2.5 Solve Inequalities

Lynn Marecek; MaryAnne Anthony-Smith

2. Solving Linear Equations and Inequalities

2.5 Solve Inequalities

Lynn Marecek and MaryAnne Anthony-Smith

Learning Objectives

By the end of this section, you will be able to:

Graph inequalities on the number line
Solve inequalities using the Subtraction and Addition Properties of inequality
Solve inequalities using the Division and Multiplication Properties of inequality
Solve inequalities that require simplification
Translate to an inequality and solve

Graph Inequalities on the Number Line

Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

What about the solution of an inequality? What number would make the inequality $x$ > $3$ true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality $x$ > $3$ .

We show the solutions to the inequality $x$ > $3$ on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of $x$ > $3$ is shown in (Figure). Please note that the following convention is used: light blue arrows point in the positive direction and dark blue arrows point in the negative direction.

The inequality $x$ > $3$ is graphed on this number line.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 3 is graphed on the number line, with an open parenthesis at x equals 3, and a red line extending to the right of the parenthesis.

The graph of the inequality $x\ge 3$ is very much like the graph of $x$ > $3$ , but now we need to show that 3 is a solution, too. We do that by putting a bracket at $x=3$ , as shown in (Figure).

The inequality $x\ge 3$ is graphed on this number line.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than or equal to 3 is graphed on the number line, with an open bracket at x equals 3, and a red line extending to the right of the bracket.

Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.

EXAMPLE 1

Graph on the number line:

a) $x\le 1$ b) $x$ < $5$ c) $x$ > $-1$

Solution

a) $x\le 1$
This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at $x=1$ to show that it is included.
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to 1 is graphed on the number line, with an open bracket at x equals 1, and a red line extending to the left of the bracket.

b) $x$ < $5$
This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put a parenthesis at $x=5$ to show it is not included.
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than 5 is graphed on the number line, with an open parenthesis at x equals 5, and a red line extending to the right of the parenthesis.

c) $x$ > $-1$
This means all numbers greater than $-1$ , but not including $-1$ . We shade in all the numbers on the number line to the right of $-1$ , then put a parenthesis at $x=-1$ to show it is not included.
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than negative 1 is graphed on the number line, with an open parenthesis at x equals negative 1, and a red line extending to the right of the parenthesis.

TRY IT 1

Graph on the number line: a) $x\le -1$ b) $x$ > $2$ c) $x$ < $3$

Show answer

a)
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to negative 1 is graphed on the number line, with an open bracket at x equals negative 1, and a dark line extending to the left of the bracket.

b)
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 2 is graphed on the number line, with an open parenthesis at x equals 2, and a dark line extending to the right of the parenthesis.

c)
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than 3 is graphed on the number line, with an open parenthesis at x equals 3, and a dark line extending to the left of the parenthesis.

We can also represent inequalities using interval notation. As we saw above, the inequality $x$ > $3$ means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation, we express $x$ > $3$ as $\left(3,\infty \right).$ The symbol $\infty$ is read as ‘infinity’. It is not an actual number. (Figure) shows both the number line and the interval notation.

The inequality $x$ > $3$ is graphed on this number line and written in interval notation.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 3 is graphed on the number line, with an open parenthesis at x equals 3, and a red line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, 3 comma infinity, parenthesis.

The inequality $x\le 1$ means all numbers less than or equal to 1. There is no lower end to those numbers. We write $x\le 1$ in interval notation as $\left(-\infty ,1\right]$ . The symbol $-\infty$ is read as ‘negative infinity’. (Figure) shows both the number line and interval notation.

The inequality $x\le 1$ is graphed on this number line and written in interval notation.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to 1 is graphed on the number line, with an open bracket at x equals 1, and a red line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 1, bracket.

Inequalities, Number Lines, and Interval Notation

Did you notice how the parenthesis or bracket in the interval notation matches the symbol at the endpoint of the arrow? These relationships are shown in (Figure).

The notation for inequalities on a number line and in interval notation use similar symbols to express the endpoints of intervals.

