2.7 Solve Linear Inequalities

Learning objectives

Learning objectives

• 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 2.7. Please note that the following convention is used: thick arrows point in the positive direction and thin arrows point in the negative direction.

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 2.8.

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 2.66

Solution

1. ⓐ $x\le{}1$ 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.

2. ⓑ $x<5$ 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.

3. ⓒ $x>-1$ 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.

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 $(3,\infty).$ The symbol $\infty$ is read as ‘infinity’. It is not an actual number. Figure 2.9 shows both the number line and the interval notation.

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 $(-\infty,1]$. The symbol $-\infty$ is read as ‘negative infinity’. Figure 2.10 shows both the number line 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 2.11.

Example 2.67

Solution

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

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

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

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.

Similar properties hold true for inequalities.

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

This leads us to the Subtraction and Addition 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:

Table

2.6

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

Example 2.68

Solution

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).

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.

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

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.

When we divide or multiply an inequality by a:

• positive number, the inequality stays the same.

• negative number, the inequality reverses.

Example 2.69

Solution

Example 2.70

Solution

Example 2.71

Solution

Example 2.72

Solution

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 2.73

Solution

Example 2.74

Solution

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 2.75

Solution

Example 2.76

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’.

Table 2.7 shows some common phrases that indicate inequalities.

Table

2.7

Example 2.77

Solution

Example 2.78

Solution

Section 2.7 Exercises

Section 2.7 Exercises

Graph Inequalities on the Number Line

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

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

Solve Inequalities using the Subtraction and Addition Properties of Inequality

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

Solve Inequalities using the Division and Multiplication Properties of Inequality

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

Solve Inequalities That Require Simplification

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

Mixed practice

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

Translate to an Inequality and Solve

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

Everyday Math

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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

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

3. Graph on the number line and write in interval notation:

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

4. Graph on the number line and write in interval notation:

ⓐ $x\le-4$ ⓑ $x\ge{}0.5$ ⓒ $x<-\frac{2}{3}$

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

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

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

$r-\frac{1}{3}\le\frac{7}{12}$

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

$c-8>0$

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

$12d\le{}60$

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

$-8q<32$

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

$-7r\le-70$

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

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

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

$-24<\frac{4}{3}n$

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

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

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

$\frac{u}{-4}\ge-16$

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

1. Solve the inequality $6x<10x+19$, graph the solution on the number line, and write the solution in interval notation.

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

3. Solve the inequality $6u+8(u-1)>10u+32$, graph the solution on the number line, and write the solution in interval notation.

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

5. Solve the inequality $9h-7(2-h)<8(h+11)+8h$, graph the solution on the number line, and write the solution in interval notation.

1. $m-45\le{}62$

2. $v+12>3$

3. $b+\frac{7}{8}\ge\frac{1}{6}$

4. $g-\frac{11}{12}<-\frac{5}{18}$

5. $6y<48$

1. $9s\ge{}81$

2. $-8v\le{}96$

3. $-7d>105$

4. $40<\frac{5}{8}k$

5. $\frac{9}{4}g\le{}36$

1. $\frac{b}{-10}\ge{}30$

2. $-18>\frac{q}{-6}$

3. $7s<-28$

4. $\frac{3}{5}x\le-45$

5. $5u\le{}8u-21$

1. $9p>14p-18$

2. $9y+5(y+3)<4y-35$

3. $4k-(k-2)\ge{}7k-26$

4. $6n-12(3-n)\le{}9(n-4)+9n$

5. $9u+5(2u-5)\ge{}12(u-1)+7u$

1. $\frac{5}{6}a-\frac{1}{4}a>\frac{7}{12}a+\frac{2}{3}$

2. $12v+3(4v-1)\le{}19(v-2)+5v$

3. $35k\ge-77$

4. $18q-4(10-3q)<5(6q-8)$

5. $-\frac{21}{8}y\le-\frac{15}{28}$

1. $d+29>-61$

2. $\frac{n}{13}\le-6$

3. Ninety times c is less than 450.

4. Ten times y is at most $-110$.

5. Six more than k exceeds 25.

1. Twelve less than x is no less than 21.

2. Negative two times s is lower than 56.

3. Fifteen less than a is at least $-7$.

4. Fighter pilots The maximum height, h, of a fighter pilot is 77 inches. Write this as an inequality.

5. Shopping The number of items, n, a shopper can have in the express check-out lane is at most 8. Write this as an inequality.

1. 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.

2. Solve the inequality $\frac{2}{5}z-\frac{1}{3}z<\frac{1}{15}z+\frac{3}{5}$, graph the solution on the number line, and write the solution in interval notation.

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

Twenty times y is at most 100

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

Nine times z is no less than 135

5. 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

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

Four more than a is at most 15.

2. ⓐ $x>1$ⓑ $x<-2$ⓒ $x\ge-3$

3. ⓐ $x\le{}0$ⓑ $x>-4$ⓒ $x\ge-1$

4. ⓐ $x>3$ⓑ $x\le-0.5$ⓒ $x\ge\frac{1}{3}$

5. ⓐ $x\le{}5$ⓑ $x\ge-1.5$ⓒ $x<-\frac{7}{3}$

1. Give an example from your life using the phrase ‘at most’.

2. Explain why it is necessary to reverse the inequality when solving $\frac{n}{-3}<12$.