = 7

2x – 3 + 3 = 7 + 3

Now solve for the negative version of x by multiplying 7 by -1

2 x  – 3 = 7→2 x  – 3 = -1(7)

Add 3 to both sides.

2x – 3 + 3 = – 7 + 3

x = – 2

Therefore, x  = –2, 5

Solve for all real numbers of x: | x + 2 | = 7

Already the absolute value expression is isolated, therefore assume the absolute symbols and solve.

| x + 2 | = 7 → x + 2 = 7

Subtract 2 from both sides.

x + 2 – 2 = 7 -2

Multiply 7 by -1 to solve for the negative version of the equation.

x + 2 = -1(7) → x + 2 = -7

Subtract by 2 on both sides.

x + 2 – 2 = – 7 – 2

Therefore, x = -9, 5

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solving linear equations with absolute value

Absolute Value Equations

More specific topics in absolute value equations.

  • Graphing Absolute Value Functions

Popular Tutorials in Absolute Value Equations

How Do You Write an Equation for a Translation of an Absolute Value Function?

How Do You Write an Equation for a Translation of an Absolute Value Function?

Want to write an equation to translate the graph of an absolute value equation? This tutorial takes you through that process step-by-step! Take an absolute value equation and perform a vertical and horizontal translation to create a new equation. Watch it all in this tutorial.

How Do You Graph an Absolute Value Function?

How Do You Graph an Absolute Value Function?

Graphing an absolute value equation can be complicated, unless you know how to dissect the equation to find and use the slope and translations. Follow along as this tutorial shows you how to identify the necessary parts of the equation and use them to graph the absolute value equation.

What Does the Constant 'k' do in y = |x|+k?

What Does the Constant 'k' do in y = |x|+k?

When you're learning about translating absolute value equations, learning about vertical translations is a MUST! Check out this tutorial and see what it takes to translate an absolute value equation vertically.

What Does the Constant 'h' do in y = |x-h|?

What Does the Constant 'h' do in y = |x-h|?

When you're learning about translating absolute value equations, learning about horizontal translations is a MUST! Check out this tutorial and see what it takes to translate an absolute value equation horizontally.

What is an Absolute Value Function?

What is an Absolute Value Function?

An absolute value function is just a function that contains absolute values. This tutorial gives a great introduction to this very useful function!

Related Topics

Other topics in solving and graphing linear inequalities :.

  • Solving Inequalities Using Addition and Subtraction
  • Solving Inequalities Using Multiplication and Division
  • Multi-Step Inequalities
  • Compound Inequalities
  • Absolute Value Inequalities
  • Graphing Inequalities in Two Variables
  • Terms of Use

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2.8: Linear Inequalities and Absolute Value Inequalities

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Learning Objectives

In this section, you will:

  • Use interval notation
  • Use properties of inequalities.
  • Solve inequalities in one variable algebraically.
  • Solve absolute value inequalities.

Several red winner’s ribbons lie on a white table.

It is not easy to make the honor roll at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities.

Using Interval Notation

Indicating the solution to an inequality such as x ≥ 4 x ≥ 4 can be achieved in several ways.

We can use a number line as shown in Figure 2 . The blue ray begins at x = 4 x = 4 and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.

A number line starting at zero with the last tick mark being labeled 11.  There is a dot at the number 4 and an arrow extends toward the right.

We can use set-builder notation : { x | x ≥ 4 } , { x | x ≥ 4 } , which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.

The third method is interval notation , in which solution sets are indicated with parentheses or brackets. The solutions to x ≥ 4 x ≥ 4 are represented as [ 4 , ∞ ) . [ 4 , ∞ ) . This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval , or a set of numbers in which a solution falls, are [ −2 , 6 ) , [ −2 , 6 ) , or all numbers between −2 −2 and 6 , 6 , including −2 , −2 , but not including 6 ; 6 ; ( − 1 , 0 ) , ( − 1 , 0 ) , all real numbers between, but not including −1 −1 and 0 ; 0 ; and ( − ∞ , 1 ] , ( − ∞ , 1 ] , all real numbers less than and including 1. 1. Table 1 outlines the possibilities.

