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### Section 2.15 : Absolute Value Inequalities

In the previous section we solved equations that contained absolute values. In this section we want to look at inequalities that contain absolute values. We will need to examine two separate cases.

#### Inequalities Involving < and \( \le \)

As we did with equations let’s start off by looking at a fairly simple case.

\[\left| p \right| \le 4\]This says that no matter what \(p\) is it must have a distance of no more than 4 from the origin. This means that \(p\) must be somewhere in the range,

\[ - 4 \le p \le 4\]We could have a similar inequality with the < and get a similar result.

In general, we have the following formulas to use here,

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{If }}\left| p \right| \le b,\,\,\,b > 0\,\,\,\,\,\,\,\,\,\,\,{\mbox{then }} - b \le p \le b\\ {\mbox{If }}\left| p \right| < b,\,\,\,b > 0\,\,\,\,\,\,\,\,\,\,\,{\mbox{then }} - b < p < b\end{align*}}\]Notice that this does **require** \(b\) to be positive just as we did with equations.

Let’s take a look at a couple of examples.

- \(\left| {2x - 4} \right| < 10\)
- \(\left| {9m + 2} \right| \le 1\)
- \(\left| {3 - 2z} \right| \le 5\)

There really isn’t much to do other than plug into the formula. As with equations \(p\) simply represents whatever is inside the absolute value bars. So, with this first one we have,

\[ - 10 < 2x - 4 < 10\]Now, this is nothing more than a fairly simple double inequality to solve so let’s do that.

\[\begin{array}{c} - 6 < 2x < 14\\ - 3 < x < 7\end{array}\]The interval notation for this solution is \(\left( { - 3,7} \right)\).

b \(\left| {9m + 2} \right| \le 1\) Show Solution

Not much to do here.

\[\begin{array}{c} - 1 \le 9m + 2 \le 1\\ - 3 \le 9m \le - 1\\ \displaystyle - \frac{1}{3} \le m \le - \frac{1}{9}\end{array}\]The interval notation is \(\left[ { - \frac{1}{3}, - \frac{1}{9}} \right]\).

c \(\left| {3 - 2z} \right| \le 5\) Show Solution

We’ll need to be a little careful with solving the double inequality with this one, but other than that it is pretty much identical to the previous two parts.

\[\begin{array}{c} -5 \le 3 - 2z \le 5\\ - 8 \le - 2z \le 2\\ 4 \ge z \ge - 1\end{array}\]In the final step don’t forget to switch the direction of the inequalities since we divided everything by a negative number. The interval notation for this solution is \(\left[ { - 1,4} \right]\).

#### Inequalities Involving > and \( \ge \)

Once again let’s start off with a simple number example.

\[\left| p \right| \ge 4\]This says that whatever \(p\) is it must be at least a distance of 4 from the origin and so \(p\) must be in one of the following two ranges,

\[p \le - 4\hspace{0.25in}{\mbox{or}}\hspace{0.25in}p \ge 4\]Before giving the general solution we need to address a common mistake that students make with these types of problems. Many students try to combine these into a single double inequality as follows,

\[ - 4 \ge p \ge 4\]While this may seem to make sense we can’t stress enough that THIS IS NOT CORRECT!! Recall what a double inequality says. In a double inequality we require that both of the inequalities be satisfied simultaneously. The double inequality above would then mean that \(p\) is a number that is simultaneously smaller than -4 and larger than 4. This just doesn’t make sense. There is no number that satisfies this.

These solutions must be written as two inequalities.

Here is the general formula for these.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{If }}\left| p \right| \ge b,\,\,\,b > 0\,\,\,\,\,\,\,\,\,\,\,{\mbox{then }}p \le - b{\mbox{ or }}p \ge b\\ {\mbox{If }}\left| p \right| > b,\,\,\,b > 0\,\,\,\,\,\,\,\,\,\,\,{\mbox{then }}p < - b{\mbox{ or }}p > b\end{align*}}\]Again, we will *require* that \(b\) be a positive number here. Let’s work a couple of examples.

