Math Core

A machine is supposed to cut bolts $50$ millimeters long, and a bolt is acceptable if it is within $0.2$ mm of that target. "Within $0.2$ of $50$ " is a statement about distance, and absolute value measures distance. Combining absolute value with an inequality lets you describe tolerances, margins of error, and "close enough" in one short statement.

Absolute value as distance

Recall that $|x|$ is the distance from $x$ to $0$ on the number line. You used this in the last unit to solve equations like $|x| = 3$ , which has two solutions, $3$ and $-3$ , because both are exactly $3$ units from $0$ .

Now ask a different question: which numbers are less than $3$ units from $0$ ? Those are the numbers between $-3$ and $3$ .

|x| < 3 means -3 < x < 3

And which numbers are more than $3$ units from $0$ ? Those are the numbers beyond $-3$ on the left or beyond $3$ on the right, two rays going outward. Pictured one piece at a time:

first piece: x < -3

second piece: x > 3

So $|x| < 3$ is an and inequality (a segment), while $|x| > 3$ is an or inequality (two rays). This pattern is the whole lesson.

Absolute value inequalities

For any expression $A$ and any positive number $b$ :

inequality	meaning	rewrite as
$\lvert A \rvert < b$	within $b$ of zero	$-b < A < b$
$\lvert A \rvert \le b$	within $b$ of zero, edges included	$-b \le A \le b$
$\lvert A \rvert > b$	farther than $b$ from zero	$A < -b$ or $A > b$
$\lvert A \rvert \ge b$	at least $b$ from zero	$A \le -b$ or $A \ge b$

A handy memory aid: less than leads to and, greater than leads to or.

Distance from a number other than zero

The expression $|x - c|$ is the distance between $x$ and $c$ . So $|x - 4| \le 6$ says " $x$ is within $6$ units of $4$ ." Starting at $4$ and going $6$ units each way lands you at $-2$ and $10$ .

Worked example: A less-than inequality

Solve $|x - 4| \le 6$ and graph the solutions.

Rewrite it as a chain, then solve:

\begin{aligned} -6 &\le x - 4 \le 6 \\ -2 &\le x \le 10 && \text{add } 4 \text{ to all three parts} \end{aligned}

-2 ≤ x ≤ 10

This matches the distance picture: the solutions are all numbers within $6$ of $4$ , and $4$ is exactly in the middle of the segment.

Worked example: A greater-than inequality

Solve $|2x + 1| > 7$ .

The expression inside must be more than $7$ units from zero, so split it into two cases joined by or:

2x + 1 < -7 \qquad \text{or} \qquad 2x + 1 > 7

Left case: $2x < -8$ , so $x < -4$ .
Right case: $2x > 6$ , so $x > 3$ .

The solution is $x < -4$ or $x > 3$ .

Check a number in each region. $x = -5$ : $|2(-5) + 1| = |-9| = 9 > 7$ , true. $x = 0$ : $|1| = 1 > 7$ , false. $x = 4$ : $|9| = 9 > 7$ , true. Only the outer regions work, as expected.

Common mistake

For a greater-than inequality, don't try to write a chain. The answer $x < -4$ or $x > 3$ is not $-4 > x > 3$ , which would claim that $x$ is less than $-4$ and greater than $3$ at the same time. Keep the word "or."

Also be careful with the left case: it is $2x + 1 < -7$ , not $2x + 1 < 7$ . The $-b$ side must have the negative sign and the symbol pointing outward.

Isolate the absolute value first

Just like with absolute value equations, the absolute value must be by itself before you split into cases.

Worked example: Isolating first

Solve $3|x + 2| - 5 < 7$ .

\begin{aligned} 3|x + 2| - 5 &< 7 \\ 3|x + 2| &< 12 && \text{add } 5 \\ |x + 2| &< 4 && \text{divide by } 3 \end{aligned}

Now rewrite as a chain: $-4 < x + 2 < 4$ . Subtract $2$ from all three parts: $-6 < x < 2$ .

Special cases

The rewriting rules in the Key Idea need $b$ to be positive. When the number on the other side is zero or negative, think about what absolute value can and can't do: it is never negative.

$|x - 1| < -2$ : an absolute value can never be less than a negative number. No solution.
$|x + 5| \ge -3$ : every absolute value is at least $0$ , so it is certainly at least $-3$ . All real numbers.
$|x| \le 0$ : an absolute value can't be below $0$ , so it must equal $0$ . Only $x = 0$ works.
$|x - 3| > 0$ : every number except $3$ gives a positive distance. The solution is all real numbers except $3$ .

Tip

Before splitting into cases, glance at the sign of the number on the other side. If it's negative, you can answer "no solution" or "all real numbers" immediately, with no algebra at all.

Writing absolute value inequalities

To describe a segment like $1 < x < 7$ with absolute value, find its center (the midpoint) and its radius (the distance from the center to either end).

Center: $\dfrac{1 + 7}{2} = 4$ .
Radius: $7 - 4 = 3$ .

The numbers between $1$ and $7$ are exactly those less than $3$ units from $4$ : $|x - 4| < 3$ .

Worked example: Manufacturing tolerance

A bolt must be $50$ mm long, give or take $0.2$ mm. Write and solve an absolute value inequality for the acceptable lengths $L$ .

The distance between $L$ and $50$ can be at most $0.2$ :

|L - 50| \le 0.2.

Rewrite: $-0.2 \le L - 50 \le 0.2$ . Add $50$ : $49.8 \le L \le 50.2$ . Any bolt from $49.8$ mm to $50.2$ mm passes inspection.

Practice

Practice 1

Which inequality is equivalent to $|x| \le 5$ ?

Practice 2

Solve $|x + 3| < 5$ . The answer has the form $a < x < b$ . Enter $a$ and $b$ , smaller first.

Separate answers with commas, e.g. 2, -5

Practice 3

Solve $|3x - 6| > 9$ .

Practice 4

Solve $2|x - 1| + 3 \le 11$ . The answer has the form $a \le x \le b$ . Enter $a$ and $b$ , smaller first.

Separate answers with commas, e.g. 2, -5

Practice 5

Solve $|x + 7| + 4 < 3$ .

Practice 6

Which absolute value inequality has the solutions $-2 \le x \le 8$ ?

Practice 7

Solve $|5 - 2x| \le 9$ . The answer has the form $a \le x \le b$ . Enter $a$ and $b$ , smaller first.

Separate answers with commas, e.g. 2, -5

Practice 8

A cereal box is labeled $16$ ounces. A box passes inspection if its actual weight $w$ satisfies $|w - 16| \le 0.4$ . What is the lightest weight, in ounces, that passes?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.