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# 6.3: Compound Inequalities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

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

## Introduction

In this section, we’ll solve compound inequalities—inequalities with more than one constraint on the possible values the solution can have.

There are two types of compound inequalities:

1. Inequalities joined by the word “and,” where the solution is a set of values greater than a number and less than another number. We can write these inequalities in the form “$x>a$ and $x,” but usually we just write “$a < x < b$.” Possible values for $x$ are ones that will make both inequalities true.
2. Inequalities joined by the word “or,” where the solution is a set of values greater than a number or less than another number. We write these inequalities in the form “$x>a$ or $x.” Possible values for $x$ are ones that will make at least one of the inequalities true.

You might wonder why the variable $x$ has to be greater than one number and/or less than the other number; why can’t it be greater than both numbers, or less than both numbers? To see why, let’s take an example.

Consider the compound inequality “$x>5$ and $x>3$.” Are there any numbers greater than 5 that are not greater than 3? No! Since 5 is greater than 3, everything greater than 5 is also greater than 3. If we say $x$ is greater than both 5 and 3, that doesn’t tell us any more than if we just said $x$ is greater than 5. So this compound inequality isn’t really compound; it’s equivalent to the simple inequality $x > 5$. And that’s what would happen no matter which two numbers we used; saying that $x$ is greater than both numbers is just the same as saying that $x$ is greater than the bigger number, and saying that $x$ is less than both numbers is just the same as saying that $x$ is less than the smaller number.

Compound inequalities with “or” work much the same way. Every number that’s greater than 3 or greater than 5 is also just plain greater than 3, and every number that’s greater than 3 is certainly greater than 3 or greater than 5—so if we say “$x>5$ or $x>3$,” that’s the same as saying just “$x>3$.” Saying that $x$ is greater than at least one of two numbers is just the same as saying that $x$ is greater than the smaller number, and saying that $x$ is less than at least one of two numbers is just the same as saying that $x$ is less than the greater number.

## 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 $x \ge -40$ and $x<60$.

This is usually written as $-40 \le x < 60$.

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: $x \ge 4$ or $x \le -1$.

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

This is usually written as $-25 < x < 25$.

Example 2

Graph the following compound inequalities on a number line.

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

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

c) $x \ge -8$ or $x \le -20$

d) $-15 < x \le 85$

Solution

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

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

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

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

## Solve a Compound Inequality With “and” or “or”

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) $-2 < 4x-5 \le 11$

b) $3x-5 < x + 9 \le 5x+13$

Solution

a) First we re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately.

$& -2 < 4x-5 \qquad \qquad 4x-5 \le 11\\& \quad 3<4x \qquad \quad and \qquad \quad 4x \le 16\\& \quad \frac{3}{4} < x \qquad \qquad \qquad \qquad \ x \le 4$

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

b) Re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately.

$& 3x-5 < x+9 \qquad \quad \ \ x+9 \le 5x+13\\& \quad \ \ 2x < 14 \qquad and \qquad -4 \le 4x\\& \qquad x < 7 \qquad \qquad \qquad \ -1 \le x$

Answer: $x < 7$ and $x \ge -1$. This can be written as: $-1 \le x < 7$.

Example 4

Solve the following compound inequalities and graph the solution set.

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

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

Solution

a) Solve each inequality separately:

$& 9-2x \le 3 \qquad \qquad 3x+10 \le 6-x\\& \ -2x \le -6 \qquad or \qquad \ \ 4x \le -4\\& \qquad x \ge 3 \qquad \qquad \qquad \ \ \ x \le -1$

Answer: $x \ge 3$ or $x \le -1$

b) Solve each inequality separately:

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

Answer: $x \ge 2$ or $x < -6.4$

The video at http://www.math-videos-online.com/solve-compound-inequality.html shows the process of solving and graphing compound inequalities in more detail. One thing you may notice in this video is that in the second problem, the two solutions joined with “or” overlap, and so the solution ends up being the set of all real numbers, or $(-\infty,\infty)$. This happens sometimes with compound inequalities that involve “or”; for example, if the solution to an inequality ended up being “$x<5$ or $x>1$,” the solution set would be all real numbers. This makes sense if you think about it: all real numbers are either a) less than 5, or b) greater than or equal to 5, and the ones that are greater than or equal to 5 are also greater than 1—so all real numbers are either a) less than 5 or b) greater than 1.

