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# 6.4: Compound Inequalities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write and graph compound inequalities on a number line.
• Solve a compound inequality with “and”.
• Solve a compound inequality with “or”.
• Solve compound inequalities using a graphing calculator (TI family).
• Solve real-world problems using compound inequalities

## Introduction

In this section, we will solve compound inequalities. In previous sections, we obtained solutions that gave the variable either as greater than or as less than a number. In this section we are looking for solutions where the variable can be in two or more intervals on the number line.

There are two types of compound inequalities:

1. Inequalities joined by the word “and”.

The solution is a set of values greater than a number and less than another number.

$a

In this case we want values of the variable for which both inequalities are true.

2. Inequalities joined by the word “or”.

The solution is a set of values greater than a number or less than another number.

$x or $x>b$

In this case, we want values for the variable in which at least one of the inequalities is true.

## Write and Graph Compound Inequalities on a Number Line

Example 1

Write the inequalities represented by the following number line graphs.

a)

b)

c)

d)

Solution

a) The solution graph shows that the solution is any value between -40 and 60, including -40 but not 60. Any value in the solution set satisfies both inequalities.

$x \geq -40$ and $x<60$

This is usually written as the following compound inequality.

$-40 \leq x <60$

b) The solution graph shows that the solution is any value greater than 1 (not including 1) or any value less than -2 (not including -2). You can see that there can be no values that can satisfy both these conditions at the same time. We write:

$x >1$ or $x <-2$

c) The solution graph shows that the solution is any value greater than 4 (including 4) or any value less than -1 (including -1). We write:

$x \geq 4$ or $x \leq -1$

d) The solution graph shows that the solution is any value less than 25 (not including 25) and any value greater than -25 (not including -25). Any value in the solution set satisfies both conditions.

$x>-25$ and $x <25$

This is usually written as $-25.

Example 2

Graph the following compound inequalities on the number line.

a) $-4 \leq x \leq 6$

b) $x <0$ or $x>2$

c) $x \geq -8$ or $x \leq -20$

d) $-15

Solution

a) The solution is all numbers between -4 and 6 including both -4 and 6.

b) The solution is either numbers less than 0 or numbers greater than 2 not including 0 or 2.

c) The solution is either numbers greater than or equal to -8 or less than or equal to -20.

d) The solution is numbers between -15 and 85, not including -15 but including 85.

## Solve a compound Inequality With “and”

When we solve compound inequalities, we separate the inequalities and solve each of them separately. Then, we combine the solutions at the end.

Example 3

Solve the following compound inequalities and graph the solution set.

a) $-2<4x-5\leq 11$

b) $3x-5

Solution

a) First, we rewrite the compound inequality as two separate inequalities with and. Then solve each inequality separately.

$-2 & <4x-5 & & & & 4x-5 \leq 11\\3 & < 4x & & \text{and} & & 4x \leq 16\\\frac{3}{4}&

Answer $\frac{3} {4} and $x\leq 4$. This can be written as $\frac{3} {4}x\leq 4$.

b) Rewrite the compound inequality as two separate inequalities by using and. Then solve each inequality separately.

$3x-5&

Answer $x<7>$ and $x\geq -1$. This can be written as $-1 \leq x <7$.

## Solve a Compound Inequality With “or”

Consider the following example.

Example 4

Solve the following compound inequalities and graph the solution set.

a) $9-2x \leq 3$ or $3x+10 \leq 6-x$

b) $\frac{x-2} {6}\leq2x-4$ or $\frac{x-2} {6}>x+5$

Solution

a) Solve each inequality separately.

$9-2x & \leq 3 & & & & 3x+10 \leq 6-x\\-2x & \leq -6 & & \text{or} & & 4x \leq -4\\x & \geq 3 & & & & x \leq -1$

Answer $x \geq 3$ or $x \leq -1$

b) Solve each inequality separately.

