Lesson Objectives
• Demonstrate an understanding of absolute value
• Learn how to solve nested absolute value equations
• Learn how to solve advanced absolute value equations

## How to Solve Advanced Absolute Value Equations

In this lesson, we want to discuss how to solve advanced absolute value equations. Up to this point, we have seen absolute value equations that can be solved pretty easily. In most cases, we just needed to isolate the absolute value operation and set up a compound equation with "or". We then solve the two resulting equations and have our answer. Now, we will push a little further and look at some examples that are much more tedious. Let’s begin by thinking about how to solve an absolute value equation where one absolute value operation is nested inside of another.

### How to solve Nested Absolute Value Equations

Example #1: Solve each equation. $$||2x - 1| + 5|=14$$ To solve this problem, let's first consider the larger absolute value expression: $$|2x - 1| + 5=14$$ $$\text{or}$$ $$|2x - 1| + 5=-14$$ Now, we can just solve each as a normal absolute value equation. $$|2x - 1| + 5=14$$ $$|2x - 1|=9$$ $$2x - 1=9$$ $$2x=10$$ $$x=5$$ $$\text{or}$$ $$2x - 1=-9$$ $$2x=-8$$ $$x=-4$$ So far, our solutions are: $$x=-4, 5$$ Now, let's look at the other part: $$|2x - 1| + 5=-14$$ $$|2x - 1|=-19$$ This part doesn't have a solution. We can conclude that we just have two solutions: $$x=-4, 5$$

### Solving Advanced Absolute Value Equations

Here, we want to look at an example where we have two absolute value operations and a loose number. The loose number stops us from setting the two absolute value operations equal to each other.
Example #2: Solve each equation. $$|x - 4| + |x - 1|=12$$ First, we need to find the values that make the expressions inside of each absolute value operation equal to zero. This is where a sign change would occur, from positive to negative or from negative to positive. This can only happen at a value of zero. $$x - 4=0$$ $$x=4$$ $$x - 1=0$$ $$x=1$$ Next, 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.
$$(-\infty, 1)$$ $$(1, 4)$$ $$(4, \infty)$$
$$-(x-1)$$$$(x-1)$$$$(x-1)$$
$$-(x-4)$$$$-(x-4)$$$$(x-4)$$
We will set up three different equations 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: $$(-\infty, 1)$$ $$|x - 4| + |x - 1|=12$$ Both expressions are negative in this interval: $$-(x - 4) - (x - 1)=12$$ $$-x + 4 - x + 1=12$$ $$-2x + 5=12$$ $$-2x=7$$ $$x=-\frac{7}{2}$$ Since -7/2 or -3.5 is in our interval, we can accept this as part of our solution set. Let's move on to the next interval: $$(1,4)$$ $$|x - 4| + |x - 1|=12$$ One expression is positive and the other is negative in this interval: $$-(x - 4) + (x - 1)=12$$ $$-x + 4 + x - 1=12$$ $$3=12$$ This leads to a false statement, so there won't be a solution in this interval. Let's move on to the final interval: $$(4, \infty)$$ Both expressions are positive, so just drop the absolute value bars: $$|x - 4| + |x - 1|=12$$ $$x - 4 + x - 1=12$$ $$2x - 5=12$$ $$2x=17$$ $$x=\frac{17}{2}$$ Since 17/2 or 8.5 is in our interval, we can accept this as part of our solution set. We can state our solution as: $$x=-\frac{7}{2}, \frac{17}{2}$$

#### Skills Check:

Example #1

Solve each equation. $$||2x + 7| - 3|=20$$

A
$$x=-7, -5, 3, 1$$
B
$$x=13, \frac{1}{2}$$
C
$$x=-15, 8$$
D
$$x=15, -8$$
E
$$x=-13, \frac{1}{2}$$

Example #2

Solve each equation. $$|2x - 3| + |x - 7|=25$$

A
$$x=-\frac{5}{3}, -1$$
B
$$x=-1, \frac{3}{2}$$
C
$$x=-5, \frac{35}{3}$$
D
$$x=-1, 3, 7, 9$$
E
$$x=-4, -2$$

Example #3

Solve each equation. $$|9x - 1| + |x + 2|=8$$

A
$$x=-\frac{5}{8}, \frac{7}{10}$$
B
$$x=-4, \frac{1}{2}$$
C
$$x=-7, 10$$
D
$$x=-19, \frac{3}{5}$$
E
$$x=-8, 7$$