EXAMPLE 2

Graph on the number line and write in interval notation.

a) $x\ge -3$ b) $x$ < $2.5$ c) $x\le -\frac{3}{5}$

Solution

a)


Shade to the right of $-3$ , and put a bracket at $-3$ .
Write in interval notation.

b)


Shade to the left of $2.5$ , and put a parenthesis at $2.5$ .
Write in interval notation.

c)


Shade to the left of $-\frac{3}{5}$ , and put a bracket at $-\frac{3}{5}$ .
Write in interval notation.

TRY IT 2

Graph on the number line and write in interval notation:

a) $x$ > $2$ b) $x\le -1.5$ c) $x\ge \frac{3}{4}$

Show answer

a)
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 2 is graphed on the number line, with an open parenthesis at x equals 2, and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, 2 comma infinity, parenthesis.

b)
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to negative 1.5 is graphed on the number line, with an open bracket at x equals negative 1.5, and a dark line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma negative 1.5, bracket.

c)
This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than or equal to 3/4 is graphed on the number line, with an open bracket at x equals 3/4, and a dark line extending to the right of the bracket. The inequality is also written in interval notation as bracket, 3/4 comma infinity, parenthesis.

Solve Inequalities using the Subtraction and Addition Properties of Inequality

The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.

Properties of Equality

$\begin{array}{cccc}\mathbf{\text{Subtraction Property of Equality}}\hfill & & & \mathbf{\text{Addition Property of Equality}}\hfill \\ \text{For any numbers}\phantom{\rule{0.2em}{0ex}}a,b,\text{and}\phantom{\rule{0.2em}{0ex}}c,\hfill & & & \text{For any numbers}\phantom{\rule{0.2em}{0ex}}a,b,\text{and}\phantom{\rule{0.2em}{0ex}}c,\hfill \\ \begin{array}{cccc}\text{if}\hfill & \hfill a& =\hfill & b,\hfill \\ \text{then}\hfill & \hfill a-c& =\hfill & b-c.\hfill \end{array}\hfill & & & \begin{array}{cccc}\text{if}\hfill & \hfill a& =\hfill & b,\hfill \\ \text{then}\hfill & \hfill a+c& =\hfill & b+c.\hfill \end{array}\hfill \end{array}$

Similar properties hold true for inequalities.

For example, we know that −4 is less than 2.
If we subtract 5 from both quantities, is the left side still less than the right side?
We get −9 on the left and −3 on the right.
And we know −9 is less than −3.
	The inequality sign stayed the same.

Similarly we could show that the inequality also stays the same for addition.

This leads us to the Subtraction and Addition Properties of Inequality.

Properties of Inequality

We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality $x+5$ > $9$ , the steps would look like this:

	$x+5$ > $9$
Subtract 5 from both sides to isolate $x$ .	$x+5-5$ > $9-5$
Simplify.	$x$ > $4$

Any number greater than 4 is a solution to this inequality.

EXAMPLE 3

Solve the inequality $n-\frac{1}{2}\le \frac{5}{8}$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Add $\frac{1}{2}$ to both sides of the inequality.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

TRY IT 3

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

$p-\frac{3}{4}\ge \frac{1}{6}$

Show answer

This figure shows the inequality p is greater than or equal to 11/12. Below this inequality is the inequality graphed on a number line ranging from 0 to 4, with tick marks at each integer. There is a bracket at p equals 11/12, and a dark line extends to the right from 11/12. Below the number line is the solution written in interval notation: bracket, 11/12 comma infinity, parenthesis.

Solve Inequalities using the Division and Multiplication Properties of Inequality

The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).

