Absolute Value Equation Calculator
This absolute value equation calculator solves |ax + b| = c, the same with an expression on the right, and equations with bars on both sides — and it writes the condition down before splitting rather than checking for extraneous answers afterwards.
Type |x − 3| = 5 and you get x = −2 and x = 8, with the working underneath.
What the Bars Mean
|t| is the distance from t to zero. That is the whole definition, and almost everything else on this page follows from it rather than from a rule to memorise.
Two consequences are worth holding onto. A distance is never negative. And every positive distance is reached from two directions — which is why the bars come off two ways.
How Many Answers, Before You Start
Once the bars stand alone with a number on the right, one glance settles the count:
| Negative | no solution | |2x − 3| = −10 |
| Zero | exactly one | |x + 1| = 0 gives x = −1 |
| Positive | two | |x − 3| = 5 gives −2 and 8 |
The middle row is where marks go missing. Students split |x + 1| = 0 into two equations out of habit and hand in the same answer twice. Only zero has an absolute value of zero, so there is nothing to split.
If something sits outside the bars, isolate first. For 3|x − 2| + 4 = 19: take the 4 off, divide by 3, and you have |x − 2| = 5, giving −3 and 7. Splitting before the bars are alone is the commonest way to go wrong here, because the sign you attach belongs to the whole right-hand side, and until the left is bare you do not know what that side is.
Why You Are Told to Check
Every textbook says to check for extraneous solutions. Few say what you are checking. Here it is:
The moment you write A = −B, you have assumed that B is not negative.
The left-hand side is a distance and cannot be negative, so the right-hand side cannot be either. That assumption never appears in your working, which is exactly why it has to be tested at the end — or, better, written down before you split.
Take |x − 1| = 2x − 5. The right side is 2x − 5, so before doing anything, note that it must satisfy 2x − 5 ≥ 0.
x − 1 = 2x − 5 gives x = 4. There the right side is 3, which is fine.
x − 1 = −(2x − 5) gives x = 2. There the right side is −1 — negative, so the assumption fails and this is not a solution.
The answer is x = 4 alone. Put x = 2 back in and you get 1 on the left and −1 on the right, which is why it never worked. It is not a mysterious quirk of the method; it is the assumption showing itself.
When the Answer Is a Whole Range
A branch can collapse entirely. In |x − 3| = x − 3, one branch becomes 0 = 0 — true no matter what x is. That branch does not give one answer, it gives all of them, limited only by the condition that the right side stays non-negative.
So the answer is every x of 3 or more, written x ≥ 3. It is a half-line, not a point and not a pair of points.
This is worth knowing because a calculator that can only print a list of numbers has no way to say it. Ask one of those for this equation and you may get a single value, or nothing at all, and neither is right.
Bars on Both Sides Need No Check
With |2x − 1| = |x + 4| there is no hidden assumption at all. Neither side can be negative whatever x is, so both branches are legitimate and no answer can turn out extraneous.
2x − 1 = x + 4 gives x = 5; 2x − 1 = −(x + 4) gives x = −1. Both stand, and neither needs testing.
The same shape can also come out with a single answer. In |x + 1| = |x + 2| the first branch reads x + 1 = x + 2, which is 0 = 1 and impossible, so only the second survives: x = −3/2, the point exactly halfway between −1 and −2. Whenever the two insides climb at the same rate, one branch always cancels itself out like that, and one answer is all you can get.
Notice how much that differs from the section above. The same equation with a bare x + 4 on the right would need the check; with bars around it, it does not. Recognising which of the two you are looking at is most of the skill.
Filling In the Equation
| = a number | |ax + b| = c, the ordinary case |
| x on the right | |ax + b| = cx + d, where the check matters |
| bars both sides | |ax + b| = |px + q|, where it does not |
Type the numbers from inside the bars and whatever is on the right. If your equation has a multiplier or a constant outside the bars, tap the line underneath and two more boxes appear. Fractions like 3/4 and decimals both work, and answers stay exact.
Open the working and you get the isolation if there was any, the count or the condition before any splitting, each branch with its own verdict, and the answer with anything discarded listed beneath it.
Absolute Value Equations FAQ
How do you solve an absolute value equation?
Why does an absolute value equation have no solution?
What is an extraneous solution?
How many solutions does an absolute value equation have?
Can an absolute value equation have infinitely many solutions?
Do you check when there are bars on both sides?
A distance is never negative — and once you have said that out loud, the check stops being a ritual.