Using Interval Notation to Express All Real Numbers Greater Than or Equal to a

Use interval notation to indicate all real numbers greater than or equal to −2. −2.

Use a bracket on the left of −2 −2 and parentheses after infinity: [ −2 , ∞ ) . [ −2 , ∞ ) . The bracket indicates that −2 −2 is included in the set with all real numbers greater than −2 −2 to infinity.

Use interval notation to indicate all real numbers between and including −3 −3 and 5. 5.

Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b

Write the interval expressing all real numbers less than or equal to −1 −1 or greater than or equal to 1. 1.

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at − ∞ − ∞ and ends at −1 , −1 , which is written as ( − ∞ , −1 ] . ( − ∞ , −1 ] .

The second interval must show all real numbers greater than or equal to 1 , 1 , which is written as [ 1 , ∞ ) . [ 1 , ∞ ) . However, we want to combine these two sets. We accomplish this by inserting the union symbol, ∪ , ∪ , between the two intervals.

( − ∞ , −1 ] ∪ [ 1 , ∞ ) ( − ∞ , −1 ] ∪ [ 1 , ∞ )

Express all real numbers less than −2 −2 or greater than or equal to 3 in interval notation.

Using the Properties of Inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

Properties of Inequalities

A d d i t i o n P r o p e r t y If a < b , then a + c < b + c . M u l t i p l i c a t i o n P r o p e r t y If a < b and c > 0 , then a c < b c . If a < b and c < 0 , then a c > b c . A d d i t i o n P r o p e r t y If a < b , then a + c < b + c . M u l t i p l i c a t i o n P r o p e r t y If a < b and c > 0 , then a c < b c . If a < b and c < 0 , then a c > b c .

These properties also apply to a ≤ b , a ≤ b , a > b , a > b , and a ≥ b . a ≥ b .

Demonstrating the Addition Property

Illustrate the addition property for inequalities by solving each of the following:

  • ⓐ x − 15 < 4 x − 15 < 4
  • ⓑ 6 ≥ x − 1 6 ≥ x − 1
  • ⓒ x + 7 > 9 x + 7 > 9

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

  • ⓐ x − 15 < 4 x − 15 + 15 < 4 + 15 Add 15 to both sides . x < 19 x − 15 < 4 x − 15 + 15 < 4 + 15 Add 15 to both sides . x < 19
  • ⓑ 6 ≥ x − 1 6 + 1 ≥ x − 1 + 1 Add 1 to both sides . 7 ≥ x 6 ≥ x − 1 6 + 1 ≥ x − 1 + 1 Add 1 to both sides . 7 ≥ x
  • ⓒ x + 7 > 9 x + 7 − 7 > 9 − 7 Subtract 7 from both sides . x > 2 x + 7 > 9 x + 7 − 7 > 9 − 7 Subtract 7 from both sides . x > 2

Solve: 3 x −2 < 1. 3 x −2 < 1.

Demonstrating the Multiplication Property

Illustrate the multiplication property for inequalities by solving each of the following:

  • ⓐ 3 x < 6 3 x < 6
  • ⓑ −2 x − 1 ≥ 5 −2 x − 1 ≥ 5
  • ⓒ 5 − x > 10 5 − x > 10
  • ⓐ 3 x < 6 1 3 ( 3 x ) < ( 6 ) 1 3 x < 2 3 x < 6 1 3 ( 3 x ) < ( 6 ) 1 3 x < 2
  • ⓑ − 2 x − 1 ≥ 5 − 2 x ≥ 6 ( − 1 2 ) ( − 2 x ) ≥ ( 6 ) ( − 1 2 ) Multiply by − 1 2 . x ≤ − 3 Reverse the inequality . − 2 x − 1 ≥ 5 − 2 x ≥ 6 ( − 1 2 ) ( − 2 x ) ≥ ( 6 ) ( − 1 2 ) Multiply by − 1 2 . x ≤ − 3 Reverse the inequality .
  • ⓒ 5 − x > 10 − x > 5 ( − 1 ) ( − x ) > ( 5 ) ( − 1 ) Multiply by − 1. x < − 5 Reverse the inequality . 5 − x > 10 − x > 5 ( − 1 ) ( − x ) > ( 5 ) ( − 1 ) Multiply by − 1. x < − 5 Reverse the inequality .

