Algebra 1 · A.REI.B.3

Absolute value inequalities

A number has to land within a set distance of a target, or stay more than that distance away. Which numbers do?

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Absolute value equationsAlgebra 1 Compound inequalities: and, orAlgebra 1

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Transforming the absolute value graphAlgebra 2

Where you are 7 in Inequalities

Algebra9

Inequalities7

▴ 2 above Writing an inequality for a condition Solving linear inequalities3 Compound inequalities: and, or

Absolute value inequalities

Polynomial and rational inequalities with sign charts

When you reach for it

“The jar must weigh within 8 grams of 340 grams” is a statement about distance, and absolute value measures distance: |w - 340| is how far w is from 340 on the number line. So the condition is |w - 340| \le 8. Reach for this lesson whenever an absolute value sits on one side of <, \le, > or \ge.

You have already met the equation version in Absolute value equations. There |x| = 4 has exactly two solutions, x = 4 and x = -4, the two numbers at distance 4 from 0; in general, for a positive number a, |u| = a means u = a or u = -a. An inequality asks instead for the numbers nearer than that, or farther. The answer is no longer two points but a whole stretch of the line, or two stretches.

This lesson takes two days. Part 1 is “less than”: the numbers trapped between two ends, an “and”. Part 2 is “greater than”, which gives an “or”, then what happens when the right side is negative or zero, and how to write an absolute value inequality from a description or a graph.

The procedure

Part 1: less than, an “and”

|x| < 4 asks for the numbers whose distance from 0 is less than 4. Walk away from 0 in either direction: on the right you must stop before 4, on the left before -4.

Figure 1Less than 4 from 0, in either direction. The ends are open because a distance of exactly 4 is not less than 4.

|x| < 4 \iff -4 < x < 4

The algebra agrees. If x \ge 0 then |x| = x, and the condition is x < 4. If x < 0 then |x| = -x, and the condition -x < 4 becomes x > -4 after multiplying by -1, which reverses the sign. Every number in -4 < x < 4 passes its own case, and every number outside fails it. The solution is one stretch, both ends at once: an “and”, written as the chain from Compound inequalities: and, or.

The same reasoning works with any expression u in place of x. For a positive number a,

|u| < a \iff -a < u < a

|u| \le a \iff -a \le u \le a

The \le version adds the two ends, where the distance is exactly a.

Distance from a number other than 0. |x - 3| is the distance between x and 3. So |x - 3| \le 2 asks for the numbers within 2 of 3, which by eye is everything from 1 to 5, ends included. The rule, with u = x - 3, gives the same thing:

-2 \le x - 3 \le 2

Adding 3 to all three parts keeps every inequality true, so 1 \le x \le 5.

Figure 2Within 2 of 3. The center is 3, the reach is 2, and both ends are included.

Think of the answer as a center, 3, and a reach, 2. To graph it, put filled dots at 1 and 5 (filled because \le includes them) and shade between. In Interval and set-builder notation it is [1, 5]. A strict < gets open circles and round brackets instead.

Isolate the absolute value first. The rule splits |u| < a, with the absolute value alone on one side. In 2|x + 1| - 3 < 7 the left side is not the absolute value of anything, so the rule does not apply yet. Add 3 to both sides, then divide both sides by 2, which is positive and so keeps the sign:

|x + 1| < 5

Now split: -5 < x + 1 < 5, and subtracting 1 from all three parts gives -6 < x < 4.

A check. The two ends cut the line into three pieces, and one number from each tells you whether the right piece was kept. At x = -10, 2|-9| - 3 = 15, not less than 7. At x = 0, 2|1| - 3 = -1, which is less than 7. At x = 10, 2|11| - 3 = 19, not less than 7. Only the middle piece works, as the answer says.

Part 2: greater than, an “or”

|x| > 4 asks for the numbers more than 4 from 0. They lie to the right of 4 or to the left of -4, and no number does both.

Figure 3More than 4 from 0. The solution is two separate pieces, so it is an “or”.

|x| > 4 \iff x < -4 \text{ or } x > 4

The cases confirm it: for x \ge 0 the condition is x > 4, and for x < 0 it is -x > 4, that is x < -4. For a positive number a and any expression u,

|u| > a \iff u < -a \text{ or } u > a

|u| \ge a \iff u \le -a \text{ or } u \ge a

A chain such as p < x < q always describes one unbroken stretch, so a solution in two pieces can only be written with “or”. In interval notation the two pieces are joined by \cup: (-\infty, -4) \cup (4, \infty). From a center other than 0 it reads the same way: |x - 3| > 2 means more than 2 from 3, so x < 1 or x > 5.

