Absolute Value — Equations

Pick a number. Any number. Now ask: how far is it from zero?

If you picked 7, the answer is 7. If you picked -7, the answer is also 7. Both numbers sit exactly seven units from zero on the number line — one to the right, one to the left. The direction is different, but the distance is the same.

That distance — stripped of direction, always non-negative — is the absolute value. It is written with vertical bars: |7| = 7 and |-7| = 7. The bars act like a filter that forgets sign and keeps only magnitude.

Now suppose someone hands you the equation |x| = 5 and asks you to solve it. They are really asking: which numbers on the number line are exactly 5 units from zero? There are exactly two such numbers — one on each side. So x = 5 or x = -5. That is the entire solution, and it follows directly from what absolute value means.

This two-answer pattern is the engine behind every absolute value equation. The bars hide a fork in the road: whatever is inside could be positive or negative, and each possibility is a separate equation to solve.

Definition and properties

Three properties follow immediately and are worth holding in your head. For all real numbers a and b:

1. Non-negativity. |a| \ge 0, with equality only when a = 0.
2. Product rule. |ab| = |a| \cdot |b|. The absolute value of a product is the product of the absolute values.
3. Triangle inequality. |a + b| \le |a| + |b|. You met this in Operations and Properties — it says that distances can cancel when signs disagree, but they can never amplify.

The product rule is the one that matters most for equations. It tells you that |ab| = |a| \cdot |b|, so you can pull factors outside the bars one at a time. The triangle inequality is less about solving equations and more about bounding them — it becomes central in Absolute Value — Inequalities.

Equations of the form |x| = a

The simplest type. The equation |x| = a asks: which numbers are a units from zero?

• If a > 0: two solutions, x = a and x = -a.
• If a = 0: one solution, x = 0.
• If a < 0: no solution. Distance is never negative, so |x| can never equal a negative number.

That third case is worth pausing on. If you see |x| = -3 on a test, you can immediately write "no solution" without any algebra at all. The bars guarantee non-negativity.

Now replace the bare x with a linear expression. The equation |2x - 6| = 10 asks: which values of x make 2x - 6 exactly 10 units from zero? The expression inside the bars is either +10 or -10:

Case 1. 2x - 6 = 10. Then 2x = 16, so x = 8.

Case 2. 2x - 6 = -10. Then 2x = -4, so x = -2.

Both answers should be checked in the original equation. |2(8) - 6| = |10| = 10. |2(-2) - 6| = |-10| = 10. Both check out.

There is a pattern here worth naming. For any equation |f(x)| = a with a > 0, the procedure is:

1. Write two cases: f(x) = a and f(x) = -a.
2. Solve each case as an ordinary equation.
3. Check every answer in the original equation (this matters when f is nonlinear or when the equation has extra terms outside the bars).

Equations of the form |f(x)| = |g(x)|

When absolute values appear on both sides, the distance interpretation shifts. The equation |f(x)| = |g(x)| says: the distance of f(x) from zero equals the distance of g(x) from zero. Two numbers have the same absolute value precisely when they are either equal or negatives of each other. So:

That is the whole rule. No new machinery — just two equations, each solved on its own.

Take |3x - 1| = |x + 5|.

Case 1. 3x - 1 = x + 5. Then 2x = 6, so x = 3.

Case 2. 3x - 1 = -(x + 5). Then 3x - 1 = -x - 5, so 4x = -4, giving x = -1.

Check: |3(3) - 1| = |8| = 8 and |3 + 5| = |8| = 8. Good. |3(-1) - 1| = |-4| = 4 and |-1 + 5| = |4| = 4. Also good.

There is an alternative approach worth knowing. Since |f(x)| = |g(x)| is the same as |f(x)|^2 = |g(x)|^2, and |a|^2 = a^2 for all real a, you can square both sides to get [f(x)]^2 = [g(x)]^2. Rearranging:

This uses the difference-of-squares identity. The advantage: it avoids case-splitting entirely and lands you at the same two equations (f(x) = g(x) and f(x) = -g(x)) in one algebraic move. The disadvantage: when f and g are not linear, squaring can introduce higher-degree equations. For linear f and g, the squaring route is clean and fast.

Equations with multiple absolute values

When more than one set of absolute-value bars appears in the same equation, the case-splitting idea extends — but you need to organise the cases carefully.

Consider |x - 1| + |x - 4| = 5. There are two absolute-value expressions, and each flips sign at a different point: x - 1 changes sign at x = 1, and x - 4 changes sign at x = 4. These critical points divide the number line into three intervals:

• Region 1: x < 1. Both x - 1 < 0 and x - 4 < 0. So |x - 1| = -(x - 1) = 1 - x and |x - 4| = -(x - 4) = 4 - x. The equation becomes (1 - x) + (4 - x) = 5, which simplifies to 5 - 2x = 5, giving x = 0. Check: 0 < 1, so the answer is in Region 1. And |0 - 1| + |0 - 4| = 1 + 4 = 5. Valid.

• Region 2: 1 \le x \le 4. Here x - 1 \ge 0 and x - 4 \le 0. So |x - 1| = x - 1 and |x - 4| = 4 - x. The equation becomes (x - 1) + (4 - x) = 5, which simplifies to 3 = 5. The left side collapses to 3 for every x in this interval, so the equation has no solution in this region.

