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# 6.2: Inequalities Using Multiplication and Division

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Solve an inequality using multiplication.
• Solve an inequality using division.
• Multiply or divide an inequality by a negative number.

## Introduction

In this section, we consider problems where we find the solution of an inequality by multiplying or dividing both sides of the inequality by a number.

## Solve an Inequality Using Multiplication

Consider the problemTo find the solution we multiply both sides by 5.We obtainx55x5x33.515\begin{align*}\text{Consider the problem} & & \frac{x} {5} & \le 3\\ \text{To find the solution we multiply both sides by} \ 5. & & 5\cdot \frac{x} {5} &\le 3.5\\ \text{We obtain} & & x & \le 15\end{align*}

The answer of an inequality can be expressed in four different ways:

1. Inequality notation The answer is simply expressed as x<15\begin{align*}x<15\end{align*}.
2. Set notation The answer is {x|x<15}\begin{align*}\left \{ x| x<15 \right \}\end{align*}. You read this as “the set of all values of x\begin{align*}x\end{align*}, such that x\begin{align*}x\end{align*} is a real number less than 15”.
3. Interval notation uses brackets to indicate the range of values in the solution.

Square or closed brackets “[” and “]” indicate that the number next to the bracket is included in the solution set.

Round or open brackets “(” and “)” indicate that the number next to the bracket is not included in the solution set.

Interval notation also uses the concept of infinity \begin{align*}\infty \end{align*} and negative infinity \begin{align*}-\infty\end{align*}.

The interval notation solution for our problem is (,15)\begin{align*}(-\infty, 15)\end{align*}.

1. Solution graph shows the solution on the real number line. A closed circle on a number indicates that the number is included in the solution set. While an open circle indicates that the number is not included in the set. For our example, the solution graph is drawn here.

Example 1

a) [-4, 6] says that the solutions is all numbers between -4 and 6 including -4 and 6.

b) (8, 24) says that the solution is all numbers between 8 and 24 but does not include the numbers 8 and 24.

c) [3, 12) says that the solution is all numbers between 3 and 12, including 3 but not including 12.

d) (-10, ) says that the solution is all numbers greater that -10, not including -10.

e) ( , ) says that the solution is all real numbers.

## Solving an Inequality Using Division

Consider the problem.To find the solution we multiply both sides by 2.We obtain.2x2x2x151226\begin{align*}\text{Consider the problem}. & & 2x & \geq 15\\ \text{To find the solution we multiply both sides by} \ 2. & & \frac{2x} {2} & \geq \frac{12} {2}\\ \text{We obtain}. & & x & \geq 6\end{align*}

Let’s write the solution in the four different notations you just learned:

Inequality notationSet notationInterval notationx6x|x6[6,)\begin{align*}& \text{Inequality notation} & & x \geq 6\\ & \text{Set notation} & & {x|x \geq 6 }\\ & \text{Interval notation} & & [6,\infty)\end{align*}

Solution graph

## Multiplying and Dividing an Inequality by a Negative Number

We solve an inequality in a similar way to solving a regular equation. We can add or subtract numbers on both sides of the inequality. We can also multiply or divide positive numbers on both sides of an inequality without changing the solution.

Something different happens if we multiply or divide by negative numbers. In this case, the inequality sign changes direction.

For example, to solve 3x<9\begin{align*}-3x <9\end{align*}

We divide both sides by –3. The inequality sign changes from < to > because we divide by a negative number. 3x3>93\begin{align*} \frac{-3x} {-3} > \frac{9} {-3}\end{align*}

x>3\begin{align*} x > -3\end{align*}

We can explain why this happens with a simple example. We know that two is less than three, so we can write the inequality.

2<3\begin{align*}2 < 3\end{align*}

If we multiply both numbers by -1 we get -2 and -3 but we know that -2 is greater than -3.

2>3\begin{align*}-2>-3\end{align*}

You see that multiplying both sides of the inequality by a negative number caused the inequality sign to change direction. This also occurs if we divide by a negative number.

Example 2

Solve each inequality. Give the solution in inequality notation and interval notation.

a) 4x<24\begin{align*}4x < 24\end{align*}

b) 9x35\begin{align*} -9x \geq -\frac{3}{5}\end{align*}

c) 5x21\begin{align*}-5x \leq 21\end{align*}

d) 12>30\begin{align*}12 > -30\end{align*}

Solution:

a) Original problem.Divide both sides by 4.Simplify4x4x4x<24<244<6 or (,6) Answer\begin{align*}\text{Original problem}.& & 4x & < 24\\ \text{Divide both sides by} \ 4. & & \frac{4x} {4} & < \frac{24} {4}\\ \text{Simplify} & & x & < 6 \ \text{or} \ (\infty, 6) \ \text{Answer}\end{align*}

b) Original problem:Divide both sides by 9.Simplify.9x99x3535193 Direction of the inequality is changed115 or [115,) Answer\begin{align*}\text{Original problem}: & & -9x & \geq \frac{-3} {5}\\ \text{Divide both sides by} \ -9. & & \frac{-9}{-9} & \le \frac{\cancel{-3}}{5}\cdot\frac{1}{\cancel{-9}_3} \ \text{Direction of the inequality is changed}\\ \text{Simplify}. & & x & \geq \frac{1} {15} \ \text{or} \ \left [ \frac{1} {15}, \infty \right)\ \text{Answer}\end{align*}

\begin{align*}\text{Original problem}: & & -5x & \leq 21\\ \text{Divide both sides by} \ -5. & & \frac{-5x} {-5} & \geq \frac{21} {-5} \ \text{Direction of the inequality is changed}\\ \text{Simplify}. & & x & \geq -\frac{21} {5} \ \text{or} \ \left [ \frac{21} {5}, \infty \right)\ \text{Answer}\end{align*}

d) \begin{align*}\text{Original problem} & & 12x & >-30\\ \text{Divide both sides by} \ 12. & & \frac{12x} {12}& >\frac{-30} {12}\\ \text{Simplify}. & & x & > -\frac{5} {5}\ \text{or} \left ( \frac{-5} {2}, \infty \right) \ \text{Answer}\end{align*}

