The first thing we must do is get the absolute value alone: $\displaystyle 5\left | x-7\right |+14< 39$

$\displaystyle 5\left | x-7\right |< 25$

$\displaystyle \left | x-7\right |< 5$

When we're working with absolute values, we are actually solving two equations:

$\displaystyle x-7< 5$     and $\displaystyle x-7> -5$

Fortunately, these can be written as one equation:

$\displaystyle -5< x-7< 5$

If you feel more comfortable solving the equations separately then go ahead and do so.

$\displaystyle 2< x< 12$

              To get $\displaystyle x$ alone, we added $\displaystyle 7$ on both sides of the inequality sign

$\displaystyle \left | y\right | + 18 \leq 11.3$

$\displaystyle \left | y\right |\leq -6.7$

Because Absolute Value must be a non-negative number, there is no solution to this Absolute Value inequality.

The following absolute value inequality can be used to assess the bowling balls that are tolerable:

$\displaystyle \left | w-8\right |\geq3$

$\displaystyle 4\left |(x+0.3) \right | - 3< 11$

$\displaystyle 4x + 1.2 - 3 < 11$

$\displaystyle 4x -1.8< 11$

$\displaystyle 4x -1.8 +1.8< 11 + 1.8$

$\displaystyle 4x< 12.8$

$\displaystyle \frac{4x}{4} < \frac{12.8}{4}$

$\displaystyle x< 3.2$

$\displaystyle 4x + 1.2 - 3> -11$

$\displaystyle 4x - 1.8> -11$

$\displaystyle 4x - 1.8 + 1.8> -11 + 1.8$

$\displaystyle 4x > -9.2$

$\displaystyle \frac{4x}{4}>\frac{-9.2}{4}$

$\displaystyle x> -2.3$

The correct answer is $\displaystyle x< 3.2$ and $\displaystyle x>2.3$

$\displaystyle \left | 6+x\right | -4 < 0$

$\displaystyle 6 + x -4 < 0$

$\displaystyle 6 + x < 4$

$\displaystyle 6-6 + x < 4-6$

$\displaystyle x< -2$

$\displaystyle 6 + x > -4$

$\displaystyle 6-6 + x > -6-4$

$\displaystyle x>-10$

The correct answer is $\displaystyle x< -2$ and $\displaystyle x>-10.$

$\displaystyle \left | \frac{x-6}{2} \right |\leq3$

$\displaystyle \frac{2}{1} \left | \frac{x-6}{2} \right | \leq 2\times3$

$\displaystyle x-6 \leq 6$

$\displaystyle x\leq 12$

$\displaystyle \frac{2}{1} \left | \frac{x-6}{2} \right | \geq 2 \times -3$

$\displaystyle x-6 \geq -6$

$\displaystyle x\geq 0$

The correct answer is $\displaystyle x\leq12$  and $\displaystyle x\geq 0.$

$\displaystyle \left | x-5\right |\leq 7$

$\displaystyle x-5 \leq 7$

$\displaystyle x-5 + 5 \leq 7 +5$

$\displaystyle x\leq 12$

$\displaystyle x-5 \geq -7$

$\displaystyle x-5 +5 \geq -7 +5$

$\displaystyle x\geq -2$

The correct answer is $\displaystyle x\leq12$ and $\displaystyle x\geq -2.$

The Absolute Value Inequality that can assess which cell phones are tolerable is:

$\displaystyle \left | w-0.4\right | \leq 9$

Step 1: Separate the equation into two equations:

First Equation: $\displaystyle |x-1|\le 6$
Second Equation: $\displaystyle -(x-1)\ge -5$

Step 2: Solve the first equation

$\displaystyle x-1\le 5$
$\displaystyle x-1+1\le 5+1$
$\displaystyle x \le 6$

Step 3: Solve the second equation

$\displaystyle -(x+1)\ge -5$
$\displaystyle -x+1\ge -5$
$\displaystyle -x+1-1\ge -5-1$

$\displaystyle -x \ge -6$

$\displaystyle \rightarrow x \le 6$

The solution is $\displaystyle x\le 6$

Before expanding the quantity within absolute value brackets, it is best to simplify the "actual values" in the problem. Thus $\displaystyle \left | x-10 \right |+3< 12$ becomes:

$\displaystyle \left | x-10 \right |< 9$

From there, note that the absolute value means that one of two things is true: $\displaystyle x-10< 9$ or $\displaystyle x-10>-9$. You can therefore solve for each possibility to get all possible solutions. Beginning with the first:

$\displaystyle x-10< 9$ means that:

$\displaystyle x< 19$

For the second:

$\displaystyle x-10>-9$ means that:

$\displaystyle x>1$

Note that the two solutions can be connected by putting the inequality signs in the same order:

$\displaystyle 1< x< 19$