# Open Resources for Community College Algebra

## Section1.6Solving One-Step Inequalities

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

### Subsection1.6.1Solving Linear Inequalities

For the most part, the properties from Fact 1.5.12 apply to inequalities too, not just equations. Here are some numerical examples to consider.
Add to both sides: if $$2\lt4\text{,}$$ then $$2\addright{1}\confirm{\lt}4\addright{1}\text{.}$$
Subtract from both sides: if $$2\lt4\text{,}$$ then $$2\subtractright{1}\confirm{\lt}4\subtractright{1}\text{.}$$
Multiply on both sides by a positive number: if $$2\lt4\text{,}$$ then $$\multiplyleft{3}2\confirm{\lt}\multiplyleft{3}4\text{.}$$
Divide on both sides by a positive number: if $$2\lt4\text{,}$$ then $$\divideunder{2}{2}\confirm{\lt}\divideunder{4}{2}\text{.}$$

#### Example1.6.2.

Solve the inequality $$t+7\lt5\text{.}$$
Explanation.
There is not much difference between the steps to solve this inequality and the steps to solve the equation $$t+7=5\text{.}$$ We can subtract $$7$$ from each side.
\begin{align*} t+7\amp\lt5\\ t+7\subtractright{7}\amp\lt5\subtractright{7}\\ t\amp\lt-2 \end{align*}
When we solve a linear inequality, there are usually infinitely many solutions. The solution set has infintely many numbers in it. This is unlike when we solve a linear equation, where there is usually only one solution and one number in the solution set. For this example, any number less than $$-2$$ is a solution.
There are at least three ways to represent an inequality’s solution set: graphically, with set-builder notation, and with interval notation. (Interval notation and set-builder notation are discussed in Section 3.) Graphically, the solution set is part of a number line:
Using interval notation, we write the solution set by reading the number line from left to right. The solution set is $$(-\infty,-2)\text{.}$$
Using set-builder notation, we write the solution with generic set braces, declaring $$t$$ to be the variable, and writing the condition that $$t$$ needs to meet: $$\{t\mid t\lt-2\}\text{.}$$
As with equations, we should check solutions to catch human mistakes. Since there are infinitely many solutions, it’s impossible to literally check them all. So we settle for something that gives us confidence our solution set is correct, but does not take forever to do.
According to our solution, all values of $$t$$ for which $$t\lt-2$$ are solutions and all values of $$t$$ for which $$t\geq2$$ are not solutions. So an approach we can use is to check if one number less than $$-2$$ (any number, your choice) satisfies the inequality. And also that $$-2$$ itself does not satisfy the inequality. And also that one number greater than $$-2$$ (any number, your choice) does not satsify the inequality.
Here we will test $$-3\text{,}$$ $$-2\text{,}$$ and $$0$$ in the original inequality.
\begin{align*} \amp\amp t+7\amp\lt5\amp\amp\\ \substitute{-3}+7\amp\wonder{\lt}5\amp \substitute{-2}+7\amp\wonder{\lt}5\amp \substitute{0}+7\amp\wonder{\lt}5\\ 4\amp\confirm{\lt}5\amp 5\amp\reject{\lt}5\amp 7\amp\reject{\lt}5 \end{align*}
It worked! The number $$-3$$ is a solution, and both $$-2$$ and $$0$$ are not. This is all as we expected. This is evidence that our solution set is correct, and we can feel more secure that we did not make a human mistake earlier when we were solving. While it certainly takes time and space to make three checks like this, it’s worth it.

#### Checkpoint1.6.3.

Solve the inequality $$x-5\gt-4\text{.}$$ Graph the solution set on a number line. State the solution set using both interval notation and set-builder notation.
Explanation.
To solve this inequality, we add $$5$$ to each side.
Graphically, we represent this solution set as:
Using interval notation, we write the solution set as $$(1, \infty)\text{.}$$ Using set-builder notation, we write it as $$\{x \mid x \gt 1\}\text{.}$$
We should check that some number less than $$1$$ is not a solution, that $$1$$ itself is not a solution, and that some number greater than $$1$$ is a solution.
\begin{equation*} \begin{aligned} \amp\amp x-5\amp\gt-4\amp\amp\\ \substitute{0}-5\amp\wonder{\gt}-4\amp \substitute{1}-5\amp\wonder{\gt}-4\amp \substitute{10}-5\amp\wonder{\gt}-4\\ -5\amp\reject{\gt}-4\amp -4\amp\reject{\gt}-4\amp 5\amp\confirm{\gt}-4 \end{aligned} \end{equation*}
Everything worked out as expected, so our solution is reasonably checked.

#### Checkpoint1.6.4.

Solve the inequality $$12\leq4p\text{.}$$ Graph the solution set on a number line. State the solution set using interval notation and using set-builder notation.
Explanation.
To solve this inequality, we dived each side by $$4\text{.}$$
\begin{equation*} \begin{aligned} 12\amp\leq4p\\ \divideunder{12}{4}\amp\leq\divideunder{4p}{4}\\ 3\amp\leq p \end{aligned} \end{equation*}
The last line says that “$$3$$ is less than or equal to $$p$$”. It may feel more natural to say the same thing but in the opposite order with the opposite comparison: “$$p$$ is greater than or equal to $$3$$”. So you could write
\begin{equation*} p\geq 3 \end{equation*}
Graphically, we represent this solution set as:
Using interval notation, we write the solution set as $$[3, \infty)\text{.}$$ Using set-builder notation, we write it as $$\{p \mid p \geq 3\}\text{.}$$
We should check that some number less than $$3$$ is not a solution, that $$3$$ itself is a solution, and that some number greater than $$3$$ is a solution.
\begin{equation*} \begin{aligned} \amp\amp 12\amp\leq4p\amp\amp\\ 12\amp\wonder{\leq}4(\substitute{0})\amp 12\amp\wonder{\leq}4(\substitute{3})\amp 12\amp\wonder{\leq}4(\substitute{5})\\ 12\amp\reject{\leq}0\amp12\amp\confirm{\leq}12\amp12\amp\confirm{\leq}20 \end{aligned} \end{equation*}
Everything worked out as expected, so our solution is reasonably checked.

