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1.08 Compound inequalities

Introduction

Inequalities have extended our mathematical reasoning to solve a problem with multiple solutions. When one inequality cannot represent the solution set to a situation, we have a tool that will allow us to do that: compound inequalities.

Compound inequalities

A compound inequality is a conjunction of two or more inequalities. The set of solutions for a compound inequality are the values that make all of the inequalities true.

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  • We use "and" to indicate that a value must satisfy both inequalities in order to be in the solution set. For example: x \lt 3 \text{ and } x \geq -2
  • We can also write this compound inequality more simply as -2\leq x <3
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  • We use "or" to indicate that a value need only satisfy at least one inequality in order to be in the solution set. For example: x \gt 3 \text{ or } x \leq -2
Interval

A set of numbers that lie between two values

Interval notation

A way to represent a solution set or interval as a pair of numbers using a combination of square brackets and parentheses.

Example:

Inequality notation

3 \leq x

Interval notation

\left[ 3 , \infty \right)

We use square brackets if the endpoint is included and parentheses if the endpoint is not included. We always use parentheses for infinity. We can join two sets together using the union symbol \cup.

A number line ranging from negative 5 to 5 in steps of 1. A filled circle is on negative 2 while the unfilled circle is on 3. The segment is from negative 2 to 3.
The solution in interval notation is written as \left[-2,3\right)
A number line ranging from negative 5 to 5 in steps of 1. A filled circle is on negative 2 while the unfilled circle is on 3. An arrow is from negative 2 to the left of negative 2 while another arrow is from 3 to the right of 3.
The solution in interval notation is written as \left(-\infty, -2\right] \cup \left(3, \infty\right)

Examples

Example 1

Consider the graphed solution of a compound inequality on the number line shown:

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a

Write a compound inequality to represent the solution set.

Worked Solution
Create a strategy

We want to describe the compound inequality and then write the solution to the compound inequality algebraically.

Description: x is less than -1 or x is greater than or equal to 2

Apply the idea

Compound inequality: x < -1 or x \geq 2

b

Write the solution to the compound inequality in interval notation.

Worked Solution
Create a strategy

Since the solution to the inequality occurs in two intervals, we should write the two intervals connected with the union symbol.

Apply the idea

\left(-\infty, -1\right) \cup \left[2, \infty\right)

Reflect and check

The unfilled circle at -1 indicates that -1 is not in the solution set of the compound inequality, which is stated with the inequality x<-1.

The filled circle at 2 indicates that 2 is in the solution set of the compound inequality, which is stated with the inequality x \geq 2.

Example 2

Solve and graph each compound inequality.

a

-2.9 \leq -2.4x + 7.3 < 3.7

Worked Solution
Create a strategy

An "and" compound inequality can be solved simultaneously as long as we apply each operation to all parts of the inequality.

Apply the idea
\displaystyle -2.9\displaystyle \leq\displaystyle -2.4x+7.3<3.7Original inequality
\displaystyle -2.9-7.3\displaystyle \leq\displaystyle -2.4x+7.3-7.3<3.7-7.3Subtraction property of inequality
\displaystyle -10.2\displaystyle \leq\displaystyle -2.4x<-3.6Evaluate the subtraction
\displaystyle \dfrac{-10.2}{-2.4}\displaystyle \geq\displaystyle \dfrac{-2.4x}{-2.4} > \dfrac{-3.6}{-2.4}Division property of inequality
\displaystyle 4.25\displaystyle \geq\displaystyle x > 1.5Evaluate the division

By the division property of inequality, division by a negative number requires reversing the inequality symbols.

We can write the inequality in order from its minimum value to its maximum value by the symmetric property of inequality: 1.5<x \leq 4.25.

Finally we can graph it on the number line with an unfilled endpoint at 1.5 because the inequality is not an equal to inequality and a filled endpoint at 4.25 because the inequality is an equal to inequality.

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Reflect and check

We could have also written the compound inequality as two inequalities and solved them separately.

Solve the first inequality:

\displaystyle -2.9\displaystyle \leq\displaystyle -2.4x+7.3First inequality
\displaystyle -2.9-7.3\displaystyle \leq\displaystyle -2.4x+7.3-7.3Subtraction property of inequality
\displaystyle -10.2\displaystyle \leq\displaystyle -2.4xEvaluate the subtraction
\displaystyle \dfrac{-10.2}{-2.4}\displaystyle \geq\displaystyle \dfrac{-2.4x}{-2.4}Division property of inequality
\displaystyle 4.25\displaystyle \geq\displaystyle xEvaluate the division

Solve the second inequality:

\displaystyle -2.4x+7.3\displaystyle <\displaystyle 3.7Second inequality
\displaystyle -2.4x+7.3-7.3\displaystyle <\displaystyle 3.7-7.3Subtraction property of inequality
\displaystyle -2.4x\displaystyle <\displaystyle -3.6Evaluate the subtraction
\displaystyle \dfrac{-2.4x}{-2.4}\displaystyle >\displaystyle \dfrac{-3.6}{-2.4}Division property of inequality
\displaystyle x\displaystyle >\displaystyle 1.5Evaluate the division

When writing the final compound inequality, we can graph the solutions on the number line and then write the compound inequality to match the solution set.

First, plot a filled circle at 4.25 and an unfilled circle at 1.5. Then, determine values of x that make both 4.25 \geq x and x>1.5 true.

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Based on the number line, the values of x that satisfy both inequalities are between 1.5, (exclusive) and 4.25, (inclusive). The solution to the inequality is 1.5 < x \leq 4.25.

b

-\dfrac{1}{2}x + 8 < 2 or 12x + 9 \leq -15

Worked Solution
Create a strategy

We can solve the two inequalities separately using properties of inequality and graph the solution on the same number line.

