4.4: Absolute Value Inequalities (2024)

Solution

The graph of the left-hand side of |x| < −5 is the “V” of Figure \(\PageIndex{1}\)(a). The graph of the right-hand side of |x| < −5 is a horizontal line located 5 units below the x-axis. This is the situation shown in Figure \(\PageIndex{1}\)(a). The graph of y = |x| is therefore never below the graph of y = −5. Thus, the inequality |x| < −5 has no solution.

An alternate approach is to consider the fact that the absolute value of x is always nonnegative and can never be less than −5. Thus, the inequality |x| < −5 has no solution.

Solution

This is the case shown in Figure \(\PageIndex{1}\)(b). The graph of y = |x| is never strictly below the x-axis. Thus, the inequality |x| < 0 has no solution.

Solution

The graph of the left-hand side of |x| < 8 is the “V” of Figure \(\PageIndex{1}\)(c). The graph of the right-hand side of |x| < 8 is a horizontal line located 8 units above the x-axis. This is the situation depicted in Figure \(\PageIndex{1}\)(c). The graphs intersect at (−8, 8) and (8, 8) and the graph of y = |x| lies strictly below the graph of y = 8 for values of x in-between −8 and 8. Thus, the solution of |x| < 8 is −8 < x < 8.

It helps the intuition if you check the results of the last example. Note that numbers between −8 and 8, such as −7.75, −3 and 6.8 satisfy the inequality,

\[|-7.75|<8 \qquad \text { and } \quad|-3|<8 \quad \text { and } \quad|6.8|<8 \nonumber \]

while values that do not lie between −8 and 8 do not satisfy the inequality. For example, none of the numbers −9.3, 8.2, and 11.7 lie between −8 and 8, and each of the following is a false statement.

\[|-9.3|<8 \quad \text { and } \qquad|8.2|<8 \qquad \text { and } \qquad|11.7|<8 \quad \text { (all are false) } \nonumber \]

If you reflect upon these results, they will help cement the notion that the solution of |x| < 8 is all values of x satisfying −8 < x < 8.

Solution

If the inequality were |x| < −3, we would not hesitate. This is the situation depicted in Figure \(\PageIndex{1}\)(a) and the inequality |x| < −3 has no solutions. The reasoning applied to |x| < −3 works equally well for the inequality |5 − 2x| < −3. The left-hand side of this inequality must be nonnegative, so its graph must lie on or above the x-axis. The right-hand side of |5 − 2x| < −3 is a horizontal line located 3 units below the x-axis. Therefore, the graph of y = |5 − 2x| can never lie below the graph of y = −3 and the inequality |5 − 2x| < −3 has no solution.

We can verify this result with the graphing calculator. Load the left- and right-hand sides of |5 − 2x| < −3 into Y1 and Y2, respectively, as shown in Figure \(\PageIndex{2}\)(a). From the ZOOM menu, select 6:ZStandard to produce the image shown in Figure \(\PageIndex{2}\)(b).

As predicted, the graph of y = |5 − 2x| never lies below the graph of y = −3, so the inequality |5 − 2x| < −3 has no solution.

Solution

We know that the left-hand side of the inequality |5 − 2x| < 0 has the “V” shape indicated in Figure \(\PageIndex{1}\)(b). The graph “touches” the x-axis when |5 − 2x| = 0, or when

\[\begin{aligned} 5-2 x &=0 \\-2 x &=-5 \\ x &=\frac{5}{2} \end{aligned} \nonumber \]

However, the graph of y = |5 − 2x| never falls below the x-axis, so the inequality |5 − 2x| < 0 has no solution.

Intuitively, it should be clear that the inequality |5−2x| < 0 has no solution. Indeed, the left-hand side of this inequality is always nonnegative, and can never be strictly less than zero.

Solution

In this example, the graph of the right-hand side of the inequality |5 − 2x| < 3 is a horizontal line located 3 units above the x-axis. The graph of the left-hand side of the inequality has the “V” shape shown in Figure \(\PageIndex{3}\)(b) and (c). You can use the intersect utility on the graphing calculator to find the points of intersection of the graphs of y = |5 − 2x| and y = 3, as we have done in Figures \(\PageIndex{3}\)(b) and (c). Note that the calculator indicates two points of intersection, one at x = 1 and a second at x = 4.

The graph of y = |5 − 2x| falls below the graph of y = 3 for all values of x between 1 and 4. Hence, the solution of the inequality |5 − 2x| < 3 is the set of all x satisfying 1 < x < 4; i.e. {x : 1 < x < 4}.

Expectations:

We need a way of summarizing this graphing calculator approach on our homework paper. First, draw a reasonable facsimile of your calculator’s viewing window on your homework paper. Use a ruler to draw all lines. Complete the following checklist.

