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Section 2.6 Solving One-Step Inequalities

We have learned how to check whether a specific number is a solution to an equation or inequality. In this section, we will begin learning how to find the solution(s) to basic inequalities ourselves.

With one small complication, we can use very similar properties to Fact 2.5.11 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 2.6.1, where the numbers \(2\) and \(4\) are multiplied by the negative number \(-1\text{.}\)

a number line with -4, -2, 2, and 4 marked; text indicated that 2 < 4; arrows swing from 2 to -2 and 4 to -4, indicating multiplication by -1; text indicates that -2 > -4
Figure 2.6.1. Multiplying two numbers by a negative number, and how their relationship changes

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

In general, we must apply the following property when solving an inequality.

Example 2.6.3.

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 1.7.

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 equation, there is usually exactly one solution. When we solve a linear inequality, there are usually infinitely many solutions. For this example, any number smaller than \(-6\) 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

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 \(-6\) satisfies the inequality, and also that one number less than \(-6\) (any number, your choice) is a solution.

\begin{align*} -2x\amp\geq12\amp -2x\amp\ge12\\ -2(\substitute{-6})\amp\stackrel{?}{\geq}12\amp -2(\substitute{-7})\amp\stackrel{?}{\ge}12\\ 12\amp\stackrel{\checkmark}{\geq}12\amp 14\amp\stackrel{\checkmark}{\ge}12 \end{align*}

Thus, both \(-6\) and \(-7\) are solutions. It's important to note this doesn't directly verify that all solutions to this inequality check. But it is evidence that our solution is correct, and it's valuable in that making these two checks would likely help us catch an error if we had made one. Consult your instructor to see if you're expected to check your answer in this manner.

Example 2.6.4.

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

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 \(-2\) is not a solution, but that some number less than \(-2\) is a solution.

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

So our solution is reasonably checked.

Checkpoint 2.6.5.
Checkpoint 2.6.6.

Exercises Exercises

Review and Warmup
1.

Add the following.

  1. \(-10+(-3)\)

  2. \(-6+(-3)\)

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

2.

Add the following.

  1. \(-10+(-1)\)

  2. \(-5+(-4)\)

  3. \(-1+(-7)\)

3.

Add the following.

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

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

  3. \(8+(-8)\)

4.

Add the following.

  1. \(3+(-10)\)

  2. \(8+(-4)\)

  3. \(8+(-8)\)

5.

Add the following.

  1. \(-7+4\)

  2. \(-1+10\)

  3. \(-4+4\)

6.

Add the following.

  1. \(-9+4\)

  2. \(-2+7\)

  3. \(-4+4\)

7.

Evaluate the following.

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

  2. \(\frac{16}{-4}\)

  3. \(\frac{-32}{8}\)

8.

Evaluate the following.

  1. \(\frac{-14}{-2}\)

  2. \(\frac{70}{-7}\)

  3. \(\frac{-64}{8}\)

9.

Do the following multiplications.

  1. \(15 \cdot \frac{2}{5}\)

  2. \(20 \cdot \frac{2}{5}\)

  3. \(25 \cdot \frac{2}{5}\)

10.

Do the following multiplications.

  1. \(18 \cdot \frac{2}{9}\)

  2. \(27 \cdot \frac{2}{9}\)

  3. \(36 \cdot \frac{2}{9}\)

11.

Evaluate the following.

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

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

  3. \(\frac{150}{-150}\)

  4. \(\frac{-19}{-19}\)

  5. \(\frac{13}{0}\)

  6. \(\frac{0}{-8}\)

12.

Evaluate the following.

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

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

  3. \(\frac{190}{-190}\)

  4. \(\frac{-11}{-11}\)

  5. \(\frac{13}{0}\)

  6. \(\frac{0}{-4}\)

Solving One-Step Inequalities using Addition/Subtraction

For each problem below, solve the inequality and, in addition to writing both set-builder and interval notations for each solution set, draw a graph of each solution set on a number line.

13.

\({x+3} > {10}\)

;

14.

\({x+4} > {8}\)

;

15.

\({x-4} \leq {7}\)

;

16.

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

;

17.

\({5} \leq {x+8}\)

;

18.

\({1} \leq {x+6}\)

;

19.

\({1} > {x-10}\)

;

20.

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

;

Solving One-Step Inequalities using Multiplication/Division

For each problem below, solve the inequality and, in addition to writing both set-builder and interval notations for each solution set, draw a graph of each solution set on a number line.

21.

\({3x} \leq {6}\)

;

22.

\({4x} \leq {16}\)

;

23.

\({5x} > {7}\)

;

24.

\({5x} > {8}\)

;

25.

\({-5x} \geq {20}\)

;

26.

\({-5x} \geq {15}\)

;

27.

\({4} \geq {-2x}\)

;

28.

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

;

29.

\({4} \lt {-x}\)

;

30.

\({5} \lt {-x}\)

;

31.

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

;

32.

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

;

33.

\({{\frac{8}{7}}x} > {8}\)

;

34.

\({{\frac{9}{7}}x} > {9}\)

;

35.

\({-{\frac{10}{3}}x} \leq {20}\)

;

36.

\({-{\frac{1}{10}}x} \leq {4}\)

;

37.

\({-4} \lt {{\frac{2}{7}}x}\)

;

38.

\({-9} \lt {{\frac{3}{4}}x}\)

;

39.

\({-5} \lt {-{\frac{5}{9}}x}\)

;

40.

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

;

41.

\({4x} > {-8}\)

;

42.

\({4x} > {-16}\)

;

43.

\({-15} \lt {-5x}\)

;

44.

\({-10} \lt {-5x}\)

;

45.

\({3} > {\frac{x}{14}}\)

;

46.

\({1} > {\frac{x}{8}}\)

;

47.

\({2} > {\frac{x}{-10}}\)

;

48.

\({3} > {\frac{x}{-8}}\)

;

49.

\({\frac{5}{6}} \geq {\frac{x}{18}}\)

;

50.

\({\frac{3}{8}} \geq {\frac{x}{40}}\)

;

51.

\({-\frac{z}{16}} \lt {-\frac{7}{8}}\)

;

52.

\({-\frac{z}{50}} \lt {-\frac{7}{10}}\)

;