- \(\left| {2x - 3} \right| > 7\)
- \(\left| {6t + 10} \right| \ge 3\)
- \(\left| {2 - 6y} \right| > 10\)

Again, \(p\) represents the quantity inside the absolute value bars so all we need to do here is plug into the formula and then solve the two linear inequalities.

\[\begin{align*}2x - 3 & < - 7 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} 2x - 3 & > 7\\ 2x & < - 4 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} 2x & > 10\\ x & < - 2 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} x & > 5\end{align*}\]The interval notation for these are \(\left( { - \infty , - 2} \right)\) or \(\left( {5,\infty } \right)\).

b \(\left| {6t + 10} \right| \ge 3\) Show Solution

Let’s just plug into the formulas and go here,

\[\begin{align*}6t + 10 & \le - 3 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}6t + 10 & \ge 3\\ 6t & \le - 13 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} 6t & \ge - 7\\ t & \le - \frac{{13}}{6} & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}t & \ge - \frac{7}{6}\end{align*}\]The interval notation for these are\(\left( { - \infty , - \frac{{13}}{6}} \right]\) or \(\left[ { - \frac{7}{6},\infty } \right)\).

c \(\left| {2 - 6y} \right| > 10\) Show Solution

Again, not much to do here.

\[\begin{align*}2 - 6y & < - 10 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}2 - 6y & > 10\\ - 6y & < - 12 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} - 6y & > 8\\ y & > 2 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} y & < - \frac{4}{3}\end{align*}\]Notice that we had to switch the direction of the inequalities when we divided by the negative number! The interval notation for these solutions is \(\left( {2,\infty } \right)\) or \(\left( { - \infty , - \frac{4}{3}} \right)\).

Okay, we next need to take a quick look at what happens if \(b\) is zero or negative. We’ll do these with a set of examples and let’s start with zero.

- \(\left| {3x + 2} \right| < 0\)
- \(\left| {x - 9} \right| \le 0\)
- \(\left| {2x - 4} \right| \ge 0\)
- \(\left| {3x - 9} \right| > 0\)

These four examples seem to cover all our bases.

a \(\left| {3x + 2} \right| < 0\) Show Solution

Now we know that \(\left| p \right| \ge 0\) and so can’t ever be less than zero. Therefore, in this case there is no solution since it is impossible for an absolute value to be strictly less than zero (*i.e.* negative).

b \(\left| {x - 9} \right| \le 0\) Show Solution

This is almost the same as the previous part. We still can’t have absolute value be less than zero, however it can be equal to zero. So, this will have a solution only if

\[\left| {x - 9} \right| = 0\]and we know how to solve this from the previous section.

\[x - 9 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = 9\]c \(\left| {2x - 4} \right| \ge 0\) Show Solution

In this case let’s again recall that no matter what \(p\) is we are guaranteed to have \(\left| p \right| \ge 0\). This means that no matter what \(x\) is we can be assured that \(\left| {2x - 4} \right| \ge 0\) will be true since absolute values will always be positive or zero.

The solution in this case is all real numbers, or all possible values of \(x\). In inequality notation this would be \( - \infty < x < \infty \).

d \(\left| {3x - 9} \right| > 0\) Show Solution

This one is nearly identical to the previous part except this time note that we don’t want the absolute value to ever be zero. So, we don’t care what value the absolute value takes as long as it isn’t zero. This means that we just need to avoid value(s) of \(x\) for which we get,

\[\left| {3x - 9} \right| = 0\hspace{0.25in} \Rightarrow \hspace{0.25in} 3x - 9 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in} x = 3\]The solution in this case is all real numbers except \(x = 3\).

Now, let’s do a quick set of examples with negative numbers.

- \(\left| {4x + 15} \right| < - 2\) and \(\left| {4x + 15} \right| \le - 2\)
- \(\left| {2x - 9} \right| \ge - 8\) and \(\left| {2x - 9} \right| > - 8\)

Notice that we’re working these in pairs, because this time, unlike the previous set of examples the solutions will be the same for each.

Both (all four?) of these will make use of the fact that no matter what \(p\) is we are guaranteed to have \(\left| p \right| \ge 0\). In other words, absolute values are always positive or zero.

a \(\left| {4x + 15} \right| < - 2\) and \(\left| {4x + 15} \right| \le - 2\) Show Solution

Okay, if absolute values are always positive or zero there is no way they can be less than or equal to a negative number.

Therefore, there is no solution for either of these.

b \(\left| {2x - 9} \right| \ge - 8\) and \(\left| {2x - 9} \right| > - 8\) Show Solution

In this case if the absolute value is positive or zero then it will always be greater than or equal to a negative number.

The solution for each of these is then all real numbers.