In finding the values for $\displaystyle x$ where $\displaystyle \left | x -3 \right | > \left |x+12 \right |$, break the comparison of these two absolute value expressions into the four possible ways this could potentially be satisfied.

The first possibility is described by the inequality:

$\displaystyle x-3 > x+12$

If you think of a number line, it is evident that there is no solution to this inequality since there will never be a case where subtracting from $\displaystyle x$ will lead to a greater number than adding to $\displaystyle x$.

The second possibility, wherein $\displaystyle x-3$ is negative and converted to its opposite to being an absolute value expression but $\displaystyle x+12$ is positive and requires no conversion, can be represented by the inequality (where the sign is inverted due to multiplication by a negative):

$\displaystyle -(x-3) < x+12$

We can simplify this inequality to find that $\displaystyle x< -4.5$ satisfies the conditions where $\displaystyle \left | x -3 \right | > \left |x+12 \right |$.

The third possibility can be represented by the following inequality (where the sign is inverted due to multiplication by a negative):

$\displaystyle x-3 < -(x+12)$

This is again simplified to $\displaystyle x< -4.5$ and is redundant with the above inequality.

The final possibility is represented by the inequality

$\displaystyle -(x-3) > -(x+12)$

This inequality simplifies to $\displaystyle 3 - x > -12 - x$. Rewriting this as $\displaystyle - x + 3 > - x -12$ makes it evident that this inequality is true of all real numbers. This does not provide any additional conditions on how to satisfy the original inequality.

The only possible condition that satisfies the inequality is that which arises in two of the tested cases, when $\displaystyle x < -4.5$.

The absolute value is the distance from a given number to 0. In our example, we are given -3. This number is 3 units away from 0, and thus the absolute value of -3 is 3. 

If a number is negative, its absolute value will be the positive number with the same magnitude. If a number is positive, it will be its own absolute value. 

Notice that the equation $\displaystyle 2x + \left | x - 3 \right | = -12$ has an $\displaystyle x$ term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case $\displaystyle 2x$) must be negative (meaning $\displaystyle x$ must be negative). 

Since $\displaystyle x$ will be a negative number, the expression within the absolute value $\displaystyle x-3$ will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by $\displaystyle -1$, since any negative value becomes its opposite when taking the absolute value.

$\displaystyle 2x + \left | x - 3 \right | = -12$

$\displaystyle \rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for $\displaystyle x$ gives the answer:

$\displaystyle \rightarrow 2x -x+3 = -12$

$\displaystyle \rightarrow x + 3 = -12$

$\displaystyle \rightarrow x = -15$

The absolute value measures the distance from zero to the given point. 

In this case, since $\displaystyle |x|=12$, $\displaystyle x=12$ or $\displaystyle x=-12$, as both values are twelve units away from zero.

$\displaystyle \textup{Set up two equations and solve for both solutions:}$

$\displaystyle 3b-4=11\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3b-4 = -11$

$\displaystyle 3b = 15\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \; \; \; \; \; \; 3b=-7$

$\displaystyle b = 5\;\;\:\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;b=-\frac{7}{3}$

The absolute value can never be negative, so the equation is ONLY valid at zero.

The equation to solve becomes $\displaystyle x-5=0$. 

$\displaystyle x=5$

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

$\displaystyle \left | 2x-5\right |=3x$

$\displaystyle 2x-5=3x$ 

and 

$\displaystyle 2x-5=-(3x)$

This gives us:

$\displaystyle -5=x$ and $\displaystyle 5x=5$

$\displaystyle x=-5\ and\ 1$

However, this question has an $\displaystyle x$ outside of the absolute value expression, in this case $\displaystyle 3x$. Thus, any negative value of $\displaystyle x$ will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, $\displaystyle x=-5$ is an extraneous solution, as $\displaystyle \left | 2x-5\right |$ cannot equal a negative number.

Our final solution is then

$\displaystyle x=1$

$\displaystyle 3\left | x+2 \right |=18$

Divide both sides by 3.

$\displaystyle \left | x+2 \right |=6$

Consider both the negative and positive values for the absolute value term.

$\displaystyle x+2=6\ or\ x+2=-6$

Subtract 2 from both sides to solve both scenarios for $\displaystyle x$.

$\displaystyle x=-8\ or\ x=4$

To find the $\displaystyle x$-intercepts, we must set $\displaystyle y=0$.

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

$\displaystyle y=\left | 2x-6\right |-8$

$\displaystyle 0=\left | 2x-6\right |-8$

$\displaystyle \left | 2x-6\right |=8$

Now we must set up our two scenarios:

$\displaystyle 2x-6=8$ and $\displaystyle 2x-6=-8$

$\displaystyle 2x=14$ and $\displaystyle 2x=-2$

$\displaystyle x=7$ and $\displaystyle x=-1$