Absolute Value Calculator – Equations and Inequalities

About the Absolute Value Equation Calculator

Absolute value problems look small but trip up many students, because one line hides two separate equations or a pair of inequalities that must be combined correctly. A sign slip, a forgotten case or an inequality that should have been flipped all lead to an answer that is only half right.

This calculator works with the linear form $|ax + b|$ compared with a number $c$ . It first checks the special cases where $c$ is negative or zero, which decide immediately whether there is no solution, a single point or every real number. Otherwise it splits the problem into its two cases, solves each one with exact fractions, and reports the answer both as a set of values and in interval notation, followed by a check you can repeat by hand.

What is an absolute value equation?

The absolute value $|u|$ is the distance of $u$ from zero on the number line, so it is never negative: $|5| = 5$ and $|-5| = 5$ . An absolute value equation sets such a distance equal to a number. Because two points lie at any positive distance from zero, the equation $|u| = c$ with $c > 0$ has two cases, $u = c$ and $u = -c$ .

An absolute value inequality compares the distance with a number instead. $|u| < c$ says that $u$ is closer to zero than $c$ , which produces one interval, $-c < u < c$ . $|u| > c$ says that $u$ is farther away than $c$ , which produces two separate rays, $u < -c$ or $u > c$ . Remembering the picture of distance on a number line is the easiest way to keep the two types apart.

How to use the Absolute Value Equation Calculator

Rewrite your problem so the absolute value is alone on one side, in the form $|ax + b|$ compared with a number.
Enter the coefficient $a$ and the constant $b$ from inside the bars. Fractions such as 1/2 and decimals are accepted.
Choose the relation: =, <, ≤, > or ≥.
Enter the number $c$ on the right-hand side and press Solve.
Read the case split, the solution set and the interval notation, then use the check to confirm the answer.

Formulas

Equation with c > 0

|u| = c \iff u = c\ \text{or}\ u = -c

Less than

|u| < c \iff -c < u < c

Greater than

|u| > c \iff u < -c\ \text{or}\ u > c

Definition

|u| = \begin{cases} u & u \ge 0 \\ -u & u < 0 \end{cases}

The piecewise definition explains why every problem splits into a positive and a negative case.

Worked example: |2x − 3| = 7

The worked example solves $|2x - 3| = 7$ . Since $7 > 0$ , the inside is either $7$ or $-7$ . The first case gives $2x = 10$ , so $x = 5$ ; the second gives $2x = -4$ , so $x = -2$ .

Both values are checked in the original equation: $|2 \cdot 5 - 3| = |7| = 7$ and $|2 \cdot (-2) - 3| = |-7| = 7$ . The solution set is $\{-2, 5\}$ .

The solution below is generated by the same engine that powers the calculator above.

Final answer

x = -2 \quad\text{or}\quad x = 5 \qquad \{-2,\ 5\}

How we read your problem

Solve an absolute value equation by splitting it into the cases where the inside is positive or negative.

|2x - 3| = 7

Step-by-step solution

Write the problem with the absolute value isolated
The absolute value is already alone on one side, so it can be split into cases.
$|2x - 3| = 7$
Split into two cases
The inside is either c units above zero or c units below zero.
$2x - 3 = 7 \quad\text{or}\quad 2x - 3 = -7$
Solve the positive case
$2x - 3 = 7 \;\Rightarrow\; 2x = 7 + 3 = 10 \;\Rightarrow\; x = 5$
Solve the negative case
$2x - 3 = -7 \;\Rightarrow\; 2x = -7 + 3 = -4 \;\Rightarrow\; x = -2$
Check each solution in the original equation
$|2\cdot (-2) - 3| = |-7| = 7 \\ |2\cdot 5 - 3| = |7| = 7$
Write the solution set in interval notation
$\{-2,\ 5\}$

Formulas used

Equation
$|u| = c\ (c > 0) \iff u = c \ \text{or}\ u = -c$
Less than
$|u| < c \iff -c < u < c$
Greater than
$|u| > c \iff u < -c \ \text{or}\ u > c$

An absolute value measures distance from zero, so |u| = c means u is exactly c units from zero on either side.