Example 3

Solve each inequality. Give the solution in inequality notation and solution graph.

a) \begin{align*} \frac{x}{2} > -40 \end{align*}

b) \begin{align*} \frac{x}{-3} \leq -12\end{align*}

c) \begin{align*} \frac{x}{25} < \frac{3}{2}\end{align*}

d) \begin{align*} \frac{x}{-7} \geq 9\end{align*}

Solution

a) \begin{align*}\text{Original problem} & & \frac{x} {2} & > 40\\ \text{Multiply both sides by} \ 2. & & 2 \cdot \frac{x} {2} & > 40 \cdot 2 \ \text{Direction of inequality is NOT changed}\\ \text{Simplify}. & & x & > -80 \ \text{Answer}\end{align*}

b) \begin{align*}\text{Original problem} & & \frac{x} {-3} & \leq -12\\ \text{Multiply both sides by} \ -3. & & -3\cdot\frac{x} {-3} & \geq -12 \cdot (-3)\ \text{Direction of inequality is changed}\\ \text{Simplify}. & & x & \geq 36 \ \text{Answer}\end{align*}

c) \begin{align*}\text{Original problem} & & \frac{x} {25}&< \frac{3} {2}\\ \text{Multiply both sides by} \ 25. & & 25 \cdot \frac{x} {25} &< \frac{3} {2} \cdot 25\ \text{direction of inequality is NOT changed}\\ \text{Simplify}. & & x &< \frac{75} {2} \ \text{or} \ x < 37.5 \ \text{Answer}\end{align*}

d) \begin{align*}\text{Original problem} & & \frac{x} {-7} & \geq 9\\ \text{Multiply both sides by} \ -7. & & -7 \cdot \frac{x} {-7} & \leq 9 \cdot (-7) \ \text{Direction of inequality is changed}\\ \text{Simplify}. & & x & \leq -63 \ \text{Answer}\end{align*}

## Lesson Summary

• There are four ways to represent an inequality:

1. Equation notation \begin{align*}x \geq 2\end{align*}

2. Set notation \begin{align*}x \geq 2\end{align*}

3. Interval notation \begin{align*}[2, \infty)\end{align*}

Closed brackets “[” and “]” mean inclusive, parentheses “(”and “)” mean exclusive.

4. Solution graph

• When multiplying or dividing both sides of an inequality by a negative number, you need to reverse the inequality.

## Review Questions

Solve each inequality. Give the solution in inequality notation and solution graph.

1. \begin{align*} 3x \leq 6\end{align*}
2. \begin{align*} \frac{x} {5}> -\frac{3} {10}\end{align*}
3. \begin{align*} -10x > 250\end{align*}
4. \begin{align*} \frac{x} {-7} \geq -5\end{align*}

Solve each inequality. Give the solution in inequality notation and interval notation.

1. \begin{align*} 9x > -\frac{3} {4}\end{align*}
2. \begin{align*} -\frac{x} {15} \leq 5\end{align*}
3. \begin{align*} 620x > 2400\end{align*}
4. \begin{align*} \frac{x} {20} \geq -\frac{7} {40}\end{align*}

Solve each inequality. Give the solution in inequality notation and set notation.

1. \begin{align*} -0.5x \leq 7.5\end{align*}
2. \begin{align*} 75x \geq 125\end{align*}
3. \begin{align*} \frac{x} {-3}>-\frac{10} {9}\end{align*}
4. \begin{align*} \frac{x} {-15} < 8\end{align*}

1. \begin{align*} x \leq 2\end{align*} or
2. \begin{align*} x > -\frac{3}{2}\end{align*} or
3. \begin{align*}x <-25\end{align*} or
4. \begin{align*} x \leq 35\end{align*} or
5. \begin{align*} x > -\frac{1}{12}\end{align*} or \begin{align*}(-\frac{1}{12}, \infty)\end{align*}
6. \begin{align*} x \geq -75\end{align*} or \begin{align*}[-75, \infty)\end{align*}
7. \begin{align*} x < 3.9 \end{align*} or \begin{align*}(-\infty, 3.9)\end{align*}
8. \begin{align*} x \geq -\frac{7}{2}\end{align*} or \begin{align*}[-\frac{7}{2}, \infty)\end{align*}
9. \begin{align*} x \geq -15 \end{align*} or \begin{align*}\left \{ x \right .\end{align*} is a real number \begin{align*}\left . \mid x \geq -15 \right \}\end{align*}
10. \begin{align*} x \geq \frac{5}{3}\end{align*} or \begin{align*}\left \{ x \right .\end{align*} is a real number \begin{align*}\left .\mid x \geq \frac{5}{3} \right \}\end{align*}
11. \begin{align*}x < -\frac{10}{3}\end{align*} or \begin{align*}\left \{ x \right .\end{align*} is a real number \begin{align*}\left . \mid x < -\frac{10}{3} \right \}\end{align*}
12. \begin{align*}x > -120 \end{align*} or \begin{align*}\left \{ x \right .\end{align*} is a real number \begin{align*}\left . \mid x > -120\right \}\end{align*}

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