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Section 1.6 Solving One-Step Inequalities

permalinkIn this section, we learn that solving basic inequalities is not that different from solving basic equations.

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Figure 1.6.1. Alternative Video Lesson

permalinkWith one small complication, we can use very similar properties to Fact 1.5.13 when we solve inequalities (as opposed to equations). Here are some numerical examples.

Add to both sides If 2<4, then 2+1<βœ“4+1.
Subtract from both sides If 2<4, then 2βˆ’1<βœ“4βˆ’1.
Multiply on both sides by a positive number If 2<4, then 3β‹…2<βœ“3β‹…4.
Divide on both sides by a positive number If 2<4, then 22<βœ“42.

permalinkHowever, something interesting happens when we multiply or divide by the same negative number on both sides of an inequality: the direction reverses! To understand why, consider Figure 1.6.2, where the numbers 2 and 4 are multiplied by the negative number βˆ’1.

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Figure 1.6.2. When two numbers are multiplied by a negative number, their relationship changes.

permalinkSo even though 2<4, if we multiply both sides by βˆ’1, we have the false inequality βˆ’2<noβˆ’4. The true inequality is βˆ’2>βˆ’4.

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Example 1.6.4.

Solve the inequality βˆ’2xβ‰₯12. State the solution set graphically, using interval notation, and using set-builder notation. (Interval notation and set-builder notation are discussed in Section 1.3.)

Explanation

To solve this inequality, we will divide each side by \(-2\text{:}\)

\begin{align*} -2x\amp\geq12\\ \divideunder{-2x}{-2}\amp\highlight{{}\leq{}}\divideunder{12}{-2}\amp\amp\text{Note the change in direction.}\\ x\amp\leq-6 \end{align*}

Note that the inequality sign changed direction at the step where we divided each side of the inequality by a negative number.

When we solve a linear inequality, there are usually infinitely many solutions. (Unlike when we solve a linear equation and only have one solution.) For this example, any number less than or equal to \(-6\) is a solution.

There are at least three ways to represent the solution set for the solution to an inequality: graphically, with set-builder notation, and with interval notation. Graphically, we represent the solution set as:

Using interval notation, we write the solution set as \((-\infty,-6]\text{.}\) Using set-builder notation, we write the solution set as \(\{x\mid x\leq-6\}\text{.}\)

As with equations, we should check solutions to catch both human mistakes as well as for possible extraneous solutions (numbers which were possible solutions according to algebra, but which actually do not solve the inequality).

Since there are infinitely many solutions, it's impossible to literally check them all. We found that all values of \(x\) for which \(x\leq-6\) are solutions. One approach is to check that one number less than \(-6\) (any number, your choice) satisfies the inequality. And that \(-6\) satisfies the inequality. And that one number greater than \(-6\) (any number, your choice) does not satsify the inequality.

\begin{align*} -2x\amp\geq12\amp -2x\amp\ge12\amp -2x\amp\ge12\\ -2(\substitute{-7})\amp\stackrel{?}{\geq}12\amp -2(\substitute{-6})\amp\stackrel{?}{\ge}12\amp -2(\substitute{-5})\amp\stackrel{?}{\ge}12\\ 14\amp\stackrel{\checkmark}{\geq}12\amp 12\amp\stackrel{\checkmark}{\ge}12\amp 12\amp\stackrel{\text{no}}{\ge}12 \end{align*}

Thus both \(-7\) and \(-6\) are solutions, while \(-5\) is not. This is evidence that our solution set is correct, and it's valuable in that making these checks would likely help us catch an error if we had made one. While it certainly does take time and space to make three checks like this, it has its value.

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Example 1.6.5.

Solve the inequality t+7<5. State the solution set graphically, using interval notation, and using set-builder notation.

Explanation

To solve this inequality, we will subtract \(7\) from each side. There is not much difference between this process and solving the equation \(t+7=5\text{,}\) because we are not going to multiply or divide by negative numbers.

\begin{align*} t+7\amp\lt5\\ t+7\subtractright{7}\amp\lt5\subtractright{7}\\ t\amp\lt-2 \end{align*}

Note again that the direction of the inequality did not change, since we did not multiply or divide each side of the inequality by a negative number at any point.

Graphically, we represent this solution set as:

Using interval notation, we write the solution set as \((-\infty,-2)\text{.}\) Using set-builder notation, we write the solution set as \(\{t\mid t\lt-2\}\text{.}\)

We should check that some number less than \(-2\) is a solution, but that \(-2\) and some number greater than \(-2\) are not solutions.

\begin{align*} t+7\amp\lt5\amp t+7\amp\lt5\amp t+7\amp\lt5\\ \substitute{-10}+7\amp\stackrel{?}{\lt}5\amp \substitute{-2}+7\amp\stackrel{?}{\lt}5\amp \substitute{0}+7\amp\stackrel{?}{\lt}5\\ -3\amp\stackrel{\checkmark}{\lt}5\amp 5\amp\stackrel{\text{no}}{\lt}5\amp 7\amp\stackrel{\text{no}}{\lt}5 \end{align*}

So our solution is reasonably checked.

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Checkpoint 1.6.6.
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Checkpoint 1.6.7.
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Reading Questions Reading Questions

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1.

