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 non-negative (zero or some positive value) in the interval, we can simply drop the absolute value bars. If it is negative in the interval, we will also drop the absolute value bars, but we must first 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.
$$(-\infty, -1)$$ $$[-1, 3)$$ $$[3, \infty)$$
$$-(x-3)$$$$-(x-3)$$$$(x-3)$$
$$-(x+1)$$$$(x+1)$$$$(x+1)$$
We will set up three different inequalities based on the expressions in the interval. From there, we will accept the solution if it's 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 solution is within our interval, we can accept this as part of our solution set. $$[-1, 3)$$ $$|x - 3| + |x + 1| > 6$$ One expression is non-negative (zero or some positive value) 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 non-negative (zero or some positive value) in this interval: $$(x-3) + (x + 1) > 6$$ $$x - 3 + x + 1 > 6$$ $$2x - 2 > 6$$ $$2x > 8$$ $$x > 4$$ Since this solution is within our interval, we can accept this as part of our solution set. We will now combine our solutions and state our solution set for our inequality: $$x < -2 \hspace{.2em}\text{or}\hspace{.2em}x > 4$$ $$(\infty, -2) ∪ (4, \infty)$$ Example #2: Solve each inequality. $$|2x - 1| ≥ |3x + 5|$$ First, we will consider where each expression inside of the absolute value bars is equal to zero. $$2x - 1=0$$ $$x=\frac{1}{2}$$ $$3x + 5=0$$ $$x=-\frac{5}{3}$$ 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 non-negative (zero or some positive value) in the interval, we can simply drop the absolute value bars. If it is negative in the interval, we will also drop the absolute value bars, but we must first 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.
$$\left(-\infty, -\frac{5}{3}\right)$$ $$\left[-\frac{5}{3}, \frac{1}{2}\right)$$ $$\left[\frac{1}{2}, \infty\right)$$
$$-(2x - 1)$$$$-(2x - 1)$$$$(2x - 1)$$
$$-(3x + 5)$$$$(3x + 5)$$$$(3x + 5)$$
We will set up three different inequalities based on the expressions in the interval. From there, we will accept the solution if it's in the interval and reject the solution if it's outside of the interval. Let's begin with: $$\left(-\infty, -\frac{5}{3}\right)$$ $$|2x - 1| ≥ |3x + 5|$$ Both expressions are negative in this interval: $$-(2x - 1) ≥ -(3x + 5)$$ $$-2x + 1 ≥ -3x - 5$$ $$x ≥ -6$$ This solution is within our interval but only up to anything less than -5/3. $$-6 ≤ x < -\frac{5}{3}$$ Note: we have excluded the -5/3 in this interval. $$\left[-\frac{5}{3}, \frac{1}{2}\right)$$ $$|2x - 1| ≥ |3x + 5|$$ One expression is non-negative (either zero or some positive value) and the other is negative in this interval: $$-(2x - 1) ≥ 3x + 5$$ $$-2x + 1 ≥ 3x + 5$$ $$-5x ≥ 4$$ $$x ≤ -\frac{4}{5}$$ The solution is within our interval but only down to -5/3.
Note: we have included the -5/3 in this interval. $$-\frac{5}{3}≤ x ≤ -\frac{4}{5}$$ $$\left[\frac{1}{2}, \infty\right)$$ Both expressions are non-negative (either zero or some positive value) in this interval: $$2x - 1 ≥ 3x + 5$$ $$-x ≥ 6$$ $$x ≤ -6$$ Since this solution is not within our interval, we will not accept this as part of our solution set. We will now combine our solutions and state our solution set for our inequality: $$-6 ≤ x ≤ -\frac{4}{5}$$ Example #3: Solve each inequality. $$\left|\frac{x - 1}{x - 5}\right| ≤ 1$$ Here, we will use our rule: $$|u| < k$$ $$-k < u < k$$ This will setup as: $$-1 ≤ \frac{x - 1}{x - 5}≤ 1$$ We need to solve two inequalities and then look for the intersection of the solution sets: $$\frac{x - 1}{x - 5}≥ -1$$ $$\text{and}$$ $$\frac{x - 1}{x - 5}≤ 1$$ Let's start with the inequality at the top. $$\frac{x - 1}{x - 5}≥ -1$$ $$\frac{x - 1}{x - 5}+ 1 ≥ 0$$ $$\frac{x - 1}{x - 5}+ \frac{x - 5}{x - 5}≥ 0$$ $$\frac{x - 1 + x - 5}{x - 5}≥ 0$$ $$\frac{2x - 6}{x - 5}≥ 0$$ We find the critical values by setting the numerator and denominator equal to zero and solving the resulting equations. $$2x - 6=0$$ $$2x=6$$ $$x=3$$ $$x - 5=0$$ $$x=5$$ We have two critical values 3 and 5. We can split the number line up into three intervals, A where we are less than 3, B where we are between 3 and 5, and C where we are greater than 5. Let's first test in A:
We can use 0: $$\frac{2(0) - 6}{(0) - 5}≥ 0$$ $$\frac{6}{5}≥ 0$$ This is a true statement since a positive is always greater than 0.
Let's next test in B:
We can use 4: $$\frac{2(4) - 6}{(4) - 5}≥ 0$$ $$-2 ≥ 0$$ This is a false statement since a negative is never greater than or equal to 0.
Let's next test in C: We can use 6: $$\frac{2(6) - 6}{(6) - 5}≥ 0$$ $$6 ≥ 0$$ This is a true statement since a positive is always greater than 0. Our solution for this part is x ≤ 3 or x > 5. Note, 5 is not included as this violates the domain (division by zero).
Let's now work on the inequality at the bottom. $$\frac{x - 1}{x - 5}- 1 ≤ 0$$ $$\frac{x - 1}{x - 5}- \frac{x - 5}{x - 5}≤ 0$$ $$\frac{x - 1 - (x - 5)}{x - 5}≤ 0$$ $$\frac{4}{x - 5}≤ 0$$ We find the critical values by setting the numerator and denominator equal to zero and solving the resulting equations. In this case, the numerator is just 4, so we only need to set our denominator equal to zero. $$x - 5=0$$ $$x=5$$ We have one critical value of 5. We can split the number line up into two intervals, A where we are less than 5, and B where we are greater than 5. Let's first test in A:
We can use 0: $$\frac{4}{(0) - 5}≤ 0$$ $$-\frac{4}{5}≤ 0$$ This is a true statement since a negative is always less than 0.
Let's next test in B:
We can use 6: $$\frac{4}{(6) - 5}≤ 0$$ $$4 ≤ 0$$ This is a false statement since a positive is never less than or equal to 0. Our solution for this part is x < 5. Note, 5 is not included as this violates the domain (division by zero). Let's put our two solutions together using "and", recall this is the intersection of the solution sets. $$x ≤ 3 \hspace{.2em}\text{or}\hspace{.2em}x > 5$$ $$\text{and}$$ $$x < 5$$ The intersection of the two solution sets would be x ≤ 3. $$x ≤ 3$$ $$(-∞, 3]$$ #### Skills Check:

Example #1

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

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. $$|3x - 5| > |2x + 1|$$

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. $$\left|\frac{x + 3}{x - 2}\right| ≤ 4$$

A
$$-\frac{8}{5}< x < \frac{9}{5}$$
B
$$x ≤ 1 \hspace{.1em}\text{or}\hspace{.1em}x ≥ \frac{11}{3}$$
C
No Solution
D
All Real Numbers
E
$$x ≤ -1 \hspace{.1em}\text{or}\hspace{.1em}x > 2$$         