$\frac{x-2} {6} & \leq 2x-4 & & & & \frac{x-2} {6} > x+5\\x-2 & \leq 6(2x-4) & & & & x-2 > 6(x+5)\\x-2 & \leq 12x-24 & & \text{or} & & x-2 > 6x+30\\22 & \leq 11x & & & & -32>5x\\2 & \leq x & & & & -6.4>x$

Answer $x \geq 2$ or $x <-64$

## Solve Compound Inequalities Using a Graphing Calculator (TI-83/84 family)

This section explains how to solve simple and compound inequalities with a graphing calculator.

Example 5

Solve the following inequalities using the graphing calculator.

a) $5x+2(x- 3) \geq 2$

b) $7x-2<10x+1<9x+5$

c) $3x+2 \leq 10$ or $3x+2 \geq 15$

Solution

a) $5x+2(x-3) \geq 2$

Step 1 Enter the inequality.

Press the [Y=] button.

Enter the inequality on the first line of the screen.

$Y_1 = 5x+2(x-3) \geq 2$

The $\geq$ symbol is entered by pressing [TEST] [2nd] [MATH] and choose option 4.

Press the [GRAPH] button.

Because the calculator translates a true statement with the number 1 and a false statement with the number 0, you will see a step function with the $y-$value jumping from 0 to 1. The solution set is the values of $x$ for which the graph shows $y = 1$.

Note: You need to press the [WINDOW] key or the [ZOOM] key to adjust window to see full graph.

The solution is $x \geq \frac{8}{7}=1.42857\ldots$, which is why you can see the $y$ value changing from 0 to 1 at 1.14.

b) $7x-2 <10x+1<9x+5$

This is a compound inequality $7x- 2 <10x + 1$ and $10x + 1 < 9x + 5$.

To enter a compound inequality:

Press the [Y=] button.

Enter the inequality as $Y_1 = (7x-2<10x+1)$ AND $(10x+1<9x+5)$

To enter the [AND] symbol press [TEST], choose [LOGIC] on the top row and choose option 1.

The resulting graph looks as shown at the right.

The solution are the values of $x$ for which $y=1$.

In this case $-1.

c) $3x+2 \leq 10$ or $3x+2 \geq 15$

This is a compound inequality $3x + 2 \leq 10$ or $3x + 2 \geq 15$

Press the [Y=] button.

Enter the inequality as $Y_1=(3x+2 \leq 10)$ OR $(3x+2 \geq 15)$ To enter the [OR] symbol press [TEST], choose [LOGIC] on the top row and choose option 2.

The resulting graph looks as shown at the right. The solution are the values of $x$ for which $y=1$. In this case, $x \leq 2.7$ or $x \geq 4.3$.

## Solve Real-World Problems Using Compound Inequalities

Many application problems require the use of compound inequalities to find the solution.

Example 6

The speed of a golf ball in the air is given by the formula $v =-32t+ 80$, where $t$ is the time since the ball was hit. When is the ball traveling between 20 ft/sec and 30 ft/sec?

Solution

Step 1

We want to find the times when the ball is traveling between 20 ft/sec and 30 ft/sec.

Step 2

Set up the inequality $20 \leq v \leq 30$

Step 3

Replace the velocity with the formula $v=-32t+80$.

$20 \leq -32t +80 \leq 30$

Separate the compound inequality and solve each separate inequality.

$&20 \leq -32t+80 & & & & -32t+80 \leq 30\\&32t \leq 60 & & \text{and} & & 50 \leq 32t\\&t \leq 1.875 & & & & 1.56 \leq t$

Answer $1.56 \leq t \leq 1.875$

Step 4 To check plug in the minimum and maximum values of $t$ into the formula for the speed.

For $t =1.56, v=-32t+80=-32(1.56)+80=30 \ ft/sec$

For $t=1.875, v=-32t+80= -32(1.875)+ 80 = 20 \ ft/sec$

So the speed is between 20 and 30ft/sec. The answer checks out.