Solve: 4 x + 7 ≥ 2 x − 3. 4 x + 7 ≥ 2 x − 3.

Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

Solving an Inequality Algebraically

Solve the inequality: 13 − 7 x ≥ 10 x − 4. 13 − 7 x ≥ 10 x − 4.

Solving this inequality is similar to solving an equation up until the last step.

13 − 7 x ≥ 10 x − 4 13 − 17 x ≥ −4 Move variable terms to one side of the inequality . −17 x ≥ −17 Isolate the variable term . x ≤ 1 Dividing both sides by −17 reverses the inequality . 13 − 7 x ≥ 10 x − 4 13 − 17 x ≥ −4 Move variable terms to one side of the inequality . −17 x ≥ −17 Isolate the variable term . x ≤ 1 Dividing both sides by −17 reverses the inequality .

The solution set is given by the interval ( − ∞ , 1 ] , ( − ∞ , 1 ] , or all real numbers less than and including 1.

Solve the inequality and write the answer using interval notation: − x + 4 < 1 2 x + 1. − x + 4 < 1 2 x + 1.

Solving an Inequality with Fractions

Solve the following inequality and write the answer in interval notation: − 3 4 x ≥ − 5 8 + 2 3 x . − 3 4 x ≥ − 5 8 + 2 3 x .

We begin solving in the same way we do when solving an equation.

− 3 4 x ≥ − 5 8 + 2 3 x − 3 4 x − 2 3 x ≥ − 5 8 Put variable terms on one side . − 9 12 x − 8 12 x ≥ − 5 8 Write fractions with common denominator . − 17 12 x ≥ − 5 8 x ≤ − 5 8 ( − 12 17 ) Multiplying by a negative number reverses the inequality . x ≤ 15 34 − 3 4 x ≥ − 5 8 + 2 3 x − 3 4 x − 2 3 x ≥ − 5 8 Put variable terms on one side . − 9 12 x − 8 12 x ≥ − 5 8 Write fractions with common denominator . − 17 12 x ≥ − 5 8 x ≤ − 5 8 ( − 12 17 ) Multiplying by a negative number reverses the inequality . x ≤ 15 34

The solution set is the interval ( − ∞ , 15 34 ] . ( − ∞ , 15 34 ] .

Solve the inequality and write the answer in interval notation: − 5 6 x ≤ 3 4 + 8 3 x . − 5 6 x ≤ 3 4 + 8 3 x .

Understanding Compound Inequalities

A compound inequality includes two inequalities in one statement. A statement such as 4 < x ≤ 6 4 < x ≤ 6 means 4 < x 4 < x and x ≤ 6. x ≤ 6. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.

Solving a Compound Inequality

Solve the compound inequality: 3 ≤ 2 x + 2 < 6. 3 ≤ 2 x + 2 < 6.

The first method is to write two separate inequalities: 3 ≤ 2 x + 2 3 ≤ 2 x + 2 and 2 x + 2 < 6. 2 x + 2 < 6. We solve them independently.

3 ≤ 2 x + 2 and 2 x + 2 < 6 1 ≤ 2 x 2 x < 4 1 2 ≤ x x < 2 3 ≤ 2 x + 2 and 2 x + 2 < 6 1 ≤ 2 x 2 x < 4 1 2 ≤ x x < 2

Then, we can rewrite the solution as a compound inequality, the same way the problem began.

1 2 ≤ x < 2 1 2 ≤ x < 2

In interval notation, the solution is written as [ 1 2 , 2 ) . [ 1 2 , 2 ) .

The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.