• Label each axis, in this case with x and y.
• Scale each axis. To do this, press the WINDOW button on your calculator, then report the values of xmin, xmax, ymin, and ymax on the appropriate axis.
• Label each graph with its equation.
• Drop dashed vertical lines from the points of intersection to the x-axis. Shade and label the solution set of the inequality on the x-axis.

Following the guidelines in the above checklist, we obtain the image in Figure \(\PageIndex{4}\).

Algebraic Approach. Let’s now explore an algebraic solution of the inequality |5 − 2x| < 3. Much as |x| < 3 implies that −3 < x < 3, the inequality

\[|5-2 x|<3 \nonumber \]

requires that

\[-3<5-2 x<3 \nonumber \]

We can subtract 5 from all three members of this last inequality, then simplify.

\[\begin{aligned}-3-5 &<5-2 x-5<3-5 \\ &-8<-2 x<-2 \end{aligned} \nonumber \]

Divide all three members of this last inequality by −2, reversing the inequality symbols as you go.

\[4>x>1 \nonumber \]

We prefer that our inequalities read from “small-to-large,” so we write

\[1<x<4 \nonumber \]

This form matches the order of the shaded solution on the number line in Figure \(\PageIndex{4}\), which we found using the graphing calculator.

Solution

The first step is to use Property 8 to write that \[|4 x+5|<7 \nonumber \]

is equivalent to the inequality

\[-7<4 x+5<7 \nonumber \]

From here, we can solve for x by first subtracting 5 from all three members, then dividing through by 4.

\[\begin{array}{l}{-12<4 x<2} \\ {-3<x<\frac{1}{2}}\end{array} \nonumber \]

We can sketch the solution on a number line.

And we can describe the solution in both interval and set-builder notation as follows.

\[\left(-3, \frac{1}{2}\right)=\left\{x :-3<x<\frac{1}{2}\right\} \nonumber \]

Solution

At first glance, the inequality \[5-3|x-4| \geq-4 \nonumber \] has a form quite dissimilar from what we’ve done thus far. However, let’s subtract 5 from both sides of the inequality.

\[-3|x-4| \geq-9 \nonumber \]

Now, let’s divide both sides of this last inequality by −3, reversing the inequality sign.

\[|x-4| \leq 3 \nonumber \]

Aha! Familiar ground. Using Property 10, this last inequality is equivalent to

\[-3 \leq x-4 \leq 3 \nonumber \]

and when we add 4 to all three members, we have the solution.

\[1 \leq x \leq 7 \nonumber \]

We can sketch the solution on a number line

And we can describe the solution with interval and set-builder notation.

\[[1,7]=\{x : 1 \leq x \leq 7\} \nonumber \]

Solution

a. The solution of |x| > −5 is all real numbers.

b. The solution of |x| > 0 is all real numbers except zero.

c. The solution of |x| > 4 is the set of all real numbers less than −4 or greater than 4.

Solution

The left-hand side of the inequality |4 − x| > −5 is nonnegative, so the graph of y = |4 − x| must lie above or on the x-axis. The graph of the right-hand side of |4 − x| > −5 is a horizontal line located 5 units below the x-axis. Therefore, the graph of y = |4 − x| always lies above the graph of y = −5. Thus, all real numbers are solutions of the inequality |4 − x| > −5.

We can verify our thinking with the graphing calculator. Load the left- and righthand sides of the inequality |4 − x| > −5 into Y1 and Y2, respectively, as shown in Figure \(\PageIndex{6}\)(a). From the ZOOM menu, select 6:ZStandard to produce the image shown in Figure \(\PageIndex{6}\)(b).

As predicted, the graph of y = |4 − x| lies above the graph of y = −5 for all real numbers.

Intuitively, the absolute value of any number is always nonnegative, so |4−x| > −5 for all real values of x.

Solution

As we saw in Figure \(\PageIndex{6}\)(b), the graph of y = |4 − x| lies on or above the x-axis for all real numbers. It “touches” the x-axis at the “vertex” of the “V,” where \[|4-x|=0 \nonumber \]

This can occur only if

\[\begin{aligned} 4-x &=0 \\-x &=-4 \\ x &=4 \end{aligned} \nonumber \]

Thus, the graph of y = |4 − x| is strictly above the x-axis for all real numbers except x = 4. That is, the solution of |4 − x| > 0 is {x : x 6= 4}.

Solution

In this example, the graph of the right-hand side of |4 − x| > 5 is a horizontal line located 5 units above the x-axis. The graph of y = |4 − x| has the “V” shape shown in Figure \(\PageIndex{6}\)(c). You can use the intersect utility on the graphing calculator to approximate the points of intersection of the graphs of y = |4 − x| and y = 5, as we have done in Figure \(\PageIndex{7}\)(c) and (d). The calculator indicates two points of intersection, one at x = −1 and a second at x = 9.