Example 7

William’s pick-up truck gets between 18 to 22 miles per gallon of gasoline. His gas tank can hold 15 gallons of gasoline. If he drives at an average speed of 40 miles per hour how much driving time does he get on a full tank of gas?

Solution

Step 1 We know

The truck gets between 18 and 22 miles/gallon

There are 15 gallons in the truck’s gas tank

William drives at an average of 40 miles/hour

Let $t =$ driving time

Step 2 We use dimensional analysis to get from time per tank to miles per gallon.

$\frac{t \ \bcancel{hours}}{1 \ \bcancel{tank}} \times \frac{1 \ \bcancel{tank}}{15 \ gallons} \times \frac{40 \ miles}{1 \ \bcancel{hours}} = \frac{40t}{45} \ \frac{miles}{gallon}$

Step 3 Since the truck gets between 18 to 22 miles/gallon, we set up the compound inequality.

$18 \leq \frac{40t} {15} \leq 22$

Separate the compound inequality and solve each inequality separately.

$18 & \leq \frac{40t} {15} & & & & \frac{40t} {15}\leq 22\\270 & \leq 40t & & \text{and} & & 40t \leq 330\\6.75 & \leq t & & & & t \leq 8.25$

Answer $6.75 \leq t \leq 8.25$. Andrew can drive between 6.75 and 8.25 hours on a full tank of gas.

Step 4

For $t = 6.75$, we get $\frac{40t} {15} \frac{40(6.75)} {15}=18 \ miles$ per gallon.

For $t = 8.25$, we get $\frac{40t} {15} \frac{40(8.25)} {15}=18 \ miles$ per gallon.

## Lesson Summary

• Compound inequalities combine two or more inequalities with “and” or “or”.
• “And” combinations mean the only solutions for both inequalities will be solutions to the compound inequality.
• “Or” combinations mean solutions to either inequality will be solutions to the compound inequality.

## Review Questions

Write the compound inequalities represented by the following graphs.

Solve the following compound inequalities and graph the solution on a number line.

1. $-5 \leq x-4 \leq 13$
2. $1 \leq 3x+4 \leq 4$
3. $-12 \leq 2-5x \leq 7$
4. $\frac{3} {4} \leq 2x+9 \leq \frac{3} {2}$
5. $-2\frac{2x-1} {3}<-1$
6. $4x-1 \geq 7$ or $\frac{9x} {2}<3$
7. $3-x<-4$ or $3-x>10$
8. $\frac{2x+3} {4}<2$ or $-\frac{x} {5}+3\frac{2} {5}$
9. $2x-7 \leq -3$ or $2x-3>11$
10. $4x+3 \leq 9$ or $-5x+4 \leq -12$
11. To get a grade of B in her Algebra class, Stacey must have an average grade greater than or equal to 80 and less than 90. She received the grades of 92, 78, 85 on her first three tests. Between which scores must her grade fall if she is to receive a grade of B for the class?

1. $-40 \leq x \leq 70$
2. $x <-2$ or $x \geq 5$
3. $-8< x < 0$
4. $x \leq -2$ or $x> 1.5$
5. $-1 \leq x \leq 17$
6. $-\frac{4}{3} \leq x \leq -\frac{1}{3}$
7. $-1 \leq x \leq \frac{14}{5}$
8. $-\frac{33}{8} \leq x \leq -\frac{15}{4}$
9. $-\frac{5}{2} \leq x <-1$
10. $x \geq 2$ or $x < \frac{2}{3}$
11. $x>7$ or $x<-7$
12. $x< \frac{5}{2}$ or $x > 13$
13. $x \leq 2$ or $x> 7$
14. $x < \frac{3}{2}$ or $x \geq \frac{16}{5}$
15. $65 \leq x < 105$

Feb 22, 2012

Aug 22, 2014