3 ≤ 2 x + 2 < 6 1 ≤ 2 x < 4 Isolate the variable term, and subtract 2 from all three parts . 1 2 ≤ x < 2 Divide through all three parts by 2 . 3 ≤ 2 x + 2 < 6 1 ≤ 2 x < 4 Isolate the variable term, and subtract 2 from all three parts . 1 2 ≤ x < 2 Divide through all three parts by 2 .

We get the same solution: [ 1 2 , 2 ) . [ 1 2 , 2 ) .

Solve the compound inequality: 4 < 2 x − 8 ≤ 10. 4 < 2 x − 8 ≤ 10.

Solving a Compound Inequality with the Variable in All Three Parts

Solve the compound inequality with variables in all three parts: 3 + x > 7 x − 2 > 5 x − 10. 3 + x > 7 x − 2 > 5 x − 10.

Let's try the first method. Write two inequalities :

3 + x > 7 x − 2 and 7 x − 2 > 5 x − 10 3 > 6 x − 2 2 x − 2 > −10 5 > 6 x 2 x > −8 5 6 > x x > −4 x < 5 6 −4 < x 3 + x > 7 x − 2 and 7 x − 2 > 5 x − 10 3 > 6 x − 2 2 x − 2 > −10 5 > 6 x 2 x > −8 5 6 > x x > −4 x < 5 6 −4 < x

The solution set is −4 < x < 5 6 −4 < x < 5 6 or in interval notation ( −4 , 5 6 ) . ( −4 , 5 6 ) . Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See Figure 3 .

A number line with the points -4 and 5/6 labeled.  Dots appear at these points and a line connects these two dots.

Solve the compound inequality: 3 y < 4 − 5 y < 5 + 3 y . 3 y < 4 − 5 y < 5 + 3 y .

Solving Absolute Value Inequalities

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at ( − x , 0 ) ( − x , 0 ) has an absolute value of x , x , as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

| A | < B , | A | ≤ B , | A | > B , or | A | ≥ B , | A | < B , | A | ≤ B , | A | > B , or | A | ≥ B ,

Where A , and sometimes B , represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all x x - values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x- values such that the distance between x x and 600 is less than or equal to 200. We represent the distance between x x and 600 as | x − 600 | , | x − 600 | , and therefore, | x − 600 | ≤ 200 | x − 600 | ≤ 200 or

−200 ≤ x − 600 ≤ 200 −200 + 600 ≤ x − 600 + 600 ≤ 200 + 600 400 ≤ x ≤ 800 −200 ≤ x − 600 ≤ 200 −200 + 600 ≤ x − 600 + 600 ≤ 200 + 600 400 ≤ x ≤ 800

This means our returns would be between $400 and $800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

Absolute Value Inequalities

For an algebraic expression X, and k > 0 , k > 0 , an absolute value inequality is an inequality of the form

| X | < k is equivalent to  − k < X < k | X | > k is equivalent to  X < − k or  X > k | X | < k is equivalent to  − k < X < k | X | > k is equivalent to  X < − k or  X > k

These statements also apply to | X | ≤ k | X | ≤ k and | X | ≥ k . | X | ≥ k .

Determining a Number within a Prescribed Distance

Describe all values x x within a distance of 4 from the number 5.

We want the distance between x x and 5 to be less than or equal to 4. We can draw a number line, such as in Figure 4 , to represent the condition to be satisfied.

A number line with one tick mark in the center labeled: 5.  The tick marks on either side of the center one are not marked.  Arrows extend from the center tick mark to the outer tick marks, both are labeled 4.

The distance from x x to 5 can be represented using an absolute value symbol, | x − 5 | . | x − 5 | . Write the values of x x that satisfy the condition as an absolute value inequality.

| x − 5 | ≤ 4 | x − 5 | ≤ 4

We need to write two inequalities as there are always two solutions to an absolute value equation.

x − 5 ≤ 4 and x − 5 ≥ − 4 x ≤ 9 x ≥ 1 x − 5 ≤ 4 and x − 5 ≥ − 4 x ≤ 9 x ≥ 1

If the solution set is x ≤ 9 x ≤ 9 and x ≥ 1 , x ≥ 1 , then the solution set is an interval including all real numbers between and including 1 and 9.