Properties of Equality

$\begin{array}{cccc}\mathbf{\text{Division Property of Equality}}\hfill & & & \mathbf{\text{Multiplication Property of Equality}}\hfill \\ \text{For any numbers}\phantom{\rule{0.2em}{0ex}}a,b,c,\text{and}\phantom{\rule{0.2em}{0ex}}c\ne 0,\hfill & & & \text{For any real numbers}\phantom{\rule{0.2em}{0ex}}a,b,c,\hfill \\ \phantom{\rule{1em}{0ex}}\begin{array}{cccc}\text{if}\hfill & a\hfill & =\hfill & b,\hfill \\ \text{then}\hfill & \frac{a}{c}\hfill & =\hfill & \frac{b}{c}.\hfill \end{array}\hfill & & & \phantom{\rule{1em}{0ex}}\begin{array}{cccc}\text{if}\hfill & a\hfill & =\hfill & b,\hfill \\ \text{then}\hfill & ac\hfill & =\hfill & bc.\hfill \end{array}\hfill \end{array}$

Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?

Consider some numerical examples.


Divide both sides by 5.		Multiply both sides by 5.
Simplify.
Fill in the inequality signs.

$\mathbf{\text{The inequality signs stayed the same.}}$

Does the inequality stay the same when we divide or multiply by a negative number?


Divide both sides by −5.		Multiply both sides by −5.
Simplify.
Fill in the inequality signs.

$\mathbf{\text{The inequality signs reversed their direction.}}$

When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.

Here are the Division and Multiplication Properties of Inequality for easy reference.

Division and Multiplication Properties of Inequality

When we divide or multiply an inequality by a:

positive number, the inequality stays the same.
negative number, the inequality reverses.

EXAMPLE 4

Solve the inequality $7y$ < $42$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Divide both sides of the inequality by 7. Since $7$ > $0$ , the inequality stays the same.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

TRY IT 4

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

$\left(8,\infty \right)$

Show answer

$c$ > $8$

This figure is a number line ranging from 6 to 10 with tick marks for each integer. The inequality c is greater than 8 is graphed on the number line, with an open parenthesis at c equals 8, and a dark line extending to the right of the parenthesis.

EXAMPLE 5

Solve the inequality $-10a\ge 50$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Divide both sides of the inequality by −10. Since $-10$ < $0$ , the inequality reverses.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

TRY IT 5

Solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$-8q$ < $32$

Show answer

$q$ > $-4$

This figure is a number line ranging from negative 6 to negative 3 with tick marks for each integer. The inequality q is greater than negative 4 is graphed on the number line, with an open parenthesis at q equals negative 4, and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, negative 4 comma infinity, parenthesis.

Solving Inequalities

Sometimes when solving an inequality, the variable ends up on the right. We can rewrite the inequality in reverse to get the variable to the left.

Think about it as “If Xavier is taller than Alex, then Alex is shorter than Xavier.”

EXAMPLE 6

Solve the inequality $-20$ < $\frac{4}{5}u$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides of the inequality by $\frac{5}{4}$ . Since $\frac{5}{4}$ > $0$ , the inequality stays the same.
Simplify.
Rewrite the variable on the left.
Graph the solution on the number line.
Write the solution in interval notation.

TRY IT 6

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

$24\le \frac{3}{8}m$

Show answer

EXAMPLE 7

Solve the inequality $\frac{t}{-2}\ge 8$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides of the inequality by $-2$ . Since $-2$ < $0$ , the inequality reverses.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

TRY IT 7

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

$\frac{k}{-12}\le 15$

Show answer

Solve Inequalities That Require Simplification

Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.

EXAMPLE 8

Solve the inequality $4m\le 9m+17$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Subtract $9m$ from both sides to collect the variables on the left.
Simplify.
Divide both sides of the inequality by −5, and reverse the inequality.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

TRY IT 8

Solve the inequality $3q\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}7q\text{\hspace{0.17em}}-\text{\hspace{0.17em}}23$ , graph the solution on the number line, and write the solution in interval notation.

Show answer

EXAMPLE 9

Solve the inequality $8p+3\left(p-12\right)$ > $7p-28$ , graph the solution on the number line, and write the solution in interval notation.