When the right side is negative or zero. The rules above need a > 0. When it is not, read the inequality as a statement about distance, which is never negative.

|u| < a with a negative or zero has no solution: no distance is below 0.
|u| > a or |u| \ge a with a negative holds for every real number: every distance is at least 0, which is already more than a. So does |u| \ge 0.
|u| \le 0 holds only when |u| = 0, so only when u = 0: usually a single number.
|u| > 0 holds everywhere except where u = 0.

The sign of a is read after isolating. |x| + 5 < 2 shows a positive 2, but it is |x| < -3, which has no solution.

Writing an inequality from a stretch. Any stretch from p to q is the set of numbers within half its width of its midpoint. Take the center and the reach

c = \frac{p + q}{2}, \qquad r = \frac{q - p}{2}

Then c - r = p and c + r = q, so p \le x \le q is the same statement as |x - c| \le r. For 1 \le x \le 5 this gives c = 3 and r = 2, which is |x - 3| \le 2 again. The numbers outside the stretch are those farther than r from c: |x - c| > r when the two ends are left out, and |x - c| \ge r when the two ends are included.

Worked examples

Examples 1 to 4 go with Part 1, and Examples 5 to 7 with Part 2.

Example 1

Within a distance of zero

Solve each inequality, describe its graph, and write the solution in interval notation.

(a) |x| \le 6 \qquad (b) |y| < 2.5

Solution

(a) The numbers at most 6 from 0: -6 \le x \le 6. The graph has filled dots at -6 and 6 with the segment between shaded. In interval notation, [-6, 6].

(b) The numbers less than 2.5 from 0: -2.5 < y < 2.5. The ends are open circles, since a distance of exactly 2.5 fails, and the interval is (-2.5, 2.5). A test number: y = -3 gives |-3| = 3, not less than 2.5, and -3 is indeed outside.

Example 2

A coefficient on x

Solve |2x - 5| < 9. Graph the solution and write it in interval notation.

Solution

The absolute value is already alone and 9 is positive, so the rule applies:

-9 < 2x - 5 < 9

Add 5 to all three parts, then divide all three by 2, which is positive and keeps both signs:

-4 < 2x < 14 \qquad\Longrightarrow\qquad -2 < x < 7

The graph has open circles at -2 and 7 and the segment between them shaded; the interval is (-2, 7). As a distance, \displaystyle |2x - 5| = 2\left|x - \frac{5}{2}\right|, so the inequality says x is less than \displaystyle \frac{9}{2} from \displaystyle \frac{5}{2}. That center is the midpoint of -2 and 7, as it should be. Check x = 0: |-5| = 5 < 9, and 0 is inside.

Example 3

Isolating the absolute value first

Solve 3|x + 4| - 5 \le 7 and write the solution in interval notation.

Solution

The rule needs the absolute value alone. Add 5 to both sides, then divide both sides by 3:

3|x + 4| \le 12 \qquad\Longrightarrow\qquad |x + 4| \le 4

Now split, and subtract 4 from all three parts:

-4 \le x + 4 \le 4 \qquad\Longrightarrow\qquad -8 \le x \le 0

So the solution is [-8, 0], filled dots at both ends. Since |x + 4| = |x - (-4)|, this is the set of numbers within 4 of -4. Check the end x = 0: 3|4| - 5 = 7, and 7 \le 7 is true, so 0 belongs.

Example 4

A negative coefficient

Solve |4 - 3x| < 10.

Solution

Split first:

-10 < 4 - 3x < 10

Subtract 4 from all three parts: -14 < -3x < 6. Dividing by -3 reverses both inequality signs, for the reason given in Solving linear inequalities: multiplying by a negative number swaps the order of any two numbers.

\frac{14}{3} > x > -2

Read from the smaller end, \displaystyle -2 < x < \frac{14}{3}, which is \displaystyle \left(-2, \frac{14}{3}\right).

There is a second route. A number and its negative have the same absolute value, so |4 - 3x| = |3x - 4|, and -10 < 3x - 4 < 10 gives -6 < 3x < 14 with no sign to reverse. The answer is the same. Check x = 5, just past \displaystyle \frac{14}{3}: |4 - 15| = 11, not less than 10.

Example 5

Farther than a distance

Solve each inequality and write the solution in interval notation.

(a) |2x + 1| \ge 7 \qquad (b) 5 - 2|x - 3| < -1

Solution

(a) The absolute value is alone and 7 is positive, so split into an “or”:

2x + 1 \le -7 \quad\text{or}\quad 2x + 1 \ge 7

The first gives 2x \le -8, so x \le -4; the second gives 2x \ge 6, so x \ge 3. The graph has filled dots at -4 and 3 with arrows shaded outward, and the solution is (-\infty, -4] \cup [3, \infty). Check a number in the gap: x = 0 gives |1| = 1, not at least 7, so the gap is rightly empty.