So | x − 5 | ≤ 4 | x − 5 | ≤ 4 is equivalent to [ 1 , 9 ] [ 1 , 9 ] in interval notation.

Describe all x- values within a distance of 3 from the number 2.

Solving an Absolute Value Inequality

Solve | x − 1 | ≤ 3 | x − 1 | ≤ 3 .

| x − 1 | ≤ 3 −3 ≤ x − 1 ≤ 3 −2 ≤ x ≤ 4 [ −2 , 4 ] | x − 1 | ≤ 3 −3 ≤ x − 1 ≤ 3 −2 ≤ x ≤ 4 [ −2 , 4 ]

Using a Graphical Approach to Solve Absolute Value Inequalities

Given the equation y = − 1 2 | 4 x − 5 | + 3 , y = − 1 2 | 4 x − 5 | + 3 , determine the x -values for which the y -values are negative.

We are trying to determine where y < 0 , y < 0 , which is when − 1 2 | 4 x − 5 | + 3 < 0. − 1 2 | 4 x − 5 | + 3 < 0. We begin by isolating the absolute value.

− 1 2 | 4 x − 5 | < − 3 Multiply both sides by –2, and reverse the inequality . | 4 x − 5 | > 6 − 1 2 | 4 x − 5 | < − 3 Multiply both sides by –2, and reverse the inequality . | 4 x − 5 | > 6

Next, we solve for the equality | 4 x − 5 | = 6. | 4 x − 5 | = 6.

4 x − 5 = 6 4 x − 5 = − 6 4 x = 11 or 4 x = − 1 x = 11 4 x = − 1 4 4 x − 5 = 6 4 x − 5 = − 6 4 x = 11 or 4 x = − 1 x = 11 4 x = − 1 4

Now, we can examine the graph to observe where the y- values are negative. We observe where the branches are below the x- axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at x = − 1 4 x = − 1 4 and x = 11 4 , x = 11 4 , and that the graph opens downward. See Figure 5 .

A coordinate plan with the x-axis ranging from -5 to 5 and the y-axis ranging from -4 to 4.  The function y = -1/2|4x – 5| + 3 is graphed.  An open circle appears at the point -0.25 and an arrow

Solve − 2 | k − 4 | ≤ − 6. − 2 | k − 4 | ≤ − 6.

Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities.

  • Interval notation
  • How to solve linear inequalities
  • How to solve an inequality
  • Absolute value equations
  • Compound inequalities
  • Absolute value inequalities

2.7 Section Exercises

When solving an inequality, explain what happened from Step 1 to Step 2:

Step 1 − 2 x > 6 Step 2 x < − 3 Step 1 − 2 x > 6 Step 2 x < − 3

When solving an inequality, we arrive at:

x + 2 < x + 3 2 < 3 x + 2 < x + 3 2 < 3

Explain what our solution set is.

When writing our solution in interval notation, how do we represent all the real numbers?

x + 2 > x + 3 2 > 3 x + 2 > x + 3 2 > 3

Describe how to graph y = | x − 3 | y = | x − 3 |

For the following exercises, solve the inequality. Write your final answer in interval notation.

4 x − 7 ≤ 9 4 x − 7 ≤ 9

3 x + 2 ≥ 7 x − 1 3 x + 2 ≥ 7 x − 1

−2 x + 3 > x − 5 −2 x + 3 > x − 5

4 ( x + 3 ) ≥ 2 x − 1 4 ( x + 3 ) ≥ 2 x − 1

− 1 2 x ≤ − 5 4 + 2 5 x − 1 2 x ≤ − 5 4 + 2 5 x

−5 ( x − 1 ) + 3 > 3 x − 4 − 4 x −5 ( x − 1 ) + 3 > 3 x − 4 − 4 x

−3 ( 2 x + 1 ) > −2 ( x + 4 ) −3 ( 2 x + 1 ) > −2 ( x + 4 )

x + 3 8 − x + 5 5 ≥ 3 10 x + 3 8 − x + 5 5 ≥ 3 10

x − 1 3 + x + 2 5 ≤ 3 5 x − 1 3 + x + 2 5 ≤ 3 5

For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation.

