**2.5 Solve Linear Inequalities**

Topics covered in this section are:

- Graph inequalities on the number line
- Solve linear inequalities
- Translate words to an inequality and solve
- Solve applications with linear inequalities

**2.5.1 Graph Inequalities on the Number Line**

What number would make the inequality $x>3$ true? Are you thinking, “$x$ could be four”? That’s correct, but $x$ could be $6$, too, or $37$, or even $3.001$. Any number greater than three is a solution to the inequality $x>3$.

We show all the solutions to the inequality $x>3$ on the number line by shading in all the numbers to the right of three, to show that all numbers greater than three are solutions. Because the number three itself is not a solution, we put an open parenthesis at three.

We can also represent inequalities using **interval notation**. There is no upper end to the solution to this inequality. In interval notation, we express $x>3$ as ($3$, $∞$). The symbol $∞$ is read as “**infinity**.” It is not an actual number.

Figure 1 shows both the number line and the interval notation.

We use the left parenthesis symbol, (, to show that the endpoint of the inequality is not included. The left bracket symbol, [, shows that the endpoint is included.

The inequality $x≤1$ means all numbers less than or equal to one. Here we need to show that one is a solution, too. We do that by putting a bracket at $x=1$. We then shade in all the numbers to the left of one, to show that all numbers less than one are solutions. See Figure 2.

There is no lower end to those numbers. We write $x≤1$ in interval notation as $(-∞,1]$. The symbol $-∞$ is read as “negative infinity.” Figure 2 shows both the number line and interval notation.

**INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION**

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

**Example 1**

Graph each inequality on the number line and write in interval notation.

- $x≥-3$
- $x<2.5$
- $x≤-\frac{3}{5}$

**Solution**

**Part 1.**

$x≥3$ | |

Shade to the right of $-3$, and put a bracket at $-3$. | |

Write in interval notation. | $[-3,∞)$ |

**Part 2.**

$x<2.5$ | |

Shade to the left of $2.5$, and put a parenthesis at $2.5$. | |

Write in interval notation. | $(-∞,2.5)$ |

**Part 3.**

$x≤-\frac{3}{5}$ | |

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

Write in interval notation. | $(-∞,-\frac{3}{5}]$ |

What numbers are greater than two but less than five? Are you thinking say, $2.5,3,3\frac{2}{3},4,4,99$? We can represent all the numbers between two and five with the inequality $2<x<5$. We can show $2<x<5$ on the number line by shading all the numbers between two and five. Again, we use the parentheses to show the numbers two and five are not included. See Figure 4.

**Example 2**

Graph each inequality on the number line and write in interval notation.

- $-3<x<4$
- $-6≤x<-1$
- $0≤x≤2.5$

**Solution**

**Part 1.**

$-3<x<4$ | |

Shade between $-3$ and $4$. Put parentheses at $-3$ and $4$. | |

Write in interval notation. | $(-3,4)$ |

**Part 2.**

$-6≤x<-1$ | |

Shade between $-6$ and $-1$. Put a bracket at $-6$ and a parenthesis at $-1$. | |

Write in interval notation. | $[-6,-1)$ |

**Part 3.**

$0≤x≤2.5$ | |

Shade between $0$ and $2.5$. Put brackets at $0$ and $2.5$. | |

Write in interval notation. | $[0,2.5]$ |

**2.5.2 Solve Linear Inequalities**

A linear inequality is much like a linear equation—but the equal sign is replaced with an inequality sign. A **linear inequality** is an inequality in one variable that can be written in one of the forms, $ax+b<c$, $ax+b≤c$, $ax+b>c$, or $ax+b≥c$.

**LINEAR INEQUALITY**

A linear inequality is an inequality in one variable that can be written in one of the following forms where $a$, $b$, and $c$ are real numbers and $a≠0$:

$ax+b<c$, | $ax+b≤c$, | $ax+b>c$, | $ax+b≥c$. |

When we solved linear equations, we were able to use the properties of equality to add, subtract, multiply, or divide both sides and still keep the equality. Similar properties hold true for inequalities.

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality. For example:

$-4<2$ | $-4<2$ |

$-4-5<2-5$ | $-4+7<2+7$ |

$-9<-3$ True | $3<9$ True |

Notice that the inequality sign stayed the same.

