Math Core

Every equation in this unit so far has had at most one solution (apart from identities). Absolute value equations are different: $|x| = 5$ has two solutions, $5$ and $-5$ . Learning to find both, and to spot when there are none, is the last piece of this unit and sets up absolute value inequalities in the next.

Absolute value as distance

Recall that the absolute value $|a|$ is the distance from $a$ to $0$ on the number line. Distance is never negative, so $|a| \ge 0$ for every real number $a$ .

So what does $|x| = 5$ ask? It asks for every number that is $5$ units from $0$ . There are two of them, one on each side:

Both -5 and 5 are 5 units from 0.

The solutions are $x = 5$ and $x = -5$ , often written $x = \pm 5$ .

The same idea works for $|x - 3| = 7$ . The expression $|x - 3|$ is the distance between $x$ and $3$ , so the equation asks for the numbers $7$ units from $3$ . Counting $7$ each way from $3$ gives $10$ and $-4$ .

-4 and 10 are each 7 units from 3.

The two-case method

Picturing distance is great for simple equations, but you need an algebraic method for harder ones. If the absolute value of something equals $c$ , then that something is either $c$ or $-c$ .

Solving an absolute value equation

Isolate the absolute value, so the equation looks like $|A| = c$ .
Look at $c$ $c$ :
- If $c > 0$ , write two equations, $A = c$ and $A = -c$ , and solve each. There are two solutions.
- If $c = 0$ , solve $A = 0$ . There is one solution.
- If $c < 0$ , there is no solution, because an absolute value can't be negative.
Check each answer in the original equation.

Worked example: Two cases

Solve $|x - 3| = 7$ .

The absolute value is already isolated, and $7 > 0$ , so split into two cases:

\begin{aligned} x - 3 &= 7 & \qquad\qquad x - 3 &= -7 \\ x &= 10 & x &= -4 \end{aligned}

Check: $|10 - 3| = |7| = 7$ ✓ and $|-4 - 3| = |-7| = 7$ ✓. The solutions are $x = 10$ and $x = -4$ , matching the number line above.

Isolate first

If there are other operations outside the absolute value bars, undo them first, exactly as in a two-step equation. Treat $|x + 1|$ as a single block until it is alone.

Worked example: Isolate, then split

Solve $2|x + 1| - 5 = 9$ .

\begin{aligned} 2|x + 1| - 5 &= 9 \\ 2|x + 1| &= 14 && \text{add } 5 \\ |x + 1| &= 7 && \text{divide by } 2 \end{aligned}

Now split into two cases:

\begin{aligned} x + 1 &= 7 & \qquad\qquad x + 1 &= -7 \\ x &= 6 & x &= -8 \end{aligned}

Check: $2|7| - 5 = 9$ ✓ and $2|-7| - 5 = 9$ ✓. The solutions are $6$ and $-8$ .

Common mistake

Don't split into cases before isolating the absolute value. From $2|x + 1| - 5 = 9$ , writing $2(x + 1) - 5 = -9$ as the second case is wrong: the negative belongs only to what's inside the bars, and only once $|x + 1|$ is alone. Also, you can't "distribute" into absolute value bars: $2|x + 1|$ is not $|2x + 1|$ or $2|x| + 2$ .

One solution or none

Worked example: Special cases

Solve each equation.

$|3x - 6| + 4 = 1$
$|2x + 8| = 0$

Solutions.

Isolate: $|3x - 6| = -3$ . An absolute value is a distance, and a distance can't be $-3$ . There is no solution.
The only number with absolute value $0$ is $0$ itself, so $2x + 8 = 0$ . Then $2x = -8$ and $x = -4$ . There is exactly one solution.

Tip

Always check both answers in the original equation. In this course the two-case method gives correct answers as long as you isolate first, but checking catches arithmetic slips, and in later courses (when variables appear on both sides of an absolute value equation) some "answers" really do fail the check.

Tolerance problems

Absolute value is the natural way to describe "within a certain amount of a target." If a part should be $40$ mm long, give or take $0.5$ mm, then the length $L$ satisfies $|L - 40| \le 0.5$ . The two boundary lengths come from the equation $|L - 40| = 0.5$ : $L = 40.5$ and $L = 39.5$ . You'll work with the full inequality in the next unit.

Practice

Practice 1

Solve $|x| = 12$ . Enter both solutions, separated by a comma.

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

Practice 2

Solve $|x + 4| = 9$ . Enter both solutions, separated by a comma.

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

Practice 3

Solve $|2x - 1| = 11$ . Enter both solutions, separated by a comma.

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

Practice 4

Solve $3|n - 2| + 2 = 20$ . Enter both solutions, separated by a comma.

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

Practice 5

How many solutions does $|x + 7| + 10 = 4$ have?

Practice 6

Solve $|4x + 12| = 0$ .

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

Practice 7

Solve $-2|x - 5| + 3 = -7$ . Enter both solutions, separated by a comma.

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

Practice 8

A machine fills bags of rice. A bag passes inspection if its weight $w$ is within $8$ grams of $500$ grams. Solve $|w - 500| = 8$ to find the lightest and heaviest weights, in grams, that pass. Enter both, separated by a comma.

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