| x + 9 | ≥ −6 | x + 9 | ≥ −6

| 2 x + 3 | < 7 | 2 x + 3 | < 7

| 3 x − 1 | > 11 | 3 x − 1 | > 11

| 2 x + 1 | + 1 ≤ 6 | 2 x + 1 | + 1 ≤ 6

| x − 2 | + 4 ≥ 10 | x − 2 | + 4 ≥ 10

| −2 x + 7 | ≤ 13 | −2 x + 7 | ≤ 13

| x − 7 | < −4 | x − 7 | < −4

| x − 20 | > −1 | x − 20 | > −1

| x − 3 4 | < 2 | x − 3 4 | < 2

For the following exercises, describe all the x -values within or including a distance of the given values.

Distance of 5 units from the number 7

Distance of 3 units from the number 9

Distance of 10 units from the number 4

Distance of 11 units from the number 1

For the following exercises, solve the compound inequality. Express your answer using inequality signs, and then write your answer using interval notation.

−4 < 3 x + 2 ≤ 18 −4 < 3 x + 2 ≤ 18

3 x + 1 > 2 x − 5 > x − 7 3 x + 1 > 2 x − 5 > x − 7

3 y < 5 − 2 y < 7 + y 3 y < 5 − 2 y < 7 + y

2 x − 5 < −11 or 5 x + 1 ≥ 6 2 x − 5 < −11 or 5 x + 1 ≥ 6

x + 7 < x + 2 x + 7 < x + 2

For the following exercises, graph the function. Observe the points of intersection and shade the x -axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.

| x − 1 | > 2 | x − 1 | > 2

| x + 3 | ≥ 5 | x + 3 | ≥ 5

| x + 7 | ≤ 4 | x + 7 | ≤ 4

| x − 2 | < 7 | x − 2 | < 7

| x − 2 | < 0 | x − 2 | < 0

For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y -values of the lines.

x + 3 < 3 x − 4 x + 3 < 3 x − 4

x − 2 > 2 x + 1 x − 2 > 2 x + 1

x + 1 > x + 4 x + 1 > x + 4

1 2 x + 1 > 1 2 x − 5 1 2 x + 1 > 1 2 x − 5

4 x + 1 < 1 2 x + 3 4 x + 1 < 1 2 x + 3

For the following exercises, write the set in interval notation.

{ x | −1 < x < 3 } { x | −1 < x < 3 }

{ x | x ≥ 7 } { x | x ≥ 7 }

{ x | x < 4 } { x | x < 4 }

{ x | x is all real numbers } { x | x is all real numbers }

For the following exercises, write the interval in set-builder notation.

( − ∞ , 6 ) ( − ∞ , 6 )

( 4 , ∞ ) ( 4 , ∞ )

[ −3 , 5 ) [ −3 , 5 )

[ −4 , 1 ] ∪ [ 9 , ∞ ) [ −4 , 1 ] ∪ [ 9 , ∞ )

For the following exercises, write the set of numbers represented on the number line in interval notation.

A number line with two tick marks labeled: -2 and 1 respectively.  There is an open circle around the -2 and a dot on the 1 with an arrow connecting the two.

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter y2 = the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, 1:abs(. Find the points of intersection, recall (2 nd CALC 5:intersection, 1 st curve, enter, 2 nd curve, enter, guess, enter). Copy a sketch of the graph and shade the x -axis for your solution set to the inequality. Write final answers in interval notation.

| x + 2 | − 5 < 2 | x + 2 | − 5 < 2

− 1 2 | x + 2 | < 4 − 1 2 | x + 2 | < 4

| 4 x + 1 | − 3 > 2 | 4 x + 1 | − 3 > 2

| x − 4 | < 3 | x − 4 | < 3

| x + 2 | ≥ 5 | x + 2 | ≥ 5

Solve | 3 x + 1 | = | 2 x + 3 | | 3 x + 1 | = | 2 x + 3 |

Solve x 2 − x > 12 x 2 − x > 12

x − 5 x + 7 ≤ 0 , x − 5 x + 7 ≤ 0 , x ≠ −7 x ≠ −7

p = − x 2 + 130 x − 3000 p = − x 2 + 130 x − 3000 is a profit formula for a small business. Find the set of x -values that will keep this profit positive.