Solution

Simplify each side as much as possible.	$8p+3\left(p-12\right)$ > $7p-28$
Distribute.	$\phantom{\rule{0.6em}{0ex}}8p+3p-36$ > $7p-28$
Combine like terms.	$\phantom{\rule{2.3em}{0ex}}11p-36$ > $7p-28$
Subtract $7p$ from both sides to collect the variables on the left.	$\phantom{\rule{2.2em}{0ex}}11p-36-7p$ > $7p-28-7p$
Simplify.	$\phantom{\rule{1.2em}{0ex}}4p-36$ > $-28$
Add 36 to both sides to collect the constants on the right.	$\phantom{\rule{1.3em}{0ex}}4p-36+36$ > $-28+36$
Simplify.	$\phantom{\rule{2em}{0ex}}4p$ > $8$
Divide both sides of the inequality by 4; the inequality stays the same.	$\phantom{\rule{2em}{0ex}}\frac{4p}{4}$ > $\frac{8}{4}$
Simplify.	$\phantom{\rule{2.5em}{0ex}}p$ > $2$
Graph the solution on the number line.
Write the solution in interal notation.	$\left(2,\infty \right)$

TRY IT 9

Solve the inequality $9y+2\left(y+6\right)$ > $5y-24$ , graph the solution on the number line, and write the solution in interval notation.

Show answer

Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.

EXAMPLE 10

Solve the inequality $8x-2\left(5-x\right)$ < $4\left(x+9\right)+6x$ , graph the solution on the number line, and write the solution in interval notation.

Solution

Simplify each side as much as possible.	$8x-2\left(5-x\right)$ < $4\left(x+9\right)+6x$
Distribute.	$8x-10+2x$ < $4x+36+6x$
Combine like terms.	$10x-10$ < $10x+36$
Subtract $10x$ from both sides to collect the variables on the left.	$10x-10-10x$ < $10x+36-10x$
Simplify.	$-10$ < $36\phantom{\rule{0.6em}{0ex}}$
The $x$ ’s are gone, and we have a true statement.	The inequality is an identity. The solution is all real numbers.
Graph the solution on the number line.
Write the solution in interval notation.	$\left(-\infty ,\infty \right)$

TRY IT 10

Solve the inequality $4b-3\left(3-b\right)$ > $5\left(b-6\right)+2b$ , graph the solution on the number line, and write the solution in interval notation.

Show answer

This figure shows an inequality that is an identity. Below this inequality is a number line ranging from negative 2 to 2 with tick marks for each integer. The identity is graphed on the number line, with a dark line extending in both directions. The inequality is also written in interval notation as parenthesis, negative infinity comma infinity, parenthesis.

EXAMPLE 11

Solve the inequality $\frac{1}{3}a-\frac{1}{8}a$ > $\frac{5}{24}a+\frac{3}{4}$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides by the LCD, 24, to clear the fractions.
Simplify.
Combine like terms.
Subtract $5a$ from both sides to collect the variables on the left.
Simplify.
The statement is false!	The inequality is a contradiction.
	There is no solution.
Graph the solution on the number line.
Write the solution in interval notation.	There is no solution.

TRY IT 11

Solve the inequality $\frac{1}{4}x-\frac{1}{12}x$ > $\frac{1}{6}x+\frac{7}{8}$ , graph the solution on the number line, and write the solution in interval notation.

Show answer

This figure shows an inequality that is a contradiction. Below this is a number line ranging from negative 2 to 2 with tick marks for each integer. No inequality is graphed on the number line. Below the number line is the statement: “No solution.”

Translate to an Inequality and Solve

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.

Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.

(Figure) shows some common phrases that indicate inequalities.

>

$\ge$

<

$\le$

is greater than

is greater than or equal to

is less than

is less than or equal to

is more than

is at least

is smaller than

is at most

is larger than

is no less than

has fewer than

is no more than

exceeds

is the minimum

is lower than

is the maximum

EXAMPLE 12

Translate and solve. Then write the solution in interval notation and graph on the number line.

Twelve times c is no more than 96.

Solution

Translate.
Solve—divide both sides by 12.
Simplify.
Write in interval notation.
Graph on the number line.

TRY IT 12

Translate and solve. Then write the solution in interval notation and graph on the number line.

Twenty times y is at most 100

Show answer

EXAMPLE 13

Translate and solve. Then write the solution in interval notation and graph on the number line.

Thirty less than x is at least 45.