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

**ADDITION AND SUBTRACTION PROPERTY OF INEQUALITY**

For any number $a$, $b$, and $c$, if $a<b$, then

$a+c<b+c \ \ \ \ a-c<b-c $

For any number $a$, $b$, and $c$, if $a>b$, then

$a+c>b+c \ \ \ \ a-c>b-c $

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality.

What happens to an inequality when we divide or multiply both sides by a constant?

Let’s first multiply and divide both sides by a positive number.

$10<15$ | $10<15$ |

$10(5)<15(5)$ | $\frac{10}{5}<\frac{15}{5}$ |

$50<75$ True | $2<3$ True |

The inequality signs stayed the same.

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

$10<15$ | $10<15$ |

$10(-5) \ ? \ 15(-5)$ | $\frac{10}{5} ? \frac{15}{5}$ |

$-50 \ ? \ -75$ True | $-2 \ ? \ -3$ True |

$-50>-75$ | $-2>-3$ |

Notice that when we filled in the inequality signs, 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.

This gives us the Multiplication and Division Property of Inequality.

**MULTIPLICATION AND DIVISION PROPERTY OF INEQUALITY**

For any numbers $a$, $b$, and $c$,

multiply or divide by a positive |

if $a<b$ and $c>0$, then $ac<bc$ and $\frac{a}{c} < \frac{b}{c}$. |

If $a>b$ and $c>0$, then $ac>bc$ and $\frac{a}{c}> \frac{b}{c}$. |

multiply or divide by a negative |

if $a<b$ and $c<0$, then $ac>bc$ and $\frac{a}{c} > \frac{b}{c}$. |

if $a>b$ and $c<0$, then $ac<bc$ and $\frac{a}{c} < \frac{b}{c}$. |

When we divide or multiply an inequality by $a$:

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

Sometimes when solving an inequality, as in the next example, the variable ends upon the right. We can rewrite the inequality in reverse to get the variable to the left.

$x>a$ has the same meaning as $a<x$.

Think about it as “If Xander is taller than Andy, then Andy is shorter than Xander.”

**Example 3**

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

- $x-\frac{3}{8} ≤ \frac{3}{4}$
- $9y<54$
- $-15< \frac{3}{5}z$

**Solution**

**Part 1**.

$x-\frac{3}{8} ≤ \frac{3}{4}$ | |

Add $\frac{3}{8}$ to both sides of the inequality. | $x- \frac{3}{8} + \frac{3}{8} ≤ \frac{3}{4} + \frac{3}{8}$ |

Simplify. | $x ≤ \frac{9}{8}$ |

Graph the solution on the number line. | |

Write the solution in interval notation. | $(-∞, \frac{9}{8}]$ |

**Part 2.**

$9y < 54$ | |

Divide both sides of the inequality by $9$; since $9$ is positive, the inequality stays the same. | $\frac{9y}{9} < \frac{54}{9}$ |

Simplify. | $y<6$ |

Graph the solution on the number line. | |

Write the solution in interval notation. | $(-∞, 6)$ |

**Part 3.**

$-15 < \frac{3}{5}z$ | |

Divide both sides of the inequality by $9$; since $9$ is positive, the inequality stays the same. | $\frac{5}{3}(-15) < \frac{5}{3}( \frac{3}{5}z)$ |

Simplify. | $-25<z$ |

Rewrite with the variable on the left. | $z>-25$ |

Graph the solution on the number line. | |

Write the solution in interval notation. | $(-25, ∞)$ |

Be careful when you multiply or divide by a negative number—remember to reverse the inequality sign.

**Example 4**

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

- $-13m ≥ 65$
- $\frac{n}{-2} ≥ 8$

**Solution**

**Part 1.**

$-13m ≥ 65$ | |

Divide both sides of the inequality by $-13$. Since $-13$ is a negative, the inequality reverses. | $\frac{-13m}{-13} ≤ \frac{65}{-13}$ |

Simplify. | $m≤-5$ |

Graph the solution on the number line. | |

Write the solution in interval notation. | $(-∞,-5]$ |

**Part 2.**

$\frac{n}{-2} ≥ 8$ | |

Multiply both sides of the inequality by $-2$. Since $-2$ is a negative, the inequality reverses. | $-2 (\frac{n}{-2}) ≤ -2(8)$ |

Simplify. | $n≤-16$ |

Graph the solution on the number line. | |

Write the solution in interval notation. | $(-∞,-16]$ |

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 make sure to pay close attention when we multiply or divide to isolate the variable.