Real-World Applications

In chemistry the volume for a certain gas is given by V = 20 T , V = 20 T , where V is measured in cc and T is temperature in ºC. If the temperature varies between 80ºC and 120ºC, find the set of volume values.

A basic cellular package costs $20/mo. for 60 min of calling, with an additional charge of $.30/min beyond that time.. The cost formula would be C = 20 + .30 ( x − 60 ) . C = 20 + .30 ( x − 60 ) . If you have to keep your bill no greater than $50, what is the maximum calling minutes you can use?

IMAGES

  1. Solving Linear Absolute Value Equations with variables on both sides

    solving linear equations with absolute value

  2. Solving Linear Absolute Value Equations

    solving linear equations with absolute value

  3. Equations Involving Absolute Value PT 1

    solving linear equations with absolute value

  4. Solving Absolute Value Equations (solutions, examples, videos

    solving linear equations with absolute value

  5. How To Solve Absolute Value Equations, Basic Introduction, Algebra

    solving linear equations with absolute value

  6. Solving Absolute Value Equations (Part-1)

    solving linear equations with absolute value

VIDEO

  1. HOW to solve ABSOLUTE value equations

  2. Solving Absolute Value Equations Part 2

  3. Linear and Absolute Value Equations

  4. Solving Absolute Value Equations

  5. How to Solve Absolute Value Equations [Algebra basics]

  6. Linear and Absolute Value Equations

COMMENTS

  1. Solving Absolute Value Equations

    Solving Absolute Value Equations. Solving absolute value equations is as easy as working with regular linear equations. The only additional key step that you need to remember is to separate the original absolute value equation into two parts: positive and negative (±) components.Below is the general approach on how to break them down into two equations:

  2. Intro to absolute value equations and graphs

    To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the absolute value, you're going to get 10, or x could be negative 5. Negative 5 minus 5 is negative 10.

  3. 2.6: Solving Absolute Value Equations and Inequalities

    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.

  4. 1.2: Solving Absolute Value Equations

    This page titled 1.2: Solving Absolute Value Equations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

  5. Algebra

    We can also give a strict mathematical/formula definition for absolute value. It is, |p| = {p if p ≥ 0 −p if p < 0 | p | = { p if p ≥ 0 − p if p < 0. This tells us to look at the sign of p p and if it's positive we just drop the absolute value bar. If p p is negative we drop the absolute value bars and then put in a negative in front ...

  6. Absolute value equations

    Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/algebra-home/alg-absolute-value...

  7. 6.12: Absolute Value Functions

    Definition 2.4: Absolute Value. The absolute value of a real number , denoted , is given by. In Definition 2.4, we define using a piecewise-defined function (Section 1.4). To check that this definition agrees with what we previously understood as absolute value, note that since , to find we use the rule , so . Similarly, since , we use the rule ...

  8. Solving absolute value equations (video)

    Lesson 1: Solving absolute value equations. Intro to absolute value equations and graphs. Solving absolute value equations. Absolute value equations. Worked example: absolute value equation with two solutions. Worked example: absolute value equations with one solution. Worked example: absolute value equations with no solution.

  9. 2.5 Absolute Value Equations

    This is because the variable whose absolute value is being taken can be either negative or positive, and both possibilities must be accounted for when solving equations. Example 2.5.1. Solve |x| = 7. | x | = 7. x =+7 and −7, or ±7 x = + 7 and − 7, or ± 7. When there are absolute values in a problem, it is important to first isolate the ...