Compound inequalities with “and,” meanwhile, can turn out to have no solutions. For example, the inequality “$x<3$ and $x>4$” has no solutions: no number is both greater than 4 and less than 3. If we write it as $4 < x < 3$ it’s even more obvious that it has no solutions; $4 < x < 3$ implies that $4 < 3$, which is false.

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

Graphing calculators can show you the solution to an inequality in the form of a graph. This can be especially useful when dealing with compound inequalities.

Example 5

Solve the following inequalities using a graphing calculator.

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

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

c) $3x+2 \le 10$ or $3x+2 \ge 15$

Solution

a) Press the [Y=] button and enter the inequality on the first line of the screen.

(To get the $\ge$ symbol, press [TEST] [2nd] [MATH] and choose option 4.)

Then press the [GRAPH] button.

Because the calculator uses the number 1 to mean “true” and 0 to mean “false,” 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 $y=1$—in other words, the set of $x-$values that make the inequality true.

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

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

b) This is a compound inequality: $7x-2 < 10x +1$ and $10x+1 < 9x+5$. You enter it like this:

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

The resulting graph should look like this:

The solution are the values of $x$ for which $y=1$; in this case that would be $-1 < x < 4$.

c) This is another compound inequality.

(To enter the [OR] symbol, press [TEST], choose [LOGIC] on the top row and choose option 2.)

The resulting graph should look like this:

The solution are the values of $x$ for which $y=1$--in this case, $x \le 2.7$ or $x \ge 4.3$.

## 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 $v=-32t+80$. When is the ball traveling between 20 ft/sec and 30 ft/sec?

Solution

First we set up the inequality $20 \le v \le 30$, and then replace $v$ with the formula $v=-32t+80$ to get $20 \le -32t+80 \le 30$.

Then we separate the compound inequality and solve each separate inequality:

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

Answer: $1.56 \le t \le 1.875$

To check the answer, we 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 30 ft/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

Let $t =$ driving time. We can use dimensional analysis to get from time per tank to miles per gallon:

$\frac{t \ hours}{1 \ tank} \times \frac{1 \ tank}{15 \ gallons} \times \frac{40 \ miles}{1 \ hour} \times \frac{40t}{15} \frac{miles}{gallon}$

Since the truck gets between 18 and 22 miles/gallon, we set up the compound inequality $18 \le \frac{40t}{15} \le 22$. Then we separate the compound inequality and solve each inequality separately:

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

Answer: $6.75 \le t \le 8.25$.

Andrew can drive between 6.75 and 8.25 hours on a full tank of gas.

If we plug in $t = 6.75$ we get $\frac{40t}{15} = \frac{40(6.75)}{15} = 18 \ miles \ per \ gallon$.

If we plug in $t = 8.25$ we get $\frac{40t}{15} = \frac{40(8.25)}{15} = 22 \ miles \ per \ gallon$.

## Lesson Summary

• Compound inequalities combine two or more inequalities with “and” or “or.”
• “And” combinations mean that only solutions for both inequalities will be solutions to the compound inequality.
• “Or” combinations mean solutions to either inequality will also 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. $-5 \le x-4 \le 13$
2. $1 \le 3x +5 \le 4$
3. $-12 \le 2-5x \le 7$
4. $\frac{3}{4} \le 2x+9 \le \frac{3}{2}$
5. $-2 \le \frac{2x-1}{3} < -1$
6. $4x-1 \ge 7$ 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. $2x-7 \le -3$ or $2x-3 > 11$
10. $4x+3< 9$ or $-5x+4 \le -12$
11. How would you express the answer to problem 5 as a set?
12. How would you express the answer to problem 5 as an interval?
13. How would you express the answer to problem 10 as a set?
14. .
1. Could you express the answer to problem 10 as a single interval? Why or why not?
2. How would you express the first part of the solution in interval form?
3. How would you express the second part of the solution in interval form?
15. Express the answers to problems 1 and 3 in interval notation.
16. Express the answers to problems 6 through 9 in interval notation.
17. Solve the inequality “$x \ge -3$ or $x < 1$” and express the answer in interval notation.
18. How many solutions does the inequality “$x \ge 2$ and $x \le 2$” have?
19. 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.
1. Between which scores must her grade on the final test fall if she is to receive a grade of B for the class? (Assume all four tests are weighted the same.)
2. What range of scores on the final test would give her an overall grade of C, if a C grade requires an average score greater than or equal to 70 and less than 80?
3. If an A grade requires a score of at least 90, and the maximum score on a single test is 100, is it possible for her to get an A in this class? (Hint: look again at your answer to part a.)

Feb 23, 2012

Feb 12, 2015