Topics
1. Equations and Inequalities
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United States of America
Grades 9-12

1.06 Absolute value equations

Lesson
Practice

Introduction

We learned that the absolute value of a number is the number's distance from zero on a number line in 6th grade and built upon this understanding with real-world scenarios in 7th grade. Now we will write and solve equations with absolute value expressions to model real-world problems. We will use absolute value to manipulate equations and make sense of problems in context.

Absolute value equations

The absolute value of a number is the distance of that number from 0 on a number line. Distance is expressed as a positive value, and so the absolute value of a number is always positive. Absolute value is sometimes called magnitude.

A number line ranging from negative 10 to 10. Three vertical lines line up with the negative 6 mark, 0 mark, and 6 mark. An arrow extends from the line at the 0 mark to the line at the negative 6 mark. A second arrow extends from the line at the 0 mark to the line on the 6 mark.

6 is six away from 0 and -6 is six away from 0, so the absolute value of both 6 and -6 is 6.

To denote the absolute value of a quantity, we use a single vertical bar on either side. So |6|=6 and \left|-6\right|=6.

Absolute value equation

An equation that contains a variable expression inside absolute value bars

Example:

\left|x\right|=4

Extraneous solution

A solution of an equation that emerges from the process of solving the problem, but is not a valid solution to the problem

Consider the equation \left|x-2\right|=3. We read this as the distance between some value, x, and 2 is three spaces on a number line.

A number line ranging from negative 5 to 5 with points at negative 1 and 5. Two vertical dashed lines, both labeled x, line up with the negative 1 mark and the 5 mark. Another vertical line that lines up with the 2 mark. An arrow extends from the line at the 2 mark to the dashed line at the negative 1 mark. A second arrow extends from the line at the 2 mark to the dashed line on the 5 mark.

On a number line, the values that are a distance of three from 2 are -1 and 5. The solution to the equation \left|x-2\right|=3 is x=-1 and x=5.

In order to solve an absolute value equation that results in two solutions, we begin by isolating the absolute value expression. Once the expression is isolated, we write and solve two separate equations without the absolute value bars: one equation equal to the distance in a positive direction, and one equation equal to the distance in a negative direction.

Examples

Example 1

Solve each absolute value equation.

a

\left|x\right|=10

Worked Solution
Apply the idea

The solutions to this equation will be any values of x that have an absolute value of 10.

The solutions are x=10 and x=-10.

Reflect and check

Notice that this equation has different solutions from the equation \left|x-10\right|=0 which has only a single solution of x=10. This shows that adding or subtracting terms in or out of the absolute value bars changes the equation, similar to parentheses.

b

\left|2.5x\right|=5

Worked Solution
Create a strategy

Since the absolute value expression is isolated, we create and solve the following two equations: 2.5x=5 \\ 2.5x=-5

Apply the idea

Solve the first equation:

\displaystyle 2.5x\displaystyle =\displaystyle 5First equation
\displaystyle \dfrac{2.5x}{2.5}\displaystyle =\displaystyle \dfrac{5}{2.5}Division property of equality
\displaystyle x\displaystyle =\displaystyle 2Evaluate the division

Solve the second equation:

\displaystyle 2.5x\displaystyle =\displaystyle -5Second equation
\displaystyle \dfrac{2.5x}{2.5}\displaystyle =\displaystyle \dfrac{-5}{2.5}Division property of equality
\displaystyle x\displaystyle =\displaystyle -2Evaluate the division

The solutions to this equation will be any values of x that make 2.5x have an absolute value of 5.

The solutions are x=2 and x=-2.

Reflect and check

Notice that this equation has the same solutions as the equation 2.5\left|x\right|=5. This demonstrates that multiplying or dividing a constant term in or out of the absolute value bars does not change the equation.

c

\left|4x-8\right|=-12

Worked Solution
Apply the idea

The absolute value of an expression can never be a negative number since absolute value is a distance. In this case, there are no solutions.

Reflect and check

We can attempt to solve this equation, but checking our solutions will reveal that both are extraneous.

\displaystyle 4x-8\displaystyle =\displaystyle -12First equation
\displaystyle 4x-8+8\displaystyle =\displaystyle -12+8Addition property of equality
\displaystyle 4x\displaystyle =\displaystyle -4Evaluate the addition
\displaystyle \dfrac{4x}{4}\displaystyle =\displaystyle \dfrac{-4}{4}Division property of equality
\displaystyle x\displaystyle =\displaystyle -1Evaluate the division

When we substitute x=-1 in |4x-8|=-12, we get \left|4(-1)-8\right|=\left|-4-8\right|=\left|-12\right|=12. The solution is not equal to -12, therefore this solution is extraneous.

