Exercise 5 Page 494

We are asked to find and graph the solution set for all possible values of $x$ in the given inequality. First, let's isolate the absolute value expression using the Properties of Inequality. Now, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than or equal to $32$ away from the midpoint in the positive direction and any number less than or equal to $32$ away from the midpoint in the negative direction.

$$\begin{array}{rl}\text{Absolute Value Inequality: }& |x+4|\le 32\\ \text{Compound Inequality: }& \text{-}32\le x+4\le 32\end{array}$$

We can split this compound inequality into two cases, one where $x+4$ is greater than or equal to $\text{-}32$ and one where $x+4$ is less than or equal to $32.$

$$\begin{array}{r}x+4\ge \text{-}32\text{and}x+4\le 32\end{array}$$

Let's isolate $x$ in both of these cases before graphing the solution set.

Case $1$

$x+4\ge \text{-}32$

SubIneq

$\text{LHS}-4\ge \text{RHS}-4$

$x\ge \text{-}36$

This inequality tells us that all values greater than or equal to $\text{-}36$ will satisfy the inequality.

Case $2$

$x+4\le 32$

SubIneq

$\text{LHS}-4\le \text{RHS}-4$

$x\le 28$

This inequality tells us that all values less than or equal to $28$ will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality.

$$\begin{array}{rl}\text{First Solution Set:}& \text{-}36\le x\\ \text{Second Solution Set:}& x\le 28\\ \text{Intersecting Solution Set:}& \text{-}36\le x\le 28\end{array}$$