An equation that equates two absolute value functions allows us to choose one of the absolute value functions and treat it as the constant. We then separate the equation into the "positive" version, $\displaystyle \tiny \small \frac{x-4}{3}=x+2$ , and the "negative" version,$\displaystyle \tiny \small \frac{x-4}{3} = -(x+2)$. Solving each equation, we obtain the solutions, $\displaystyle -5$ and $\displaystyle \tiny \tiny -\frac{1}{2}$, respectively.  

First, split $\displaystyle |x+3|=4$ into two possible scenarios according to the absolute value.

$\displaystyle x+3=4$

$\displaystyle x+3=-4$

Looking at $\displaystyle x+3=4$, we can solve for x by subtracting 3 from both sides, so that we get x = 1.

Looking at $\displaystyle x+3=-4$, we can solve for x by subtracting 3 from both sides, so that we get x = –7.

So therefore, the solution is x = –7, 1.

|x – 3| = 2 means that it can be separated into x – 3 = 2 and x – 3 = –2.

So both x = 5 and x = 1 work.

x – 3 = 2 Add 3 to both sides to get x = 5

x – 3 = –2 Add 3 to both sides to get x = 1

Because of the absolute value signs,

$\displaystyle \small x+2=12$ or $\displaystyle \small x+2=-12$

Subtract 2 from both sides of both equations:

$\displaystyle \small x+2-2=12-2$ or $\displaystyle \small x+2-2=-12-2$

$\displaystyle \small x=10$ or $\displaystyle \small x=-14$

$\displaystyle |x-4|=\frac{5}{8}$

There are two answers to this problem:

$\displaystyle x-4=\frac{5}{8}$

$\displaystyle x=\frac{5}{8}+4=4\frac{5}{8}$

and

$\displaystyle -(x-4)=\frac{5}{8}$

$\displaystyle 4-x=\frac{5}{8}$

$\displaystyle -x=\frac{5}{8}-4=-3\frac{3}{8}$

$\displaystyle x=-3\frac{3}{8}$

An absolute value expression differs from a normal expression only in its sign. Instead of being a positive or negative quantity, an absolute value represents a scalar distance from zero, so it does not have a sign. For example, $\displaystyle \left | -2\right |$ is the same as $\displaystyle \left | 2\right |$ because both represent a value 2 units away from zero. In this problem, $\displaystyle \left | -4x + 3\right |$ equals $\displaystyle \left | -5 \right |$, or 5. $\displaystyle \left | 3x - 14\right |$ equals 8. The final answer is $\displaystyle (5)(8)$ or 40.

$\displaystyle \textup{Set up two equations for absolute value equations:}$

$\displaystyle 2a-4=8\; \; \; \; \; \; \; \; \; \; \; \; 2a-4=-8$

$\displaystyle 2a=12\;\;\;\;\;\;\;\;\;\;\; \; \; \; \; \; \; 2a=-4$

$\displaystyle a=6\; \; \; \;\;\:\;\;\;\;\;\;\;\;\;\;\;\;\;\;a=-2$

$\displaystyle \left | 3x-7\right |+7 = -6$

To find a solution, subtract $\displaystyle 7$ first to isolate the absolute value expression.

$\displaystyle \left | 3x-7\right | = -13$

There is no value of $\displaystyle x$ that makes this true, as no number has a negative absolute value. The equation has no solution.

\(\displaystyle \left | 5x+3\right |-3=0\)

\(\displaystyle \left | 5x+3\right |=3\)

\(\displaystyle 5x+3 = 3 \;or\; 5x+3 = -3\)

\(\displaystyle First \;equation: 5x+3=3\)
\(\displaystyle 5x=0\)
\(\displaystyle x=0\)

\(\displaystyle Second\;equation:5x+3 = -3\)

\(\displaystyle 5x=-6\)

\(\displaystyle x = -6/5\)

\(\displaystyle Therefore, \;x =0, -6/5\)

$\displaystyle \left | 4x+19 \right | = 31$ can be rewritten as the compound statement:

$\displaystyle 4x+19= - 31$  or  $\displaystyle 4x+19= 31$

Solve each separately to obtain the solution set:

$\displaystyle 4x+19= - 31$

$\displaystyle 4x+19 - 19 = -31- 19$

$\displaystyle 4x = -50$

$\displaystyle 4x \div 4= -50 \div 4$

$\displaystyle x = -12.5$

$\displaystyle 4x+19= 31$

$\displaystyle 4x+19 - 19 = 31- 19$

$\displaystyle 4x = 12$

$\displaystyle 4x \div 4= 12 \div 4$

$\displaystyle x = 3$

So either $\displaystyle x = -12.5$ or $\displaystyle x = 3$