• Region 3: x > 4. Both expressions are positive. |x - 1| = x - 1 and |x - 4| = x - 4. The equation becomes (x - 1) + (x - 4) = 5, giving 2x - 5 = 5, so x = 5. Check: 5 > 4, and |5 - 1| + |5 - 4| = 4 + 1 = 5. Valid.

The solutions are x = 0 and x = 5.

The general strategy for equations with multiple absolute values:

1. Find the critical points — the values of x where each expression inside a |\cdot| equals zero.
2. Divide the number line into intervals at those critical points.
3. In each interval, every absolute-value expression has a definite sign, so you can drop the bars (inserting a minus sign where needed) and solve the resulting ordinary equation.
4. Check that each solution actually falls in the interval you assumed. Discard any that don't.

This method always works, no matter how many absolute values appear. With k absolute-value terms, you get at most k + 1 regions and at most k + 1 equations to solve.

An interactive look at |x - c| = r

The equation |x - c| = r asks: which numbers are exactly r units away from c? Drag the two controls below to change c (the centre) and r (the radius). The two solution points move in real time.

Notice that the two solutions are always symmetric around c. The "\pm" in the solutions x = c \pm r is naming exactly this symmetry — one solution sits r units to the right of c, the other r units to the left.

Two worked examples

Common confusions

A few traps that catch almost everyone the first time.

• "|x| = -5 has the solution x = -5." No. The absolute value is never negative, so |x| = -5 has no solution at all. The negative sign inside the bars and the negative sign on the right side are doing completely different things.

• "|x^2 - 4| = x^2 - 4 always." Only when x^2 - 4 \ge 0, which means x \le -2 or x \ge 2. Between -2 and 2, x^2 - 4 is negative, so |x^2 - 4| = -(x^2 - 4) = 4 - x^2. The absolute value is a piecewise definition, and the "which piece?" question depends on the sign of whatever is inside.

• "After splitting into cases, I do not need to check answers." You always need to check. The case-split introduces assumptions about sign — and sometimes a solution from one case violates the sign assumption of that case, or the original equation has restrictions you cannot see after splitting. Checking takes ten seconds and saves you from phantom solutions.

• "|a + b| = |a| + |b|." Almost never. This is the triangle inequality trap. The equality holds only when a and b have the same sign (or one is zero). In general, |a + b| \le |a| + |b|. If you treat the bars as distributing over a sum, you will get wrong answers.

• "|f(x)| = |g(x)| means f(x) = g(x)." Only half the story. It also includes f(x) = -g(x). Two numbers can have the same absolute value by being negatives of each other. Missing the second case means losing half the solutions.

Going deeper

Absolute value as a metric

The absolute value is more than a convenience — it is the distance function on the real line. The distance between any two real numbers a and b is |a - b|. This function has three properties that any distance function must have:

1. |a - b| \ge 0, with equality if and only if a = b (non-negativity).
2. |a - b| = |b - a| (symmetry — distance does not depend on direction).
3. |a - c| \le |a - b| + |b - c| (triangle inequality — going through an intermediate point never shortens the journey).

These three properties define what mathematicians call a metric. The absolute value is the simplest example — the one-dimensional metric on \mathbb{R}. Every equation |x - c| = r is asking for the set of points at distance r from c in this metric. In one dimension, that set has at most two elements. In two dimensions, the same question — "which points are at distance r from c?" — gives a circle. In three dimensions, a sphere. The pattern extends all the way up.

Why checking matters: extraneous solutions

When you solve an equation by squaring both sides — as in the alternative approach for |f(x)| = |g(x)| — you are applying a non-invertible operation. Squaring turns both 5 and -5 into 25, so it can introduce solutions that satisfy the squared equation but not the original. These are called extraneous solutions, and they are the reason the "check" step is not optional.

For absolute value equations with linear interiors, extraneous solutions rarely appear — both cases usually give genuine answers. But the moment the interiors are nonlinear (quadratics, rationals, square roots), extraneous solutions become common. The habit of checking saves you from reporting answers that do not actually work.

Equations with nested absolute values

Sometimes absolute value expressions nest: \big||x| - 3\big| = 2. The outer bars ask for |x| to be a distance 2 from 3, so |x| = 5 or |x| = 1. Each of these gives two solutions: x = \pm 5 and x = \pm 1. So the original equation has four solutions: x \in \{-5, -1, 1, 5\}. Nested absolute values peel off from the outside in — each layer doubles the number of cases, but each individual case is simple.

Where this leads next

Absolute value equations are the prerequisite for several strands.

• Absolute Value — Inequalities — what happens when = becomes < or >. The two-point answer becomes an interval.
• Intervals and Inequalities Preview — the notation and logic of intervals, which absolute value inequalities produce as answers.
• Operations and Properties — the triangle inequality and the modulus rules that underpin everything on this page.
• Quadratic Equations — Introduction — when the expression inside the bars is quadratic, the case-split produces quadratics to solve.
• Linear Equations in One Variable — the foundational equation-solving moves that each case reduces to.