1.4: Absolute value inequalities

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Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

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Using the notation from the previous section, we now solve inequalities involving the absolute value. These inequalities may be solved in three steps:

Step 1: Solve the corresponding equality. The solution of the equality divides the real number line into several subintervals.
Step 2: Using step 1, check the inequality for a number in each of the subintervals. This check determines the intervals of the solution set.
Step 3: Check the endpoints of the intervals.

Here are some examples for the above solution method.

Example $\PageIndex{1}$

Solve for $x$:

$|x+7|<2$
$|3x-5|\geq 11$
$|12-5x|\leq 1$

Solution

We follow the three steps described above. In step 1, we solve the corresponding equality, $|x+7|=2$. $x+7=2$ & $x+7=-2$ \[\begin{array}{l|l}
x+7=2 & x+7=-2 \\
\Longrightarrow x=-5 & \Longrightarrow x=-9
\end{array} \nonumber \]The solutions $x=-5$ and $x=-9$ divide the number line into three subintervals:

Now, in step 2, we check the inequality for one number in each of these subintervals.

\[ \begin{array}{c|c|c}
\text { Check: } \quad x=-10 & \text { Check: } \quad x=-7 & \text { Check: } \quad x=0 \\
|(-10)+7| \stackrel{?}{<} 2 & |(-7)+7| \stackrel{?}{<} 2 & |0+7| \stackrel{?}{<} 2 \\
|-3| \stackrel{?}{<} 2 & |0| \stackrel{?}{<} 2 & |7| \stackrel{?}{<} 2 \\
3 \stackrel{?}{<} 2 & 0 \stackrel{?}{<} 2 & 7 \stackrel{?}{<} 2 \\
\text { false } & \text { true } & \text { false }
\end{array} \nonumber \]

Since $x=-7$ in the subinterval given by $-9<x<-5$ solves the inequality $|x+7|<2$, it follows that all numbers in the subinterval given by $-9<x<-5$ solve the inequality. Similarly, since $x=-10$ and $x=0$ do not solve the inequality, no number in these subintervals will solve the inequality. For step 3, we note that the numbers $x=-9$ and $x=-5$ are not included as solutions since the inequality is strict (that is we have $<$ instead of $\leq$).The solution set is therefore the interval $S=(-9,-5)$. The solution on the number line is:

$clipboard_ea1d81614b0d8e2564fd8ec56ba182dbf.png$

We follow the steps as before. First, in step 1, we solve $|3x-5|=11$. \[\begin{array}{l|l}
3 x-5=11 & 3 x-5=-11 \\
\Longrightarrow 3 x=16 & \Longrightarrow 3 x=-6 \\
\Longrightarrow x=\dfrac{16}{3} & \Longrightarrow x=-2
\end{array} \nonumber \]

The two solutions $x=-2$ and $x=\dfrac{16}{3}=5\dfrac {1}{3}$ divide the number line into the subintervals displayed below. \[x<-2 \hspace{1in} -2<x<5\frac 1 3 \hspace{1in} 5\frac 1 3<x \nonumber\]

$clipboard_e0d7878e0eca72b0ee120cafa8166dd75.png$

For step 2, we check a number in each subinterval. This gives:

\[\begin{array}{c|c|c|c}
\text { Check: } x=-3 & \text { Check: } \quad x=1 & \text { Check: } \quad x=6 \\
|3 \cdot(-3)-5| \stackrel{?}{\geq} 11 & |3 \cdot 1-5| \stackrel{?}{\geq} 11 & |3 \cdot 6-5| \stackrel{?}{\geq} 11 \\
|-9-5| \stackrel{?}{\geq} 11 & |3-5| \stackrel{?}{\geq} 11 & |18-5| \stackrel{?}{\geq} 11 \\
|-14| \stackrel{?}{\geq} 11 & |-2| \stackrel{?}{\geq} 11 & |13| \stackrel{?}{\geq} 11 \\
14 \stackrel{?}{\geq} 11 & 2 \stackrel{?}{\geq} 11 & 13 \stackrel{?}{\geq} 11 \\
\text { true } & \text { false } & \text { true }
\end{array} \nonumber \]

For step 3, note that we include $-2$ and $5\dfrac {1}{3}$ in the solution set since the inequality is “greater than or equal to” (that is $\geq$, as opposed to $>$). Furthermore, the numbers $-\infty$ and $\infty$ are not included, since $\pm\infty$ are not real numbers.