(b) Isolate. Subtract 5 from both sides: -2|x - 3| < -6. Divide both sides by -2, which reverses the sign:

|x - 3| > 3

Now split: x - 3 < -3 or x - 3 > 3, so x < 0 or x > 6, which is (-\infty, 0) \cup (6, \infty). Without the reversal the inequality would have been a “less than” and the answer the gap instead of the two outer pieces.

Example 6

Negative and zero on the right

Solve (a) |x - 2| + 6 < 1 \qquad (b) 4 - |3x + 1| < 9 \qquad (c) |2x - 6| \le 0

Solution

(a) Subtract 6 from both sides: |x - 2| < -5. A distance is never negative, so it is never less than -5. There is no solution; the solution set is \varnothing.

(b) Subtract 4 from both sides: -|3x + 1| < 5. Multiply both sides by -1, reversing the sign: |3x + 1| > -5. Every absolute value is at least 0, and 0 > -5, so this holds for every real number: (-\infty, \infty).

(c) An absolute value cannot be less than 0, so |2x - 6| \le 0 forces |2x - 6| = 0, that is 2x - 6 = 0. The only solution is x = 3, and its graph is a single filled dot.

Example 7

Writing the inequality from a description or a graph

(a) A jar of peanut butter is labeled 340 g. The factory accepts a jar when its fill weight, w grams, is within 8 g of the label. Write the accepted weights as an absolute value inequality and solve it. Then write the condition for a rejected jar.

(b) and (c): write an absolute value inequality whose solution is each graph below.

Figure 4Example 7. Each graph is described by a center, a reach, and whether it shades inside or outside.

Solution

(a) The distance between w and 340 is |w - 340|, and “within 8” includes 8 itself:

|w - 340| \le 8

Splitting gives -8 \le w - 340 \le 8, so 332 \le w \le 348. A jar is rejected when it is more than 8 g from the label, which is |w - 340| > 8: lighter than 332 g or heavier than 348 g.

(b) The shaded stretch runs from -6 to 2. Its center is \displaystyle \frac{-6 + 2}{2} = -2 and its reach is \displaystyle \frac{2 - (-6)}{2} = 4. The dots are filled and the inside is shaded, so the inequality is |x - (-2)| \le 4, that is |x + 2| \le 4. Check: -4 \le x + 2 \le 4 gives -6 \le x \le 2.

(c) The open circles at 1 and 7 have center 4 and reach 3. The shading points outward, so the numbers are more than 3 from 4, and the open circles make it strict: |x - 4| > 3. Check: x - 4 < -3 or x - 4 > 3 gives x < 1 or x > 7.

Boundaries

The splitting rules need a positive right side. |u| < a becomes -a < u < a, and |u| > a becomes u < -a or u > a, only when a > 0. For a \le 0, read the inequality as a distance, as in Example 6.

The absolute value must stand alone. The rules split |u| itself. Anything added to it or multiplying it is moved off first, by the moves for linear inequalities, and dividing or multiplying by a negative reverses the sign, which can turn “less than” into “greater than”.

Strict or not decides the ends. < and > leave out the numbers at distance exactly a (open circles, round brackets); \le and \ge keep them (filled dots, square brackets).

Less than is one stretch; greater than is two. For a > 0, |u| < a gives an “and”, a single interval, and |u| > a gives an “or”, two rays. The two never share a number, and between them they miss only the numbers where |u| = a.

Real numbers only. Everything here orders numbers on a line, so the variable is a real number throughout and “no solution” means no real number works.

Where this goes next

Every absolute value inequality has a picture in the plane: |x - 3| \le 2 holds exactly where the V-shaped graph of y = |x - 3| lies on or below the line y = 2. That reading comes from The absolute value graph and grows into Transforming the absolute value graph.

“Within a distance” is how precision is stated. A made-up poll reported as 52\% give or take 3 points is saying the true percentage p most likely satisfies |p - 52| \le 3; that “give or take” is the Point estimates and the margin of error. And the foundation of calculus is built from exactly these inequalities: the formal definition of a limit says that f(x) can be kept within any distance \varepsilon of L, |f(x) - L| < \varepsilon, by keeping x close enough to a, 0 < |x - a| < \delta.

Exercises

Part 1: less than

Solve each inequality, describe its graph, and write the solution in interval notation.

|x| < 11
\displaystyle |m| \le \frac{1}{2}
|x + 5| \le 3
|3x - 4| < 11
|5 - 2y| \le 7
4|x - 2| + 3 < 23
7 - 2|x + 1| \ge 1

Part 1: a problem and a question

A speed camera’s reading is within 3 km/h of a car’s true speed. It reads 64 km/h. Write an absolute value inequality for the true speed s and solve it.
Explain why |x - 6| < 1 and x - 6 < 1 have different solutions, and give a number that satisfies the second but not the first.