### Subsection1.6.2Negation

Something interesting happens when we multiply or divide by a negative number on each side of an inequality: the direction reverses! To understand why, consider Figure 5, where the numbers $$2$$ and $$4$$ are each multiplied by $$-1\text{.}$$
Starting with $$2\lt4\text{,}$$ if we multiply both sides by $$-1$$ and leave the inequality direction alone, we would get the false inequality $$-2\reject{\lt}-4\text{.}$$ We should change the direction so we have the true inequality $$-2\gt-4\text{.}$$
Do not change the inequality direction when multiplying/dividing by a positive number, or when adding/subtracting by any number.

#### Example1.6.7.

Solve the inequality $$-2x\geq12\text{.}$$ State the solution set graphically, using interval notation, and using set-builder notation.
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*}
The inequality sign changed direction in the same step where we divided by a negative number.
Graphically, the solution set is part of a number line:
Using interval notation, we write the solution set as $$(-\infty,-6]\text{.}$$ Using set-builder notation, we write the solution set as $$\{x\mid x\leq-6\}\text{.}$$
We should check that some number less than $$-6$$ is a solution, that $$-6$$ itself is also a solution, and that some number greater than $$-6$$ is not a solution.
\begin{align*} \amp\amp -2x\amp\ge12\amp\amp\\ -2(\substitute{-7})\amp\wonder{\geq}12\amp -2(\substitute{-6})\amp\wonder{\ge}12\amp -2(\substitute{-5})\amp\wonder{\ge}12\\ 14\amp\confirm{\geq}12\amp 12\amp\confirm{\ge}12\amp 10\amp\reject{\ge}12 \end{align*}
Everything came out as expected, so our solution is reasonably checked.

#### Checkpoint1.6.8.

Solve the inequality $$-\frac{1}{2}z\lt2\text{.}$$ Graph the solution set on a number line. State the solution set using interval notation and using set-builder notation.
Explanation.
To solve this inequality, we need to multiply by $$-2$$ to each side. Since $$-2$$ is negative, we will need to change the direction of the inequality.
\begin{equation*} \begin{aligned} -\frac{1}{2}z\amp\lt2\\ \multiplyleft{-2}\left(-\frac{1}{2}z\right)\amp\highlight{\gt}\multiplyleft{-2}(2)\\ z\amp\gt-4 \end{aligned} \end{equation*}
Graphically, we have:
Using interval notation, we write the solution set by reading the number line from left to right. The solution set is $$(-4,\infty)\text{.}$$
Using set-builder notation, we write the solution with generic set braces, declaring $$z$$ to be the variable, and writing the condition that $$z$$ needs to meet: $$\{z\mid z\gt-4\}\text{.}$$
We should check that some number less than $$-4$$ is not a solution, that $$-4$$ itself is not a solution, and that some number greater than $$-4$$ is a solution.
\begin{equation*} \begin{aligned} \amp\amp-\frac{1}{2}z\amp\lt2\amp\amp\\ -\frac{1}{2}(\substitute{-10})\amp\wonder{\lt}2\amp-\frac{1}{2}(\substitute{-4})\amp\wonder{\lt}2\amp-\frac{1}{2}(\substitute{0})\amp\wonder{\lt}2\\ 5\amp\reject{\lt}2\amp2\amp\reject{\lt}2\amp0\amp\confirm{\lt}2 \end{aligned} \end{equation*}
Everything works out as expected, so we have checked our solution set reasonably well.

#### 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 make the process different from when you solve a similar simple linear equation?

#### 3.

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

### Exercises1.6.4Exercises

#### Skills Practice

##### Solve the Inequality.
Solve the inequality. Graph the solution set, and write the solution set using both interval notation and set-builder notation.
###### 1.
$${n+1}\gt{5}$$
###### 2.
$${u+9}\lt{8}$$
###### 3.
$${18+z}\geq{13}$$
###### 4.
$${6+F}\leq{9}$$
###### 5.
$${K-14}\lt{-15}$$
###### 6.
$${Q-3}\geq{-9}$$
###### 7.
$${12X}\leq{36}$$
###### 8.
$${19c}\gt{-38}$$
###### 9.
$${\frac{i}{2}}\gt{-2}$$
###### 10.
$${\frac{n}{10}}\gt{1}$$
###### 11.
$${-\frac{u}{2}}\gt{-3}$$
###### 12.
$${-\frac{z}{6}}\gt{2}$$
###### 13.
$${-2E}\lt{-4}$$
###### 14.
$${-10K}\leq{30}$$
###### 15.
$${-Q}\geq{5}$$
###### 16.
$${-X}\gt{1}$$
###### 17.
$${{\frac{7}{6}}c}\geq{-{\frac{14}{3}}}$$
###### 18.
$${{\frac{1}{8}}i}\leq{{\frac{5}{8}}}$$
###### 19.
$${-{\frac{2}{7}}n}\leq{{\frac{4}{7}}}$$
###### 20.
$${-{\frac{10}{3}}t}\lt{{\frac{40}{3}}}$$