The solution set is therefore the union of the two intervals: \[S=\Big(-\infty,-2\Big]\cup \Big[5\dfrac {1}{3}, \infty\Big) \nonumber \]

To solve $|12-5x|\leq 1$, we first solve the equality $|12-5x|=1$. \[\begin{array}{l|l}
12-5 x=1 & 12-5 x=-1 \\
\Longrightarrow-5 x=-11 & \Longrightarrow-5 x=-13 \\
\Longrightarrow x=\frac{-11}{-5}=2.2 & \Longrightarrow x=\frac{-13}{-5}=2.6
\end{array} \nonumber \]

This divides the number line into three subintervals, and we check the original inequality $|12-5x|\leq 1$ for a number in each of these subintervals.

\[\begin{array}{c|c|c|c}
\text {Interval: } \quad x<2.2 & \text {Interval: } \quad 2.2<x<2.6 & \text {Interval: } \quad 2.6<x\\
\text {Check: } \quad x=1 & \text {Check: } \quad x=2.4 & \text {Check: } \quad x=3 \\
|12-5 \cdot 1| \stackrel{?}{\leq} 1 & |12-5 \cdot 1| \stackrel{?}{\leq} 1 & |12-5 \cdot 3| \stackrel{?}{\leq} 1 \\
|12-5| \stackrel{?}{\leq} 1 & |12-12| \stackrel{?}{\leq} 1 & |12-15| \stackrel{?}{\leq} 1 \\
|7| \stackrel{?}{\leq} 1 & |0| \stackrel{?}{\leq} 1 & |-3| \stackrel{?}{\leq} 1 \\
7 \stackrel{?}{\leq} 1 & 0 \stackrel{?}{\leq} 1 & 3 \stackrel{?}{\leq} 1 \\
\text { false } & \text { true } & \text { false }
\end{array} \nonumber \]

The solution set is the interval $S=[2.2,2.6]$, where we included $x=2.2$ and $x=2.6$ since the original inequality “less than or equal to” ($\leq$) includes the equality.

Note

Alternatively, whenever you have an absolute value inequality you can turn it into two inequalities.

Here are a couple of examples.

Example $\PageIndex{2}$

Solve for $x$: $|12-5x|\leq 1$

Solution

Note that $|12-5x|\leq 1$ implies that

\[-1\leq 12-5x\leq1 \nonumber \]

so that

\[-13\leq -5x\leq -11 \nonumber \]

and by dividing by $-5$ (remembering to switch the direction of the inequalities when multiplying or dividing by a negative number) we see that

\[\frac{13}{5}\geq x\geq \frac{11}{5} \nonumber \]

or in interval notation, we have the solution set

\[S=\left[\frac{11}{5},\frac{13}{5}\right] \nonumber\]

Example $\PageIndex{3}$

If $|x+6|>2$ then either $x+6>2$ or $x+6<-2$ so that either $x>-4$ or $x<-8$ so that in interval notation the solution is $S=(-\infty,-8)\cup(-4,\infty)$.

Solution

There is a geometric interpretation of the absolute value on the number line as the distance between two numbers:

$clipboard_e8a95ea5bc21fc953df9681bc64a8644c.png$

distance between $a$ and $b$ is $|b-a|$ which is also equal to $|a-b|$

This interpretation can also be used to solve absolute value equations and inequalities.

Example $\PageIndex{4}$

Solve for $x$:

$|x-6|=4$
$|x-6|\leq 4$
$|x-6|\geq 4$

Solution

Consider the distance between $x$ and $6$ to be $4$ on a number line:

$clipboard_e17c977283e3ff391111c7da91e948997.png$

There are two solutions, $x=2$ or $x=10$. That is, the distance between $2$ and $6$ is $4$ and the distance between $10$ and $6$ is $4$.

Numbers inside the braces above have distance $4$ or less. The solution is given on the number line as:

$clipboard_e54e786ef1671d92c3b9291ecb0fc38f1.png$

In interval notation, the solution set is the interval $S=[2,10]$. One can also write that the solution set consists of all $x$ such that $2\leq x\leq 10$.

Numbers outside the braces above have distance $4$ or more. The solution is given on the number line as:

$clipboard_ea020fa310cf3fcba8ba44ff547e18ed9.png$

In interval notation, the solution set is the interval $(-\infty,2]$ and $[10,\infty)$, or in short it is the union of the two intervals:

\[S= (\infty,2]\cup [10,\infty) \nonumber \]

One can also write that the solution set consists of all $x$ such that $x\leq 2$ or $x\geq 10$.

Example \(\PageIndex{1}\)

Note

Example \(\PageIndex{2}\)

Example \(\PageIndex{3}\)

Example \(\PageIndex{4}\)