Solution

Translate.
Solve—add 30 to both sides.
Simplify.
Write in interval notation.
Graph on the number line.

TRY IT 13

Translate and solve. Then write the solution in interval notation and graph on the number line.

Nineteen less than p is no less than 47

Show answer

Key Concepts

Subtraction Property of Inequality
For any numbers a, b, and c,
if $a$ < $b$ then $a-c$ < $b-c$ and
if $a$ > $b$ then $a-c$ > $b-c.$
Addition Property of Inequality
For any numbers a, b, and c,
if $a$ < $b$ then $a+c$ < $b+c$ and
if $a$ > $b$ then $a+c$ > $b+c.$
Division and Multiplication Properties of Inequality
For any numbers a, b, and c,
if $a$ < $b$ and $c$ > $0$ , then $\frac{a}{c}$ < $\frac{b}{c}$ and $ac$ > $bc$ .
if $a$ > $b$ and $c$ > $0$ , then $\frac{a}{c}$ > $\frac{b}{c}$ and $ac$ > $bc$ .
if $a$ < $b$ and $c$ < $0$ , then $\frac{a}{c}$ > $\frac{b}{c}$ and $ac$ > $bc$ .
if $a$ > $b$ and $c$ < $0$ , then $\frac{a}{c}$ < $\frac{b}{c}$ and $ac$ < $bc$ .
When we divide or multiply an inequality by a:
- positive number, the inequality stays the same.
- negative number, the inequality reverses.

2.5 Exercise Set

In the following exercises, graph each inequality on the number line.

1. $x$ > $1$
2. $x$ < $-2$
3. $x \ge -3$
1. $x\le 0$
2. $x$ > $-4$
3. $x\ge -1$

In the following exercises, graph each inequality on the number line and write in interval notation.

1. $x$ > $3$
2. $x\le -0.5$
3. $x\ge \frac{1}{3}$
1. $x\le 5$
2. $x\ge -1.5$
3. $x$ < $-\frac{7}{3}$

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$m-45\le 62$
$v+12$ > $3$
$b+\frac{7}{8}\ge \frac{1}{6}$
$g-\frac{11}{12}$ < $-\frac{5}{18}$

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$6y$ < $48$
$9s\ge 81$
$-8v\le 96$
$-7d$ > $105$
$40$ < $\frac{5}{8}k$
$\frac{9}{4}g\le 36$
$\frac{b}{-10}\ge 30$
$-18$ > $\frac{q}{-6}$
$7s$ < $-28$
$\frac{3}{5}x\le -45$

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$5u\le 8u-21$
$9p$ > $14p+18$
$9y+5\left(y+3\right)$ < $4y-35$
$4k-\left(k-2\right)\ge 7k-26$
$6n-12\left(3-n\right)\le 9\left(n-4\right)+9n$
$9u+5\left(2u-5\right)\ge 12\left(u-1\right)+7u$
$\frac{5}{6}a-\frac{1}{4}a$ > $\frac{7}{12}a+\frac{2}{3}$
$12v+3\left(4v-1\right)\le 19\left(v-2\right)+5v$

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$35k\ge -77$
$18q-4\left(10-3q\right)$ < $5\left(6q-8\right)$
$-\frac{21}{8}y\le -\frac{15}{28}$
$d+29$ > $-61$
$\frac{n}{13}\le -6$

In the following exercises, translate and solve .Then write the solution in interval notation and graph on the number line.

Ninety times c is less than 450.
Ten times y is at most $-110$ .
Six more than k exceeds 25.
Twelve less than x is no less than 21.
Negative two times s is lower than 56.
Fifteen less than a is at least $-7$ .
The maximum height, h, of a fighter pilot is 77 inches. Write this as an inequality.

Answers

1. a. b. c.	2. a. b. c.	3. a. b. c.
4. a. b. c.	5.	6.
7.	8.	9.
10.	11.	12.
13.	14.	15.
16.	17.	18.
19.	20.	21.
22.	23.	24.
25.	26.	27.
28.	29.	30.
31.	32.	33.
34.	35.	36.
37.	38. $h\le 77$