Absolute Value Equation Calculator

What does absolute value mean?

Absolute value is a number's distance from zero on the number line. Distance can't be negative, so |5| = 5 and |−5| = 5.

That idea explains the whole method. |x| = 5 asks which numbers are 5 units from zero, and there are two: 5 and −5. In the same way, |x − 4| = 2 asks which numbers are 2 units from 4, which gives x = 2 or x = 6.

This distance view is a fast way to check answers. On a number line, 2 and 6 sit exactly 2 steps either side of 4. If your two answers aren't the same distance from the center point, one of them is probably wrong.

How do you solve |2x − 3| = 7?

Write two equations, 2x − 3 = 7 and 2x − 3 = −7, and solve each. The answers are x = 5 and x = −2.

The negative case is the one people skip. It's also why the inside, 2x − 3, gets a negative value of −7 when x = −2, and the bars turn that back into 7.

Why are there two cases?

Because two different numbers have the same absolute value. If |A| = 7, then A is either 7 or −7, and nothing else.

So the general rule is: for c > 0, |A| = c becomes A = c or A = −c. Each case is an ordinary linear equation, and the solver works through both with the same steps it uses on the linear equation calculator page.

When does an absolute value equation have no solution or one solution?

If the absolute value equals a negative number, there's no solution. If it equals zero, there's exactly one.

|x + 4| = −2 has no solution because a distance can't be −2. You can stop as soon as you see the negative, with no algebra needed. For |3x − 6| = 0, both cases are the same equation, 3x − 6 = 0, so they give the same single answer.

Watch out, though. The check for a negative has to happen after you isolate the absolute value. |x| − 5 = −2 looks negative at first, but adding 5 gives |x| = 3, which has two solutions.

What if there's a number outside the bars?

Get the absolute value alone on one side before you split into cases. For 3|x − 1| = 12, divide by 3 first to get |x − 1| = 4, which gives x = −3 or x = 5. For |x + 1| − 4 = 6, add 4 to get |x + 1| = 10, which gives x = −11 or x = 9.

Treat the bars like a pair of parentheses you aren't allowed to open yet. Anything outside them gets moved with ordinary inverse operations, and only then do you split into the two cases.

Do this step yourself before typing. This absolute value solver currently expects the form |linear expression| = number, so enter |x − 1| = 4, not 3|x − 1| = 12. The inside also has to be linear, so |x² − 1| = 3 isn't supported.

What are common mistakes with absolute value equations?

Solving only the positive case is the most frequent. That gives one answer to a problem with two.

Another is splitting before isolating. From 3|x − 1| = 12, writing 3x − 1 = 12 or 3x − 1 = −12 mixes up the coefficient and the bars. Divide by 3 first.

Last, some students "solve" |x + 4| = −2 and get x = −6 or x = −2. Neither checks: |−6 + 4| = 2, not −2. Always substitute your answers back in.

What if x has a negative sign inside the bars?

Nothing changes about the method. Split into two cases and solve each, dividing by −1 where needed.

For |5 − x| = 3, the cases are 5 − x = 3 and 5 − x = −3. The first gives −x = −2, so x = 2. The second gives −x = −8, so x = 8. Both check: |5 − 2| = 3 and |5 − 8| = |−3| = 3.

Notice that |5 − x| and |x − 5| are always equal, because a − b and b − a are opposites with the same distance from zero. So |5 − x| = 3 asks for the numbers 3 away from 5, which are 2 and 8.

Where do absolute value equations come up?

They show up whenever something is allowed to be a set amount above or below a target. Machined parts, thermostat ranges and test score cutoffs all work this way.

Say a bolt should be 50 mm long, and the limit is 2 mm either way. The lengths right at the limit solve |x − 50| = 2, which gives x = 48 or x = 52. The bolts that pass are the ones between those two numbers, and finding all of them is an inequality question.