Apply the idea

Solve the first inequality:

\displaystyle -\dfrac{1}{2}x + 8\displaystyle <\displaystyle 2First inequality
\displaystyle -\dfrac{1}{2}x + 8 - 8\displaystyle <\displaystyle 2-8Subtraction property of inequality
\displaystyle -\dfrac{1}{2}x \displaystyle <\displaystyle -6Evaluate the subtraction
\displaystyle -\dfrac{2}{1} \cdot -\dfrac{1}{2}x\displaystyle >\displaystyle -6 \cdot -\dfrac{2}{1}Multiplication property of inequality
\displaystyle x\displaystyle >\displaystyle 12Evaluate the multiplication

Solve the second inequality:

\displaystyle 12x+9\displaystyle \leq\displaystyle -15Second inequality
\displaystyle 12x+9-9\displaystyle \leq\displaystyle -15-9Subtraction property of inequality
\displaystyle 12x\displaystyle \leq\displaystyle -24Evaluate the subtraction
\displaystyle \dfrac{12x}{12}\displaystyle \leq\displaystyle \dfrac{-24}{12}Division property of inequality
\displaystyle x\displaystyle \leq\displaystyle -2Evaluate the division

The solution to the compound inequality is x>12 or x \leq -2. The solution set is shown on the number line:

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c

3.5x+0.75 < 13 or -6.25x-8 \geq 4.5

Worked Solution
Apply the idea

Solve the first inequality:

\displaystyle 3.5x+0.75\displaystyle <\displaystyle 13First inequality
\displaystyle 3.5x+0.75-0.75\displaystyle <\displaystyle 13-0.75Subtraction property of inequality
\displaystyle 3.5x\displaystyle <\displaystyle 12.25Evaluate the subtraction
\displaystyle \dfrac{3.5x}{3.5}\displaystyle <\displaystyle \dfrac{12.25}{3.5}Division property of inequality
\displaystyle x\displaystyle <\displaystyle 3.5Evaluate the division

Solve the second inequality:

\displaystyle -6.25x-8\displaystyle \geq\displaystyle 4.5Second inequality
\displaystyle -6.25x-8+8\displaystyle \geq\displaystyle 4.5+8Addition property of inequality
\displaystyle -6.25x\displaystyle \geq\displaystyle 12.5Evaluate the addition
\displaystyle \dfrac{-6.25x}{-6.25}\displaystyle \leq\displaystyle \dfrac{12.5}{-6.25}Division property of inequality
\displaystyle x\displaystyle \leq\displaystyle -2Evaluate the division

When graphing the solutions to the inequality, we see an overlap in the solution sets since the division property of inequality by a negative number required us to reverse the inequality symbol for the second inequality.

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Since the statement for the compound inequality is x<3.5 or x\leq-2 and the solution set is a single set of values that are strictly less than 3.5, the solution to the inequality is x < 3.5 since its solutions will satisfy either inequality.

Example 3

To earn a final grade of a B in her social studies class, Malia must have a test score average of an 82 to an 87 on 4 tests. Suppose that Malia scored 89, 91, and 80 on her first three tests.

a

Write a compound inequality to solve for the possible scores Malia can earn on her 4th test in order to earn a B.

Worked Solution
Create a strategy

Since Malia's scores must be an average of an 82 to an 87, we can write an "and" compound inequality statement to determine the range of test scores she could earn.

Apply the idea

Let x= \text{ the final test score}. Since the minimum score she needs is an 82 and the maximum score she needs is an 87, we can write our expression representing the average of the four test scores in between.

\displaystyle 82 \displaystyle \leq\displaystyle \dfrac{89 + 91 + 80 + x}{4} \leq 87
Reflect and check

After writing a compound inequality to model a real-world situation, it helps to review the inequality written and confirm that it is reasonable for the situation.

b

Solve the compound inequality from part (a) and interpret its meaning in context.

Worked Solution
Apply the idea
\displaystyle 82 \displaystyle \leq\displaystyle \dfrac{89 + 91 + 80 + x}{4} \leq 87Original inequality
\displaystyle 82 \displaystyle \leq\displaystyle \dfrac{260 + x}{4} \leq 87Evaluate the addition
\displaystyle 82 \cdot 4 \displaystyle \leq\displaystyle \dfrac{260 + x}{4} \cdot 4 \leq 87 \cdot 4Multiplication property of inequality
\displaystyle 328 \displaystyle \leq\displaystyle 260+x \leq 348Evaluate the multiplication
\displaystyle 328-260\displaystyle \leq\displaystyle 260 + x - 260 \leq 348-260Subtraction property of inequality
\displaystyle 68\displaystyle \leq\displaystyle x \leq 88Evaluate the subtraction

Since x can be anywhere from 68 to 88, (inclusive) and we know that x represents Malia's score needed on her final test, the solution means that Malia needs anywhere from a 68 to an 88 on her final test in order to earn a B in social studies.

c

Determine whether x=75 is a viable solution.

Worked Solution
Create a strategy

Substituting the given value into the inequality will determine if the solution is a solution to the inequality, then making sense of the solution in context will further our decision.

Apply the idea

If x=75, then 68 \leq 75 \leq 88 is a true statement.

x=75 means a score of 75 on the test, and since we know the score can be from a 68 to an 88, we can confirm that x=75 is a viable solution.

Idea summary

Remember both types of compound inequalities:

  • "And" statements: a \leq x \leq b written in interval notation as \left[a,b\right]
  • "Or" statements: x < a or x \geq b written in interval notation as \left(-\infty, a\right) \cup \left[b, \infty\right)

Outcomes

A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems.

A.REI.B.3

Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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