Solve Absolute Value Inequalities

Solve Absolute Value Equations

As we prepare to solve absolute value equations, we review our definition of absolute value.

We learned that both a number and its opposite are the same distance from zero on the number line. Since they have the same distance from zero, they have the same absolute value. For example:

The numbers $5$ and $-5$ are both five units away from zero.

For the equation $|x|=5$, we are looking for all numbers that make this a true statement. We are looking for the numbers whose distance from zero is 5. We just saw that both 5 and $-5$ are five units from zero on the number line. They are the solutions to the equation.

The solution can be simplified to a single statement by writing $x=\pm5$. This is read, “$x$ is equal to positive or negative 5”. We can generalize this to the following property for absolute value equations.

Example 2.68. Solve: (a) $|x|=8$ (b) $|y|=-6$ (c) $|z|=0$.

(a) Write the equivalent equations.

Thus, $x=\pm8$.

(b) Since an absolute value is always positive, there are no solutions to this equation.

(c) Write the equivalent equations: $z=-0$ or $z=0$. Since $-0=0$, $z=0$. Both equations tell us that $z=0$ and so there is only one solution.

To solve an absolute value equation, we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations.

Example 2.69. How to Solve Absolute Value Equations. Solve $|5x-4|-3=8$.

Check $x=3$:

Check $x=-\tfrac{7}{5}$:

Example 2.70. Solve $2|x-7|+5=9$.

Check:

Remember, an absolute value is always positive!

Example 2.71. Solve $\left|\tfrac{2}{3}x-4\right|+11=3$.

Some of our absolute value equations could be of the form $|u|=|v|$ where $u$ and $v$ are algebraic expressions. For example, $|x-3|=|2x+1|$.

How would we solve them? If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The property for absolute value equations says that for any algebraic expression, $u$, and a positive real number, $a$, if $|u|=a$, then $u=-a$ or $u=a$.

This tells us that if $|u|=|v|$, then $u=-v$ or $u=v$.

When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed.

Example 2.72. Solve $|5x-1|=|2x+3|$.

Check. We leave the check to you.

Solve Absolute Value Inequalities with “Less Than”

Let’s look now at what happens when we have an absolute value inequality. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work.

Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation $|x|=5$, we saw that both 5 and $-5$ are five units from zero on the number line. They are the solutions to the equation.

What about the inequality $|x|\leq5$? Where are the numbers whose distance is less than or equal to 5? We know $-5$ and 5 are both five units from zero. All the numbers between $-5$ and 5 are less than five units from zero.

On the number line, the solution is the segment from $-5$ through 5, with a closed bracket at each endpoint. This shows $-5\leq x\leq5$.

In a more general way, we can see that if $|u|\leq a$, then $-a\leq u\leq a$.

After solving an inequality, it is often helpful to check some points to see if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete check, it often helps verify the solution.

Example 2.73. Solve $|x|<7$. Graph the solution and write the solution in interval notation.

Write the equivalent inequality: $-7<x<7$.

On the number line, shade the segment between $-7$ and 7 and place an open parenthesis at each endpoint.

The solution in interval notation is $(-7,7)$.

Check: To verify, check a value in each section of the number line showing the solution. Choose numbers such as $-8$, 1, and 9.

Example 2.74. Solve $|5x-6|\leq4$. Graph the solution and write the solution in interval notation.

On the number line, shade the segment from $\tfrac{2}{5}$ through 2 and place a closed bracket at each endpoint.

The solution using interval notation is $\left[\tfrac{2}{5},2\right]$. Check: The check is left to you.

Solve Absolute Value Inequalities with “Greater Than”

What happens for absolute value inequalities that have “greater than”? Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line.

We started with the inequality $|x|\leq5$. 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.

Now we want to look at the inequality $|x|\geq5$. Where are the numbers whose distance from zero is greater than or equal to five?

Again both $-5$ and 5 are five units from zero and so are included in the solution. Numbers whose distance from zero is greater than five units would be less than $-5$ and greater than 5 on the number line.

On the number line, place closed brackets at $-5$ and 5. Shade to the left of $-5$ and to the right of 5. This shows $x\leq-5$ or $x\geq5$.

In a more general way, we can see that if $|u|\geq a$, then $u\leq-a$ or $u\geq a$.

Example 2.75. Solve $|x|>4$. Graph the solution and write the solution in interval notation.

Write the equivalent inequality: $x<-4$ or $x>4$.

On the number line, place open parentheses at $-4$ and 4. Shade to the left of $-4$ and to the right of 4.

The solution using interval notation is $(-\infty,-4)\cup(4,\infty)$.

Check: To verify, check a value in each section of the number line showing the solution. Choose numbers such as $-6$, 0, and 7. Then $|-6|>4$ is true, $|0|>4$ is false, and $|7|>4$ is true.

Example 2.76. Solve $|2x-3|\geq5$. Graph the solution and write the solution in interval notation.

On the number line, place closed brackets at $-1$ and 4. Shade to the left of $-1$ and to the right of 4.

The solution using interval notation is $(-\infty,-1]\cup[4,\infty)$. Check: The check is left to you.

Solve Applications with Absolute Value

Absolute value inequalities are often used in the manufacturing process. An item must be made with near perfect specifications. Usually there is a certain tolerance of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected.

Example 2.77. The ideal diameter of a rod needed for a machine is 60 mm. The actual diameter can vary from the ideal diameter by 0.075 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Let $x=$ the actual measurement.

The diameter of the rod can be between 59.925 mm and 60.075 mm.

Key terms. Absolute value is the distance of a number from zero on the number line. Tolerance is the allowed difference from a specification.

Adapted from Intermediate Algebra 2e, Section 2.7 by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at OpenStax. Changes: adapted the source text and examples for web presentation and converted selected Try It exercises into interactive checks.