To solve this inequality, it is best to break it up into two separate inequalities to eliminate the absolute value function:

$\displaystyle 5x-10 < 25$ or $\displaystyle 5x-10 > -25$.

Then, solve each one separately:

$\displaystyle 5x-10 < 25 \Rightarrow 5x < 35 \Rightarrow x < 7$

$\displaystyle 5x-10 > -25 \Rightarrow 5x > -15 \Rightarrow x > -3$

Combining these solutions gives: $\displaystyle -3 < x < 7$

The inequality compares an absolute value function with a negative integer. Since the absolute value of any real number is greater than or equal to 0, it can never be less than a negative number. Therefore, $\displaystyle \small \left |15x -39 \right| < -42$ can never happen. There is no solution. 

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

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 $\displaystyle x$.

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

We start by finding the midpoint of the interval, which is enclosed by 60% of 204 and 80% of 204.

$\displaystyle .6(204)=122.4\approx 122$

$\displaystyle .8(204)=163.2\approx 163$

We find the midpoint, or average, of these endpoints by adding them and dividing by two:

$\displaystyle \frac{122+163}{2}=142.5$

142.5 is exactly 20.5 units away from both endpoints, 122 and 163. Since we are looking for the range of numbers between 122 and 163, all possible values have to be within 20.5 units of 142.5. If a number is greater than 20.5 units away from 142.5, either in the positive or negative direction, it will be outside of the [122, 163] interval. We can express this using absolute value in the following way:

$\displaystyle \left | x-142.5 \right |\leq 20.5$

We start by finding the midpoint of the interval, which is enclosed by 90 and 210. We find the midpoint, or average, of these two endpoints by adding them and dividing by two:

$\displaystyle \frac{90+210}{2}=150$

150 is exactly 60 units away from both endpoints, 90 and 210. Since we are looking for the range of numbers that fall in between 90 and 210, this means that any possible value can't be more than 60 units away from 150. If a number is more than 60 units away from 150, in either the increasing or decreasing direction, it will be outside of the [90, 210] interval. We can express this using absolute value in the following way:

$\displaystyle \left | x-150 \right |\leq 60$

We start problems like these by finding the midpoint of the two endpoints, which in this case are 21 and 65. We find the midpoint, or average, by adding them and dividing by two:

$\displaystyle \frac{21+65}{2}=43$

43 is right in between 21 and 65, and is exactly 22 units away from each endpoint. Since we are looking for all numbers which fall outside of the the [21, 65] interval, we are looking for values which are further than 22 units away from 43, in both the positive and negative directions. Using absolute value, we express this as:

$\displaystyle \left | x-43 \right |\geq 22$

When you subtract 43 from any number larger than 65, the absolute value of the result will be greater than 22. Similarly, when you subtract 43 from any number less than 21, the absolute value of the result will be greater than 22.

Using the "greater than or equal to" sign is necessary in order to include the endpoints, 21 and 65, in our set of allowed ages.

Solve for positive values by ignoring the absolute value. Solve for negative values by switching the inequality and adding a negative sign to 7.

First, subtract 5 from both sides to get the absolute value expression alone.

$\displaystyle | 4x - 7 | + 5 = 86$

$\displaystyle | 4x - 7 | = 81$

Split this into two linear equations:

$\displaystyle 4x - 7 = 81$

$\displaystyle 4x = 88$

$\displaystyle x = 22$

or 

$\displaystyle 4x - 7 = -81$

$\displaystyle 4x = -74$

$\displaystyle x = -74 \div 4 = -18.5$

The solution set is $\displaystyle \begin{Bmatrix} -18.5, 22 \end{Bmatrix}$

The absolute value gives two problems to solve. Remember to switch the "less than" to "greater than" when comparing the negative term.

$\displaystyle x-2\leq 4$ or $\displaystyle x-2\geq -4$

Solve each inequality separately by adding $\displaystyle +2$ to all sides.

$\displaystyle x\leq 6$ or $\displaystyle x\geq -2$

This can be simplified to the format $\displaystyle -2 \leq x\leq 6$.

$\displaystyle \textup{Set up two inequalities. Reverse the inequality sign for the negative one.}$

$\displaystyle 2x+4\leq10\;\;\;\;\;\;\;\;\;\;\;\;\;2x+4\geq-10$

$\displaystyle 2x\leq6\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \; 2x\geq-14$

$\displaystyle x\leq3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x\geq-7$