Part 2: greater than

Solve each inequality and write the solution in interval notation.

|x| > 2
|x - 1| \ge 5
|4x + 3| > 9
3|2 - x| - 4 \ge 8

Part 2: negative and zero on the right

|x + 7| < -2
|3x - 1| + 5 \ge 2
|5x + 10| \le 0
|x - 4| > 0

Part 2: writing the inequality

Sabela keeps her pool’s chlorine level, c parts per million, between 1 and 3, both included. Write this as an absolute value inequality.
A machine checks each pill it presses. A pill’s mass, m grams, is flagged if it differs from 0.15 by more than 0.05. Write an absolute value inequality for a flagged pill and solve it.
A number line shows open circles at -9 and -1, with the segment between them shaded. Write an absolute value inequality for it.
A number line shows filled dots at -1 and 4, with arrows shaded outward from both. Write an absolute value inequality for it.

Reasoning

Let r > 0. Use distance to explain why no number satisfies both |x - c| < r and |x - c| > r, and say which numbers satisfy neither.

Answers

-11 < x < 11, open circles at -11 and 11 with the segment shaded; (-11, 11).
\displaystyle -\frac{1}{2} \le m \le \frac{1}{2}, filled dots at both ends; \displaystyle \left[-\frac{1}{2}, \frac{1}{2}\right].
-8 \le x \le -2, filled dots; [-8, -2]. From -3 \le x + 5 \le 3; these are the numbers within 3 of -5.
\displaystyle -\frac{7}{3} < x < 5, open circles; \displaystyle \left(-\frac{7}{3}, 5\right). From -11 < 3x - 4 < 11, add 4: -7 < 3x < 15.
-1 \le y \le 6, filled dots; [-1, 6]. From -7 \le 5 - 2y \le 7, subtract 5: -12 \le -2y \le 2, and dividing by -2 reverses both signs: 6 \ge y \ge -1.
-3 < x < 7, open circles; (-3, 7). Subtract 3 and divide by 4: |x - 2| < 5.
-4 \le x \le 2, filled dots; [-4, 2]. Subtract 7: -2|x + 1| \ge -6. Divide by -2, reversing the sign: |x + 1| \le 3, so -3 \le x + 1 \le 3.
|s - 64| \le 3, so -3 \le s - 64 \le 3 and 61 \le s \le 67.
|x - 6| < 1 says x is less than 1 from 6, a short stretch 5 < x < 7. x - 6 < 1 only says x < 7, with no limit on how far below 6 it goes. x = 0 satisfies 0 - 6 < 1, but |0 - 6| = 6, which is not less than 1.
x < -2 or x > 2; (-\infty, -2) \cup (2, \infty).
x \le -4 or x \ge 6; (-\infty, -4] \cup [6, \infty). From x - 1 \le -5 or x - 1 \ge 5.
x < -3 or \displaystyle x > \frac{3}{2}; \displaystyle (-\infty, -3) \cup \left(\frac{3}{2}, \infty\right). From 4x + 3 < -9 or 4x + 3 > 9.
x \le -2 or x \ge 6; (-\infty, -2] \cup [6, \infty). Add 4 and divide by 3: |2 - x| \ge 4. Then 2 - x \le -4 gives x \ge 6, and 2 - x \ge 4 gives x \le -2.
No solution, \varnothing. A distance cannot be less than -2.
Every real number, (-\infty, \infty). Subtract 5: |3x - 1| \ge -3, true for every x since an absolute value is never negative.
Only x = -2. An absolute value is never below 0, so 5x + 10 = 0.
Every real number except 4; (-\infty, 4) \cup (4, \infty). |x - 4| is positive unless x - 4 = 0.
|c - 2| \le 1. The center is \displaystyle \frac{1 + 3}{2} = 2 and the reach is \displaystyle \frac{3 - 1}{2} = 1, with both ends included.
|m - 0.15| > 0.05, so m - 0.15 < -0.05 or m - 0.15 > 0.05: m < 0.1 or m > 0.2. A mass is positive, so the flagged masses are 0 < m < 0.1 or m > 0.2.
|x + 5| < 4. The center is \displaystyle \frac{-9 + (-1)}{2} = -5 and the reach is \displaystyle \frac{-1 - (-9)}{2} = 4; open circles and the inside shaded make it a strict “less than”.
\displaystyle \left|x - \frac{3}{2}\right| \ge \frac{5}{2}, or, multiplying both sides by 2, |2x - 3| \ge 5. The center is \displaystyle \frac{3}{2} and the reach is \displaystyle \frac{5}{2}; filled dots and outward shading make it “greater than or equal”.
|x - c| is one number, the distance from x to c, and one number cannot be both less than r and greater than r. Every distance is less than r, equal to r or greater than r, so the numbers that satisfy neither are those at distance exactly r from c: x = c - r and x = c + r.