Lesson Objectives

- Demonstrate an understanding of absolute value
- Learn how to solve advanced absolute value inequalities

## How to Solve Advanced Absolute Value Inequalities

In the last lesson, we learned how to solve advanced absolute value equations. Now, we will go one step further and learn how to solve advanced absolute value inequalities.

Example #1: Solve each inequality. $$|x - 3| + |x + 1| > 6$$ First, we will consider where each expression inside of the absolute value bars is equal to zero. $$x - 3=0$$ $$x=3$$ $$x + 1=0$$ $$x=-1$$ We will use these values to split the number line up into three intervals. We think about the value of the absolute value expression in each interval. If it is positive in the interval, we can simply drop the absolute value bars. If it is negative in the interval, we can drop the absolute value bars, but we must change the expression inside of the absolute value operation into its opposite. We can do this by wrapping our expression inside of parentheses and then placing a negative outside.

We will set up three different inequalities based on the expressions in the interval. From there, we will accept the solution if it lies in the interval and reject the solution if it's outside of the interval. Let's begin with: $$(-\infty, -1)$$ $$|x - 3| + |x + 1| > 6$$ Both expressions are negative in this interval: $$-(x-3) - (x + 1) > 6$$ $$-x + 3 - x - 1 > 6$$ $$-2x + 2 > 6$$ $$-2x > 4$$ $$x < -2$$ Since this is within our interval, we can accept this as part of our solution set. $$(-1, 3)$$ $$|x - 3| + |x + 1| > 6$$ One expression is positive and the other is negative in this interval: $$-(x-3) + (x + 1) > 6$$ $$-x + 3 + x + 1 > 6$$ $$4 > 6$$ This leads to a false statement, so we know that there isn't a solution in this interval. $$(3, \infty)$$ $$|x - 3| + |x + 1| > 6$$ Both expressions are positive in this interval: $$(x-3) + (x + 1) > 6$$ $$x - 3 + x + 1 > 6$$ $$2x - 2 > 6$$ $$2x > 8$$ $$x > 4$$ So our solution set for our inequality: $$x < -2 \hspace{.2em}\text{or}\hspace{.2em}x > 4$$ $$(\infty, -2) ∪ (4, \infty)$$

Example #1: Solve each inequality. $$|x - 3| + |x + 1| > 6$$ First, we will consider where each expression inside of the absolute value bars is equal to zero. $$x - 3=0$$ $$x=3$$ $$x + 1=0$$ $$x=-1$$ We will use these values to split the number line up into three intervals. We think about the value of the absolute value expression in each interval. If it is positive in the interval, we can simply drop the absolute value bars. If it is negative in the interval, we can drop the absolute value bars, but we must change the expression inside of the absolute value operation into its opposite. We can do this by wrapping our expression inside of parentheses and then placing a negative outside.

$$(-\infty, -1)$$ | $$(-1, 3)$$ | $$(3, \infty)$$ |
---|---|---|

$$-(x-3)$$ | $$-(x-3)$$ | $$(x-3)$$ |

$$-(x+1)$$ | $$(x+1)$$ | $$(x+1)$$ |

#### Skills Check:

Example #1

Solve each inequality. $$|x + 1| + |2x - 9|< 15$$

Please choose the best answer.

A

$$-\frac{16}{3}< x < \frac{27}{3}$$

B

$$-\frac{7}{5}< x < \frac{29}{3}$$

C

No Solution

D

All Real Numbers

E

$$-\frac{7}{3}< x < \frac{23}{3}$$

Example #2

Solve each inequality. $$|3x - 5| > |2x + 1|$$

Please choose the best answer.

A

No Solution

B

All Real Numbers

C

$$x < \frac{4}{5}\hspace{.1em}\text{or}\hspace{.1em}x > 6$$

D

$$x < \frac{13}{5}\hspace{.1em}\text{or}\hspace{.1em}x > 9$$

E

$$\frac{4}{5}< x < 6$$

Example #3

Solve each inequality. $$|2x - 3| + |3x - 1| < 5$$

Please choose the best answer.

A

$$-\frac{8}{5}< x < \frac{9}{5}$$

B

$$-\frac{1}{5}< x < \frac{9}{5}$$

C

No Solution

D

All Real Numbers

E

$$x < 3 \hspace{.1em}\text{or}\hspace{.1em}x > 7$$

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