**Example 5**

Solve the inequality $6y≤11y+17$ graph the solution on the number line, and write the solution in interval notation.

**Solution**

$6y≤11y+17$ | |

Subtract $11y$ from both sides to collect the variables on the left. | $6y-11y≤11y-11y+17$ |

Simplify. | $-5y≤17$ |

Divide both sides of the inequality by $-5$, and reverse the inequality. | $\frac{-5y}{-5}≥\frac{17}{-5}$ |

Simplify. | $y≥\frac{17}{-5}$ |

Graph the solution on the number line. | |

Write the solution in interval notation. | $[-\frac{17}{5}, ∞)$ |

When solving inequalities, it is usually easiest to collect the variables on the side where the coefficient of the variable is largest. This eliminates negative coefficients and so we don’t have to multiply or divide by a negative—which means we don’t have to remember to reverse the inequality sign.

**Example 6**

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

**Solution**

$8p+3(p-12) > 7p-28$ | |

Simplify each side as much as possible. | |

Distribute. | $8p+3p-36 > 7p-28$ |

Combine like terms. | $11p-36 > 7p-28$ |

Subtract $7p$ from both sides to collect the variables on the left, since $11>7$. | $11p-7p-36 > 7p-28-7p$ |

Simplify. | $4p-36>-28$ |

Add $36$ to both sides to collect the constants on the right. | $4p-36+36>-28+36$ |

Simplify. | $4p>8$ |

Divide both sides of the inequality by $4$; the inequality stays the same. | $\frac{4p}{4} > \frac{8}{4}$ |

Simplify. | $p>2$ |

Graph the solution on the number line. | |

Write the solution in interval notation. | $(2,∞)$ |

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 7**

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

**Solution**

Simplify each side as much as possible. | $8x-2(5-x) < 4(x+9)+6x$ |

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. | $-10 < 36$ |

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. | $(-∞,∞)$ |

We can clear fractions in inequalities much as we did in equations. Again, be careful with the signs when multiplying or dividing by a negative.

**Example 8**

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**

$\frac{1}{3}a – \frac{1}{8}a > \frac{5}{24}a + \frac{3}{4}$ | |

Multiply both sides by the LCD, $24$, to clear the fractions. | $24(\frac{1}{3}a – \frac{1}{8}a) > 24(\frac{5}{24}a + \frac{3}{4})$ |

Simplify. | $8a-3a > 5a +18$ |

Combine like terms. | $5a > 5a+18$ |

Subtract $5a$ from both sides to collect the variables on the left. | $5a-5a > 5a-5a+18$ |

Simplify. | $0>18$ |

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. |

**2.5.3 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. Table 2.2 shows some common phrases that indicate inequalities.

> | ≥ | < | ≤ |
---|---|---|---|

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 |

Table 2.2 |

**Example 9**

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

Twenty-seven less than $x$ is at least $48$.

**Solution**

Twenty seven less than $x$ is at least $48$. | |

Translate. | $x-27≥48$ |

Solve – add $27$ to both sides. | $x-27+27≥48+27$ |

Simplify. | $x≥75$ |

Graph on the number line. | |

Write in interval notation. | $[75,∞)$ |

**2.5.4 Solve Applications with Linear Inequalities**

Many real-life situations require us to solve inequalities. The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations.

We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality.

Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down.

**Example 10**

Dawn won a mini-grant of $\$4,000$ to buy tablet computers for her classroom. The tablets she would like to buy cost $\$254.12$ each, including tax and delivery. What is the maximum number of tablets Dawn can buy?

