Skip to main content

Section 1.6 Solving One-Step Inequalities

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

Figure 1.6.1. Alternative Video Lesson

With 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 \(2\lt4\text{,}\) then \(2\addright{1}\stackrel{\checkmark}{\lt}4\addright{1}\text{.}\)
Subtract from both sides If \(2\lt4\text{,}\) then \(2\subtractright{1}\stackrel{\checkmark}{\lt}4\subtractright{1}\text{.}\)
Multiply on both sides by a positive number If \(2\lt4\text{,}\) then \(\multiplyleft{3}2\stackrel{\checkmark}{\lt}\multiplyleft{3}4\text{.}\)
Divide on both sides by a positive number If \(2\lt4\text{,}\) then \(\divideunder{2}{2}\stackrel{\checkmark}{\lt}\divideunder{4}{2}\text{.}\)

However, 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 \(-1\text{.}\)

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

So even though \(2\lt4\text{,}\) if we multiply both sides by \(-1\text{,}\) we have the false inequality \(-2\stackrel{\text{no}}{\lt}-4\text{.}\) The true inequality is \(-2\gt-4\text{.}\)

Example 1.6.4.

Solve the inequality \(-2x\geq12\text{.}\) 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:

a number line with a mark at -6; a thick line overlays the number line to the left of -6 with an arrow pointing left; there is a right bracket at -6
epspdfpngsvgtex

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.

a number line with a mark at -6; a thick line overlays the number line to the left of -6 with an arrow pointing left; there is a right bracket at -6; three arrows point to -7, -6, and -5 on the number line, suggesting that these values will be checked as possible solutions
epspdfpngsvgtex
\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.

Example 1.6.5.

Solve the inequality \(t+7\lt5\text{.}\) 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:

a number line with a mark at -2; a thick line overlays the number line to the left of -2 with an arrow pointing left; there is a right parenthesis at -2
epspdfpngsvgtex

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.

Checkpoint 1.6.6.
Checkpoint 1.6.7.

Reading Questions Reading Questions

1.

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

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?

3.

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

Exercises Exercises

Review and Warmup
1.

Add the following.

  1. \(-9+(-3)\)

  2. \(-7+(-6)\)

  3. \(-2+(-9)\)

2.

Add the following.

  1. \(-9+(-1)\)

  2. \(-5+(-3)\)

  3. \(-2+(-7)\)

3.

Add the following.

  1. \(4+(-8)\)

  2. \(10+(-2)\)

  3. \(5+(-5)\)

4.

Add the following.

  1. \(5+(-9)\)

  2. \(7+(-3)\)

  3. \(5+(-5)\)

5.

Add the following.

  1. \(-6+5\)

  2. \(-4+9\)

  3. \(-6+6\)

6.

Add the following.

  1. \(-8+1\)

  2. \(-1+6\)

  3. \(-6+6\)

7.

Evaluate the following.

  1. \(\frac{-27}{-9}\)

  2. \(\frac{35}{-5}\)

  3. \(\frac{-35}{7}\)

8.

Evaluate the following.

  1. \(\frac{-72}{-8}\)

  2. \(\frac{28}{-7}\)

  3. \(\frac{-60}{6}\)

9.

Do the following multiplications.

  1. \(12 \cdot \frac{3}{4}\)

  2. \(16 \cdot \frac{3}{4}\)

  3. \(20 \cdot \frac{3}{4}\)

10.

Do the following multiplications.

  1. \(20 \cdot \frac{4}{5}\)

  2. \(25 \cdot \frac{4}{5}\)

  3. \(30 \cdot \frac{4}{5}\)

11.

Evaluate the following.

  1. \(\frac{-5}{-1}\)

  2. \(\frac{9}{-1}\)

  3. \(\frac{120}{-120}\)

  4. \(\frac{-15}{-15}\)

  5. \(\frac{8}{0}\)

  6. \(\frac{0}{-7}\)

12.

Evaluate the following.

  1. \(\frac{-4}{-1}\)

  2. \(\frac{7}{-1}\)

  3. \(\frac{170}{-170}\)

  4. \(\frac{-18}{-18}\)

  5. \(\frac{8}{0}\)

  6. \(\frac{0}{-2}\)

Solving One-Step Inequalities using Addition/Subtraction
13.

Solve this inequality.

\(\displaystyle{ {x+5} > {6} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

14.

