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Algebra 1 : Absolute Value Inequalities

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Example Questions

Example Question #1 : Absolute Value Inequalities

Solve for :

| 8y + 2 | < -38

Possible Answers:

$\left ( - 4 \frac{1}{2}, 5 \right )$

$\left ( -5, 4 \frac{1}{2} \right )$

The inequality has no solution.

$\left ( -5, -4 \frac{1}{2} \right )$

$\left ( 4 \frac{1}{2}, 5 \right )$

Correct answer:

The inequality has no solution.

Explanation:

The absolute value of a number must always be nonnegative, so | 8y + 2 | can never be less than . This means the inequality has no solution.

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Example Question #2 : Absolute Value Inequalities

Solve the inequality $\left | 4x-1\right |+7\leq14$ .

Possible Answers:

$x\geq-3$

$-6\leq x\leq 0.5$

$-1.5\leq x\leq2$

$-2\leq x\leq 1.5$

$x\leq1$

Correct answer:

$-1.5\leq x\leq2$

Explanation:

First, we can simplify this inequality by subtracting 7 from both sides. This gives us

$\left | 4x-1\right |\leq7$

Next, however, we need to make two separate inequalities due to the presence of an absolute value expression. What this inequality actually means is that

$4x-1\leq7$

and

$4x-1\geq -7$

(Be careful with the inequality signs here! The second sign must be switched to allow for the effect of absolute value on negative numbers. In other words, the inequality must be greater than because, after the absolute value is applied, it will be less than 7.) When we solve the two inequalities, we get two solutions:

$x\leq2$

and

$x\geq-1.5$

For the original statement to be true, both of these inequalities must be fulfilled. We're left with a final answer of

$-1.5\leq x\leq2$

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Example Question #1 : Solving Absolute Value Equations

Solve the inequality:

$\small \left |15x - 39\right | < -42$

Possible Answers:

$\small \left ( -\infty, \infty \right )$

$\small \left ( -\infty , -\frac{1}{5}\right ) \cup \left (\frac{27}{5} , \infty \right )$

$\small \varnothing$ (no solution)

$\small \left(-\frac{1}{5}, \frac{61}{5} \right )$

Correct answer:

$\small \varnothing$ (no solution)

Explanation:

The inequality compares an absolute value function with a negative integer. Since the absolute value of any real number is greater than or equal to 0, it can never be less than a negative number. Therefore, $\small \left |15x -39 \right| < -42$ can never happen. There is no solution.

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Example Question #1 : Solving Inequalities

Solve this inequality.

$\left | \frac{4x+3}{x+6}\right |\eq>1$

Possible Answers:

$x<-\frac{9}{5}, x\eq<1$

$x\eq<\frac{9}{5}, x\eq>-1$

$x\eq>-\frac{9}{5}, x \eq<1$

$x\eq>-\frac{9}{5}, x\eq>1$

$x \eq< -\frac{9}{5}, \hspace x \eq>1$

Correct answer:

$x \eq< -\frac{9}{5}, \hspace x \eq>1$

Explanation:

Split the inequality into two possible cases as follows, based on the absolute values.

First case: $\frac{4x+3}{x+6}\eq>1$

Second case: $\frac{4x+3}{x+6} \eq< -1$

Let's find the inequality of the first case.

$\frac{4x+3}{x+6} \eq>1$

Multiply both sides by x + 6.

$4x+3 \eq>x+6$

Subtract x from both sides, then subtract 3 from both sides.

$3x \eq>3$

Divide both sides by 3.

$x\eq>1$

Let's find the inequality of the second case.

$\frac{4x+3}{x+6} \eq< -1$

Multiply both sides by x + 6.

$4x+3 \eq< -1(x+6)$

Simplify.

4x+3< -x-6

Add x to both sides, then subtract 3 from both sides.

$5x\eq< -9$

Divide both sides by 5.

$x \eq<-\frac{9}{5}$

So the range of x-values is $x<-\frac{9}{5}$ and $x \eq>1$ .

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Example Question #4 : Solving Absolute Value Equations

Solve for :

$\left |2x+3 \right |>7$

Possible Answers:

x>3.5, x<-5

x<2, x>-5

x>2

x>2, x<-10

x>2, x<-5

Correct answer:

x>2, x<-5

Explanation:

Solve for positive values by ignoring the absolute value. Solve for negative values by switching the inequality and adding a negative sign to 7.