What are three ways to express the solution set to a linear inequality?

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2.

When you go through the motions of solving a simple linear inequality, what step(s) might you need to take where something special happens that you don't have to worry about when solving a simple linear equation?

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3.

Why does checking the solution set to an inequality take more effort than checking the solution set to an equation?

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Exercises Exercises

Review and Warmup
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1.

Add the following.

  1. βˆ’9+(βˆ’3)

  2. βˆ’7+(βˆ’6)

  3. βˆ’2+(βˆ’9)

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2.

Add the following.

  1. βˆ’9+(βˆ’1)

  2. βˆ’5+(βˆ’3)

  3. βˆ’2+(βˆ’7)

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3.

Add the following.

  1. 4+(βˆ’8)

  2. 10+(βˆ’2)

  3. 5+(βˆ’5)

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4.

Add the following.

  1. 5+(βˆ’9)

  2. 7+(βˆ’3)

  3. 5+(βˆ’5)

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5.

Add the following.

  1. βˆ’6+5

  2. βˆ’4+9

  3. βˆ’6+6

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6.

Add the following.

  1. βˆ’8+1

  2. βˆ’1+6

  3. βˆ’6+6

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7.

Evaluate the following.

  1. βˆ’27βˆ’9

  2. 35βˆ’5

  3. βˆ’357

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8.

Evaluate the following.

  1. βˆ’72βˆ’8

  2. 28βˆ’7

  3. βˆ’606

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9.

Do the following multiplications.

  1. 12β‹…34

  2. 16β‹…34

  3. 20β‹…34

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10.

Do the following multiplications.

  1. 20β‹…45

  2. 25β‹…45

  3. 30β‹…45

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11.

Evaluate the following.

  1. βˆ’5βˆ’1

  2. 9βˆ’1

  3. 120βˆ’120

  4. βˆ’15βˆ’15

  5. 80

  6. 0βˆ’7

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12.

Evaluate the following.

  1. βˆ’4βˆ’1

  2. 7βˆ’1

  3. 170βˆ’170

  4. βˆ’18βˆ’18

  5. 80

  6. 0βˆ’2

Solving One-Step Inequalities using Addition/Subtraction
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13.

Solve this inequality.

x+5>6

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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14.

Solve this inequality.

x+5>9

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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15.

Solve this inequality.

xβˆ’1≀7

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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16.

Solve this inequality.

xβˆ’1≀6

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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17.

Solve this inequality.

2≀x+9

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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18.

Solve this inequality.

3≀x+7

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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19.

Solve this inequality.

3>xβˆ’10

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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20.

Solve this inequality.

4>xβˆ’9

In set-builder notation, the solution set is .

In interval notation, the solution set is .

Solving One-Step Inequalities using Multiplication/Division
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21.

Solve this inequality.

5x≀10

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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22.

Solve this inequality.

5x≀20

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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23.

Solve this inequality.

7x>10

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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24.

Solve this inequality.

4x>1

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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25.

Solve this inequality.

βˆ’2xβ‰₯8

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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26.

Solve this inequality.

βˆ’3xβ‰₯9

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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27.

Solve this inequality.

6β‰₯βˆ’3x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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28.

Solve this inequality.

16β‰₯βˆ’4x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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29.

Solve this inequality.

7<βˆ’x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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30.

Solve this inequality.

8<βˆ’x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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31.

Solve this inequality.

βˆ’x≀9

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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32.

Solve this inequality.

βˆ’x≀10

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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33.

Solve this inequality.

13x>2

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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34.

Solve this inequality.

29x>8

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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35.

Solve this inequality.

βˆ’45x≀8

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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36.

Solve this inequality.

βˆ’52x≀15

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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37.

Solve this inequality.

βˆ’12<67x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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38.

Solve this inequality.

βˆ’21<75x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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39.

Solve this inequality.

βˆ’16<βˆ’87x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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40.

Solve this inequality.

βˆ’18<βˆ’98x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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41.

Solve this inequality.

5x>βˆ’15

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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42.

Solve this inequality.

2x>βˆ’8

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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43.

Solve this inequality.

βˆ’6<βˆ’2x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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44.

Solve this inequality.

βˆ’6<βˆ’3x

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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45.

Solve this inequality.

56β‰₯x12

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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46.

Solve this inequality.

76β‰₯x36

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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47.

Solve this inequality.

βˆ’z24<βˆ’58

In set-builder notation, the solution set is .

In interval notation, the solution set is .

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48.

Solve this inequality.

βˆ’z40<βˆ’98

In set-builder notation, the solution set is .

In interval notation, the solution set is .

Challenge
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49.

Choose the correct inequality or equal sign to make the relation true.

  1. Let x and y be integers, such that x<y.

    Then xβˆ’y

    • <

    • >

    • =

    yβˆ’x.

  2. Let x and y be integers, such that 1<x<y.

    Then xy

    • <

    • >

    • =

    x+y.

  3. Let x and y be rational numbers, such that 0<x<y<1.

    Then xy

    • <

    • >

    • =

    x+y.

  4. Let x and y be integers, such that x<y.

    Then x+2y

    • <

    • >

    • =

    2x+y.