  10. Absolute value equations, functions, & inequalities

    Algebra (all content) 20 units · 412 skills. Unit 1 Introduction to algebra. Unit 2 Solving basic equations & inequalities (one variable, linear) Unit 3 Linear equations, functions, & graphs. Unit 4 Sequences. Unit 5 System of equations. Unit 6 Two-variable inequalities. Unit 7 Functions. Unit 8 Absolute value equations, functions, & inequalities.

  11. PDF 1.6 Solving Linear Equations

    Absolute value can be positive or negative Two equations to solve. Now notice we have two equations to solve, each equation will give us a different solution. Both equations solve like any other two-step equation. 2x − 1 = 7 2x − 1 = − 7. + 1 + 1 + 1 + 1. 2x = 8. 2 2. x = 4. or 2x = − 6 2 2 x = − 3.

  12. Absolute value equations

    This equation has no solution, since an absolute value cannot be negative. Example 5 : Solve |2x - 6| = 0 . Solution: Since positive and negative 0 mean the same thing, we only need one equation . 2x - 6 = 0 . 2x = 6 . x = 3 . Exercise 1: Solve absolute value equations

  13. Solving Equations Containing Absolute Values

    Solving One-Step Equations Containing Absolute Values with Addition. The absolute value of a number or expression describes its distance from [latex]0[/latex] on a number line. Since the absolute value expresses only the distance, not the direction of the number on a number line, it is always expressed as a positive number or [latex]0[/latex].

  14. PDF Solving Linear Equations

    Solving Linear Equations - Absolute Value Objective: Solve linear absolute value equations. When solving equations with absolute value we can end up with more than one possible answer. This is because what is in the absolute value can be either nega-tive or positive and we must account for both possibilities when solving equations.

  15. 1.2: Absolute Value Equations

    Definition: Absolute Value. Absolute value for linear equations in one variable is given by. If | x | = a, then x = a or x = − a. 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 1.2. 2.

  16. Absolute Value Equation Calculator

    Free absolute value equation calculator - solve absolute value equations with all the steps. Type in any equation to get the solution, steps and graph ... Equations. Basic (Linear) One-Step Addition; One-Step Subtraction; One-Step Multiplication; One-Step Division; One-Step Decimals; Two-Step Integers;

  17. PDF Solving Absolute Value Equations

    38 Chapter 1 Solving Linear Equations SELF-ASSESSMENT 1 I do not understand. 2 I can do it with help. 3 I can do it on my own. 4 I can teach someone else. EXAMPLE 1 Solving Absolute Value Equations Solve each equation. Graph the solutions, if possible. a. ∣ x − 4 ∣ = 6 b. ∣ 3x + 1 ∣ = −5 SOLUTION a. Write the two related linear equations for ∣ x − 4 ∣ = 6.

  18. Solve absolute value equations (practice)

    Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

  19. Solving Absolute Value Equations

    Example 1. Solve the equation for x: |3 + x| − 5 = 4. Solution. Isolate the absolute value expression by applying the Law of equations. This means, we add 5 to both sides of the equation to obtain; | 3 + x | − 5 + 5 = 4 + 5. | 3 + x |= 9. Calculate for the positive version of the equation.

  20. Absolute value inequalities

    Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/algebra-home/alg-absolute-value...

  21. Absolute Value Equations

    An absolute value function is just a function that contains absolute values. This tutorial gives a great introduction to this very useful function! Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long.

  22. Absolute Value Equation Calculator

    About absolute value equations. Solve an absolute value equation using the following steps: Get the absolve value expression by itself. Set up two equations and solve them separately.

  23. High School Math on Instagram: "Get your hands on an Algebra 1 or

    0 likes, 1 comments - absolutevaluemath on December 10, 2023: "Get your hands on an Algebra 1 or Algebra 2 pack of 5 Digital Matching Activities! This is the UL..."

  24. 2.8: Linear Inequalities and Absolute Value Inequalities

    To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently. Absolute Value Inequalities For an algebraic expression X, and k > 0 , k > 0 , an absolute value inequality is an inequality of the form