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Example Question #4792 : Algebra Ii

Give the solution set for the following equation:

| 4x - 7 | + 5 = 86

Possible Answers:

$\begin{Bmatrix} 22, 24.5 \end{Bmatrix}$

$\begin{Bmatrix} -24.5, 22 \end{Bmatrix}$

$\begin{Bmatrix} -18.5, 22 \end{Bmatrix}$

$\begin{Bmatrix} 22 \end{Bmatrix}$

$\begin{Bmatrix} 18.5, 22 \end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} -18.5, 22 \end{Bmatrix}$

Explanation:

First, subtract 5 from both sides to get the absolute value expression alone.

| 4x - 7 | + 5 = 86

| 4x - 7 | = 81

Split this into two linear equations:

4x - 7 = 81

4x = 88

x = 22

4x - 7 = -81

4x = -74

$x = -74 \div 4 = -18.5$

The solution set is $\begin{Bmatrix} -18.5, 22 \end{Bmatrix}$

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Example Question #2 : Solving Absolute Value Equations

Solve for in the inequality below.

$|x-2| \leq 4$

Possible Answers:

$x\leq 4$

$-2 \leq x\leq 6$

All real numbers

No solutions

$x\leq -2\; \; or\; \; x\geq 6$

Correct answer:

$-2 \leq x\leq 6$

Explanation:

The absolute value gives two problems to solve. Remember to switch the "less than" to "greater than" when comparing the negative term.

$x-2\leq 4$ or $x-2\geq -4$

Solve each inequality separately by adding to all sides.

$x\leq 6$ or $x\geq -2$

This can be simplified to the format $-2 \leq x\leq 6$ .

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Example Question #6 : Solving Absolute Value Equations

$\textup{If }\left | 2x+4\right |\leq10\textup{ , what are all possible values of }x\textup{?}$

Possible Answers:

$x\leq-7\textup{ and }x\geq3$

x=-7,x=3

$-3\leq x\leq3$

$-7\leq x\leq3$

$-3\leq x\leq7$

Correct answer:

$-7\leq x\leq3$

Explanation:

$\textup{Set up two inequalities. Reverse the inequality sign for the negative one.}$

$2x+4\leq10\;\;\;\;\;\;\;\;\;\;\;\;\;2x+4\geq-10$

$2x\leq6\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \; 2x\geq-14$

$x\leq3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x\geq-7$

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Example Question #3 : Absolute Value Inequalities

Solve the inequality.

$\small \left | x+2 \right |>5$

Possible Answers:

$\small x<-3\ or\ x>7$

$\small x<-7\ or\ x>3$

$\small -3<x<7$

$\small -7<x<3$

Correct answer:

$\small x<-7\ or\ x>3$

Explanation:

$\small \left | x+2 \right |>5$

Remove the absolute value by setting the term equal to either $\small 5$ or $\small -5$ . Remember to flip the inequality for the negative term!

$\small x+2>5\ or\ x+2<-5$

Solve each scenario independently by subtracting $\small 2$ from both sides.

$\small x+2-2>5-2\ or\ x+2-2<-5-2$

$\small x>3\ or\ x<-7$

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Example Question #4 : Absolute Value Inequalities

Solve for :

| 4y + 7 | > -5

Possible Answers:

$(-\infty , -3) \cup (3, \infty )$

$\left(-\infty , -\frac{1}{2}\right) \cup \left (3, \infty \right)$

$\left(-\infty , -\frac{1}{2}\right) \cup \left ( \frac{1}{2}, \infty \right)$

$(-\infty , \infty )$

$(-\infty , -3) \cup \left ( \frac{1}{2}, \infty \right)$

Correct answer:

$(-\infty , \infty )$

Explanation:

The absolute value of any number is nonnegative, so | 4y + 7 | must always be greater than . Therefore, any value of makes this a true statement.

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Algebra 1 : Absolute Value Inequalities

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : Absolute Value Inequalities

Example Question #2 : Absolute Value Inequalities

Example Question #1 : Solving Absolute Value Equations

Example Question #1 : Solving Inequalities

Example Question #4 : Solving Absolute Value Equations

Example Question #4792 : Algebra Ii

Example Question #2 : Solving Absolute Value Equations

Example Question #6 : Solving Absolute Value Equations

Example Question #3 : Absolute Value Inequalities

Example Question #4 : Absolute Value Inequalities

All Algebra 1 Resources

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