The graph of y = |4 − x| lies above the graph of y = 5 for all values of x that lie either to the left of −1 or to the right of 9. Hence, the solution of |4 − x| > 5 is the set {x : x < −1 or x > 9}.

Following the guidelines established in Example \(\PageIndex{6}\), we create the image shown in Figure \(\PageIndex{8}\) on our homework paper. Note that we’ve labeled each axis, scaled each axis with xmin, xmax, ymin, and ymax, labeled each graph with its equation, and shaded and labeled the solution on x-axis.

Algebraic Approach. Let’s explore an algebraic solution of |4 − x| > 5. In much the same manner that |x| > 5 leads to the conditions x < −5 or x > 5, the inequality

\[|4-x|>5 \nonumber \]

requires that

\[4-x<-5 \qquad \text { or } \qquad 4-x>5 \nonumber \]

We can solve each of these independently by first subtracting 4 from each side of the inequality, then multiplying both sides of each inequality by −1, reversing each inequality as we do so.

\[\begin{array}{rllrrl}{4-x} & {<} & {-5} & {\text { or }} & {4-x} & {>} & {5} \\ {-x} & {<} & {-9} && {-x} & {>} & {1} \\ {x} & {>} & {9} && {x} & {<} & {-1}\end{array} \nonumber \]

We prefer to write this solution in the order

\[x<-1 \qquad \text { or } \qquad x>9 \nonumber \]

as it then matches the order of the graphical solution shaded in Figure \(\PageIndex{8}\). That is, the solution set is {x : x < −1 or x > 9}.

Solution

The first step is to use Property 17 to write that \[|4 x-3|>1 \nonumber \] is equivalent to

\[4 x-3<-1 \qquad \text { or } \qquad 4 x-3>1 \nonumber \]

We can now solve each inequality independently. We begin by adding 3 to both sides of each inequality, then we divide both sides of the resulting inequalities by 4.

\[\begin{array}{rrlrrl}{4 x-3} & {<} & {-1} & {\text { or }} & {4 x-3} & {>} & {1} \\ {4 x} & {<} & {2} && {4 x} & {>} & {4} \\ {x} & {<} & {\frac{1}{2}} & &{x} & {>} & {1}\end{array} \nonumber \]

We can sketch the solutions on a number line.

And we can describe the solution using interval and set-builder notation.

\[(-\infty, 1 / 2) \cup(1, \infty)=\{x : x<1 / 2 \text { or } x>1\} \nonumber \]

Solution

Again, at first glance, the inequality \[3|1-x|-4 \geq|1-x| \nonumber \]

looks unlike any inequality we’ve attempted to this point. However, if we subtract |1 − x| from both sides of the inequality, then add 4 to both sides of the inequality, we get

\[3|1-x|-|1-x| \geq 4 \nonumber \]

On the left, we have like terms. Note that 3|1−x|−|1−x| = 3|1−x|−1|1−x| = 2|1−x|. Thus,

\[2|1-x| \geq 4 \nonumber \]

Divide both sides of the last inequality by 2.

\[|1-x| \geq 2 \nonumber \]

We can now use Property 19 to write

\[1-x \leq-2 \quad \text { or } \qquad 1-x \geq 2 \nonumber \]

We can solve each of these inequalities independently. First, subtract 1 from both sides of each inequality, then multiply both sides of each resulting inequality by −1, reversing each inequality as you go.

\[\begin{array}{rllrrl}{1-x } & {\leq} & {-2} & {\text { or }} & {1-x } & {\geq} & { 2} \\ {-x } & {\leq} & {-3} && {-x } & {\geq } & {1} \\ {x} & { \geq } & {3} & &{x} & { \leq} & {-1}\end{array} \nonumber \]

We prefer to write this in the order

\[x \leq-1 \qquad \text { or } \qquad x \geq 3 \nonumber \]

We can sketch the solutions on a number line.

And we can describe the solutions using interval and set-builder notation.

\[(-\infty,-1] \cup[3, \infty)=\{x : x \leq-1 \text { or } x \geq 3\} \nonumber \]

Solution

This inequality is pronounced “the distance between x and 3 is less than 8.” Draw a number line, locate 3 on the line, then note two points that are 8 units away from 3.

Now, we need to shade the points that are less than 8 units from 3.

Hence, the solution of the inequality |x − 3| < 8 is \[(-5,11)=\{x :-5<x<11\} \nonumber \]

Solution

First, write the inequality as a difference.

\[|x-(-5)|>2 \nonumber \]

This last inequality is pronounced “the distance between x and −5 is greater than 2.” Draw a number line, locate −5 on the number line, then note two points that are 2 units from −5.

Now, we need to shade the points that are greater than 2 units from −5.

Hence, the solution of the inequality |x + 5| > 2 is

\[(-\infty,-7) \cup(-3, \infty)=\{x : x<-7 \quad \text { or } \quad x>-3\} \nonumber \]