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | the maximum number of tablets Dawn can buy |

Step 3. Name what you are looking for. | |

Choose a variable to represent that quantity. | Let $n$= the number of tablets Dawn can buy |

Step 4. Translate. Write a sentence that gives the information to find it. | $\$253.12$ times the number of tablets is no more than $\$4,000$. |

Translate into an inequality. | $254.12n ≤ 4000$ |

Step 5. Solve the inequality.But $n$ must be a whole number of tablets, so round to $15$. | $n≤15.74$ $n≤15$ |

Step 6. Check the answer in the problem and make sure it makes sense. | |

Rounding down the price to $\$250$, $15$ tablets would cost $\$3,750$, while $16$ tablets would be $\$4,000$. So a maximum of $15$ tablets at $\$254.12$ seems reasonable. | |

Step 7. Answer the question with a complete sentence. | Dawn can buy a maximum of $15$ tablets. |

**Example 11**

Taleisha’s phone plan costs her $\$28.80$ a month plus $\$0.20$ per text message. How many text messages can she send/receive and keep her monthly phone bill no more than $\$50$?

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | the number of text messages Taleisha can make |

Step 3. Name what you are looking for. | |

Choose a variable to represent that quantity. | Let $t=$ the number of text messages. |

Step 4. Translate Write a sentence that gives the information to find it. | $\$28.80$ plus $\$0.20$ times the number of text messages is less than or equal to $\$50$. |

Translate into an inequality. | $28.80+0.20t≤50$ |

Step 5. Solve the inequality. | $0.2t≤21.2$ $t≤ 106$ text messages |

Step 6. Check the answer in the problem and make sure it makes sense.Yes, $28.80+0.20(106)=50$. | |

Step 7. Write a sentence that answers the question. | Taleisha can send/receive no more than $106$ text messages to keep her bill no more than $\$50$. |

Profit is the money that remains when the costs have been subtracted from the revenue. In the next example, we will find the number of jobs a small businesswoman needs to do every month in order to make a certain amount of profit.

**Example 12**

Felicity has a calligraphy business. She charges $\$2.50$ per wedding invitation. Her monthly expenses are $\$650$. How many invitations must she write to earn a profit of at least $\$2,800$ per month?

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | the number of invitations Felicity needs to write |

Step 3. Name what you are looking for.Choose a variable to represent it. | Let $j=$ the number of invitations. |

Step 4. Translate. Write a sentence that gives the information to find it. | $\$2.50$ times the number of invitations minus $\$650$ is at least $\$2,800$. |

Translate into an inequality. | $2.50j-650≥2,800$ |

Step 5. Solve the inequality. | $2.5j≥3,450$ $j≤1,380$ invitations |

Step 6. Check the answer in the problem and make sure it makes sense. | |

If Felicity wrote $1400$ invitations, her profit would be $2.50(1400) − 650$, or $\$2,850$. This is more than $\$2800$. | |

Step 7. Write a sentence that answers the question. | Felicity must write at least $1,380$ invitations. |

There are many situations in which several quantities contribute to the total expense. We must make sure to account for all the individual expenses when we solve problems like this.

**Example 13**

Malik is planning a six-day summer vacation trip. He has $\$840$ in savings, and he earns $\$45$ per hour for tutoring. The trip will cost him $\$525$ for airfare, $\$780$ for food and sightseeing, and $\$95$ per night for the hotel. How many hours must he tutor to have enough money to pay for the trip?

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | the number of hours Malik must tutor |

Step 3. Name what you are looking for. | |

Choose a variable to represent that quantity. | Let $h=$ the number of hours. |

Step 4. Translate. Write a sentence that gives the information to find it. | The expenses must be less than or equal to the income. The cost of airfare plus the cost of food and sightseeing and the hotel bill must be less than the savings plus the amount earned tutoring. |

Translate into an inequality. | $525+780+95(6)≤840+45h$ |

Step 5. Solve the inequality. | $\begin{align*} 1,875&≤840+45h \\ 1,035&≤45h \\ 23&≤h \\ h&≥23 \end{align*}$ |

Step 6. Check the answer in the problem and make sure it makes sense.We substitute $23$ into the inequality. $\begin{align*} 1,875&≤840+45h \\ 1,875 &≤840+45(23) \\ 1,875 &≤ 1875 \end{align*}$ | |

Step 7. Write a sentence that answers the question. | Malik must tutor at least $23$ hours. |

*Licenses & Attributions*

*Licenses & Attributions**CC Licensed Content, Original*

*Revision and Adaption. Provided by: Minute Math. License: CC BY 4.0*

*CC Licensed Content, Shared Previously*

*Marecek, L., & Mathis, A. H. (2020). Solve Linear Inequalities. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/2-5-solve-linear-inequalities*.*License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/2-introduction*