\displaystyle 4x-8\displaystyle =\displaystyle 12Second equation
\displaystyle 4x-8+8\displaystyle =\displaystyle 12+8Addition property of equality
\displaystyle 4x\displaystyle =\displaystyle 20Evaluate the addition
\displaystyle \dfrac{4x}{4}\displaystyle =\displaystyle \dfrac{20}{4}Division property of equality
\displaystyle x\displaystyle =\displaystyle 5Evaluate the division

When we substitute x=5 in |4x-8|=-12, we get \left|4(5)-8\right|=\left|20-8\right|=\left|12\right|=12. The solution is not equal to -12, so this solution is also extraneous.

d

2\left|x+1\right|+3=21

Worked Solution
Create a strategy

First, we need to perform inverse arithmetic operations to isolate the absolute value expression. From there, we can create two separate equations and solve them.

Apply the idea
\displaystyle 2\left|x+1\right|+3\displaystyle =\displaystyle 21Original equation
\displaystyle 2\left|x+1\right|+3-3\displaystyle =\displaystyle 21-3Subtraction property of equality
\displaystyle 2\left|x+1\right|\displaystyle =\displaystyle 18Evaluate the subtraction
\displaystyle \dfrac{2\left|x+1\right|}{2}\displaystyle =\displaystyle \dfrac{18}{2}Division property of equality
\displaystyle \left|x+1\right|\displaystyle =\displaystyle 9Evaluate the division

Now that the absolute value expression is isolated, we can solve the following two equations: x+1=9 \\ x+1=-9

Solve the first equation:

\displaystyle x+1\displaystyle =\displaystyle 9First equation
\displaystyle x+1-1\displaystyle =\displaystyle 9-1Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle 8Evaluate the subtraction

Solve the second equation:

\displaystyle x+1\displaystyle =\displaystyle -9Second equation
\displaystyle x+1-1\displaystyle =\displaystyle -9-1Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle -10Evaluate the subtraction

The solutions are x=8 and -10.

Reflect and check

We can check that the solutions are viable solutions to the absolute value equation by substituting them into the given equation and verifying that the statement is true.

Checking x=8 \text{: }2\left|8+1\right|+3=2\left|9\right| +3=2\cdot 9+3=18+3=21. Because the original equation was also equal to 21 we know this is a valid solution.

Checking x=-10 \text{: }2\left|-10+1\right|+3=2\left|-9\right| +3=2\cdot 9+3=18+3=21. Because the original equation was also equal to 21 we know this is also a valid solution.

Example 2

A machine is used to fill each of several bags with 16 ounces of sugar. After the bags are filled, another machine weighs them. The bag can only weigh 0.3 ounces heavier or lighter than the desired weight, otherwise, the bag is rejected. Write the equation for the heaviest and lightest bag the machine will approve.

Worked Solution
Create a strategy

In general, we can determine the minimum and maximum weight allowances for a bag of sugar based on the problem. The minimum and maximum weight allowances are the solutions to the absolute value equation, so the equation we write should lead to those two solutions.

Apply the idea

A bag of sugar can weigh 16-0.3 or 15.7 ounces at its lightest, and 16+0.3 or 16.3 ounces at its heaviest. On a number line, we can see that the weights are 0.3 from 16:

15\text{ oz}15.5\text{ oz}16\text{ oz}16.5\text{ oz}17\text{ oz}

The absolute value equation that represents the difference between the bag's weight, 16 ounces, and its weight allowances is 0.3 ounces.

Therefore, the equation that represents the minimum and maximum weight allowances for a bag of sugar is |x-16|=0.3.

Reflect and check

Using a number line to identify the solutions and distance from a particular value on the number line for a problem in context helps us visualize the problem.

Idea summary

An absolute value equation is an equation where at least one expression contains an absolute value. Consider the absolute value equation

\left|ax+b\right| = k

When k\gt0, an absolute value equation has two solutions. When k=0, an absolute value equation has one solution. When k\lt0, an absolute value equation has no solutions.

Viable solutions for an absolute value equation also depend on context. If a solution is mathematically valid but does not make sense in the context then we say it is non-viable.

Outcomes

A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems.

A.REI.B.3

Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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