Absolute value inequalities involving > and \( \ge \) are solved as follows.

\[\begin{align*}\left| p \right| > d\,\,\,\,\hspace{0.25in}\,\,\,\, & \Rightarrow \hspace{0.25in}\,\,\,\,\,\,p < - d\hspace{0.25in}{\rm{or}}\hspace{0.25in}p > d\\ \left| p \right| \ge d\,\,\,\,\hspace{0.25in}\,\,\,\, & \Rightarrow \hspace{0.25in}\,\,\,\,\,\,p \le - d\hspace{0.25in}{\rm{or}}\hspace{0.25in}p \ge d\end{align*}\]

Note that you get two separate inequalities in the solution. That is the way that it must be. You can NOT put these together into a single inequality. Once I get the solution to this problem I’ll show you why that is.

Here is the solution

\[\begin{align*}\begin{aligned}4x + 5 & < - 3\\ 4x & < - 8\\ x & < - 2\end{aligned} & \hspace{0.75in}{\rm{OR}}\hspace{0.5in} & \begin{aligned}4x + 5 & > 3\\ 4x & > - 2\\ x & > - \frac{1}{2}\end{aligned}\end{align*}\]

So the solution to this inequality will be \(x\)’s that are less than -2 or greater than \( - \frac{1}{2}\).

Now, as I mentioned earlier you CAN NOT write the solution as the following double inequality.

\[ - 2 > x > - \frac{1}{2}\]

When you write a double inequality (as we have here) you are saying that \(x\) will be a number that will simultaneously satisfy both parts of the inequality. In other words, in writing this I’m saying that \(x\) is some number that is less than -2 and AT THE SAME TIME is greater than \( - \frac{1}{2}\). I know of no number for which this is true. So, this is simply incorrect. Don’t do it. This is however, a VERY common mistake that students make when solving this kind of inequality.