Absolute Value Equations

The absolute value of a number is its distance from 0 on the number line, and a distance can never be negative. That one idea explains every absolute value equation. Ask “which numbers are 6 units from 0?” and there are two answers, 6 and −6, one on each side of 0. This lesson shows how to turn an equation with absolute value bars into two ordinary equations, why you must get the bars alone first, and why some equations have only one solution or none at all.

Absolute value is distance from zero

The absolute value of a number is its distance from 0 on the number line. We write it with bars: |x|.

• |5| = 5, because 5 is 5 units from 0.
• |−5| = 5, because −5 is also 5 units from 0, on the other side.
• |0| = 0.

A distance is never negative, so |x| is never negative. Absolute value tells you how far, not which way.

The bars work like parentheses: work out what is inside first, then take the absolute value. For example, |4 − 11| = |−7| = 7. A minus sign outside the bars is applied afterward: −|13| = −13, but |−13| = 13.

The simplest case: |x| = a

Because |x| is the distance from 0, the equation |x| = a asks: which numbers are a units from 0?

• If a > 0, there are two solutions, x = a and x = −a, one on each side of 0. For example, |x| = 6 gives x = 6 or x = −6.
• If a = 0, there is one solution. Only 0 is 0 units from 0, so x = 0.
• If a < 0, there is no solution. No number is a negative distance from 0.

A solution is a single value of x. The equation has two solutions, but x takes only one of them at a time, so write “x = 6 or x = −6.”

Distance from a number other than 0

The distance between two numbers on the number line is the absolute value of their difference. So |x − c| is the distance between x and c.

For example, |x − 3| = 5 asks: which numbers are 5 units from 3? Start at 3 and go 5 to the left and 5 to the right: x = −2 or x = 8. The algebra gives the same answers: x − 3 = 5 or x − 3 = −5.

For |x + 2|, write x + 2 as x − (−2). That is the distance between x and −2. So |x + 2| = 5 asks which numbers are 5 units from −2: x = 3 or x = −7. The center is the number that x is being compared to: |x − 3| is centered at 3, and |x + 2| = |x − (−2)| is centered at −2. (A quick way to find it: the center is the value of x that makes the inside of the bars equal 0. That shortcut works when x has coefficient 1 inside the bars.)

Solving |ax + b| = c

When the bars hold something like 2x + 1, the same idea works. The expression inside the bars is c units from 0, so it equals c or −c.

The rule. Let A stand for any expression. If c > 0, then |A| = c means A = c or A = −c.

For |2x + 1| = 9, write two equations and solve each one with the steps you know from multi-step equations:

• 2x + 1 = 9 gives x = 4.
• 2x + 1 = −9 gives x = −5.

In the second equation, change the sign of the number on the right, the c. A slip to avoid: changing the sign of just one term inside the bars.

If c = 0, the two equations A = 0 and A = −0 are the same equation, so there is one solution. If c < 0, stop: there is no solution.

Isolate the absolute value first

The rule “inside equals c or −c” is about the bars standing alone on one side. In 3|x + 4| − 6 = 15, the bars have a 3 multiplying them and a 6 subtracted. The distance |x + 4| is not 15. First undo the extras, like in a multi-step equation: add 6 to both sides, then divide both sides by 3 (not zero). That gives |x + 4| = 7. Now split into two cases.

The number you end up with decides what happens. A negative coefficient is a common trap. From −2|x| = −8, dividing by −2 gives |x| = 4: two solutions. From −2|x| = 8, dividing by −2 gives |x| = −4: no solution. Decide only after the bars are alone.

Even an equation that looks harmless can hide no solution. In 2|x − 5| + 9 = 3, subtracting 9 and dividing by 2 gives |x − 5| = −3. A distance cannot be −3, so there is no solution.

Check both answers

Substitute each answer into the original equation. Work out each side, starting with the arithmetic inside the bars, then the absolute value, then the rest.

For an equation like |2x + 1| = 9, once the bars are alone and the number on the right is positive, both cases give true solutions. So if one answer fails the check, you made an arithmetic slip, or you split before isolating.

Check in the original equation, not in a line you rewrote. If the rewrite had a mistake, the check would agree with the mistake.

A preview of later: in equations where x also appears outside the bars, like |x − 1| = 2x, one of the two answers can fail. There, checking is not optional.

Tolerance and distance in real life

Many real situations say “within a certain distance of a target.” A tolerance is the greatest difference allowed from the target. If the target is T and the tolerance is d, a measurement x is exactly at a limit when |x − T| = d. That equation gives the two edge values, T − d and T + d.

The values that pass are all the numbers between the two edges, including the edges. The equation finds the edges. To describe every passing value at once, use a three-part inequality like the ones in Compound Inequalities: T − d ≤ x ≤ T + d.

For problems about positions on a straight road or trail, |x − c| = d means “x is d units from c.” Define your variable in words, and include units.