Solve this inequality.

\(\displaystyle{ {x+5} > {9} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

15.

Solve this inequality.

\(\displaystyle{ {x-1} \leq {7} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

16.

Solve this inequality.

\(\displaystyle{ {x-1} \leq {6} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

17.

Solve this inequality.

\(\displaystyle{ {2} \leq {x+9} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

18.

Solve this inequality.

\(\displaystyle{ {3} \leq {x+7} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

19.

Solve this inequality.

\(\displaystyle{ {3} > {x-10} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

20.

Solve this inequality.

\(\displaystyle{ {4} > {x-9} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

Solving One-Step Inequalities using Multiplication/Division
21.

Solve this inequality.

\(\displaystyle{ {5x} \leq {10} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

22.

Solve this inequality.

\(\displaystyle{ {5x} \leq {20} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

23.

Solve this inequality.

\(\displaystyle{ {7x} > {10} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

24.

Solve this inequality.

\(\displaystyle{ {4x} > {1} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

25.

Solve this inequality.

\(\displaystyle{ {-2x} \geq {8} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

26.

Solve this inequality.

\(\displaystyle{ {-3x} \geq {9} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

27.

Solve this inequality.

\(\displaystyle{ {6} \geq {-3x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

28.

Solve this inequality.

\(\displaystyle{ {16} \geq {-4x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

29.

Solve this inequality.

\(\displaystyle{ {7} \lt {-x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

30.

Solve this inequality.

\(\displaystyle{ {8} \lt {-x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

31.

Solve this inequality.

\(\displaystyle{ {-x} \leq {9} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

32.

Solve this inequality.

\(\displaystyle{ {-x} \leq {10} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

33.

Solve this inequality.

\(\displaystyle{ {{\frac{1}{3}}x} > {2} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

34.

Solve this inequality.

\(\displaystyle{ {{\frac{2}{9}}x} > {8} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

35.

Solve this inequality.

\(\displaystyle{ {-{\frac{4}{5}}x} \leq {8} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

36.

Solve this inequality.

\(\displaystyle{ {-{\frac{5}{2}}x} \leq {15} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

37.

Solve this inequality.

\(\displaystyle{ {-12} \lt {{\frac{6}{7}}x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

38.

Solve this inequality.

\(\displaystyle{ {-21} \lt {{\frac{7}{5}}x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

39.

Solve this inequality.

\(\displaystyle{ {-16} \lt {-{\frac{8}{7}}x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

40.

Solve this inequality.

\(\displaystyle{ {-18} \lt {-{\frac{9}{8}}x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

41.

Solve this inequality.

\(\displaystyle{ {5x} > {-15} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

42.

Solve this inequality.

\(\displaystyle{ {2x} > {-8} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

43.

Solve this inequality.

\(\displaystyle{ {-6} \lt {-2x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

44.

Solve this inequality.

\(\displaystyle{ {-6} \lt {-3x} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

45.

Solve this inequality.

\(\displaystyle{ {\frac{5}{6}} \geq {\frac{x}{12}} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

46.

Solve this inequality.

\(\displaystyle{ {\frac{7}{6}} \geq {\frac{x}{36}} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

47.

Solve this inequality.

\(\displaystyle{ {-\frac{z}{24}} \lt {-\frac{5}{8}} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

48.

Solve this inequality.

\(\displaystyle{ {-\frac{z}{40}} \lt {-\frac{9}{8}} }\)

In set-builder notation, the solution set is .

In interval notation, the solution set is .

Challenge
49.

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

  1. Let \(x\) and \(y\) be integers, such that \(x \lt y\text{.}\)

    Then \(x-y\)

    • <

    • >

    • =

    \(y-x\text{.}\)

  2. Let \(x\) and \(y\) be integers, such that \(1 \lt x \lt y \text{.}\)

    Then \(xy\)

    • <

    • >

    • =

    \(x + y\text{.}\)

  3. Let \(x\) and \(y\) be rational numbers, such that \(0 \lt x \lt y \lt 1\text{.}\)

    Then \(xy\)

    • <

    • >

    • =

    \(x + y\text{.}\)

  4. Let \(x\) and \(y\) be integers, such that \(x \lt y\text{.}\)

    Then \(x + 2y\)

    • <

    • >

    • =

    \(2x + y\text{.}\)

login