Verification

Substituting -2 and 5 gives 7 on the left, matching the right-hand side.

Detailed explanation

Why the right-hand side matters first

Before splitting into cases, look at the sign of $c$ . If $c$ is negative, the equation $|u| = c$ has no solution, since a distance cannot be negative. The inequality $|u| < c$ also has no solution, while $|u| > c$ is true for every real number. Students who skip this check often produce two numbers that fail when substituted back.

When $c = 0$ , the equation $|u| = 0$ forces $u = 0$ , so there is exactly one solution. The strict inequality $|u| < 0$ is impossible, $|u| \le 0$ again has the single solution $u = 0$ , and $|u| > 0$ holds for every value except the one that makes the inside zero.

Less-than gives an interval, greater-than gives two rays

For $|x + 4| < 6$ , the inside must lie strictly between $-6$ and $6$ . Subtracting 4 throughout gives $-10 < x < 2$ , which is the open interval $(-10, 2)$ . A single compound inequality like this is often called an and-inequality, because both conditions must hold at once.

For $|3x - 2| \ge 4$ , the inside must be at most $-4$ or at least $4$ . Solving each part gives $x \le -\frac{2}{3}$ or $x \ge 2$ , written $(-\infty, -\frac{2}{3}] \cup [2, \infty)$ . This is an or-inequality: a value only needs to satisfy one of the two parts. A quick test of a point between the endpoints, such as $x = 0$ , gives $|-2| = 2$ , which is not at least 4, confirming that the middle is excluded.

Negative coefficients and fractional answers

When $a$ is negative, dividing by it reverses each inequality sign. For $|-2x + 5| \le \frac12$ the compound inequality is $-\frac12 \le -2x + 5 \le \frac12$ . Subtracting 5 and dividing by $-2$ flips both signs and gives $\frac94 \le x \le \frac{11}{4}$ . The calculator solves each boundary equation separately and then orders the endpoints, which avoids the flip error entirely.

All arithmetic is done with exact fractions, so an answer such as $\frac{11}{4}$ is not replaced by a rounded decimal. This matters when the answer is used again, for example when graphing the solution on a number line or substituting it into another equation.

Common mistakes

Solving only the positive case $ax + b = c$ and forgetting $ax + b = -c$ .
Writing $|u| > c$ as $-c > u > c$ , which no number satisfies; it must be split into two separate inequalities joined by or.
Forgetting that $|u| = c$ has no solution when $c$ is negative.
Dividing by a negative coefficient without reversing the inequality sign.
Removing the bars before isolating them, for example treating $2|x| + 1 = 7$ as $2x + 1 = \pm 7$ instead of first getting $|x| = 3$ .

Tips

Think of $|x - k|$ as the distance between $x$ and $k$ ; then $|x - 3| < 2$ means points within 2 units of 3, the interval $(1, 5)$ .
Use the midpoint of the endpoints as a quick test value for less-than problems.
Mark open circles for < and > and closed dots for ≤ and ≥ when you draw the solution on a number line.
If the bars are not alone, move everything else away first, then enter the isolated form here.
For equations with absolute values on both sides, square both sides or compare $u = v$ and $u = -v$ ; the inequality calculator can then help with any remaining linear steps.

Frequently asked questions

How do you solve an absolute value equation?

Isolate the absolute value, check that the right-hand side is not negative, then split |u| = c into u = c and u = −c and solve each linear equation. Finally substitute both answers back to confirm them.

Why does an absolute value equation usually have two solutions?

Because two numbers lie at the same distance from zero on the number line, one on each side. For example both 5 and −5 have absolute value 5, so |u| = 5 is satisfied by u = 5 and by u = −5.

When does an absolute value equation have no solution?

When the isolated absolute value is set equal to a negative number, such as |x − 1| = −2. An absolute value is a distance and can never be negative, so no value of x works.

What is the difference between |x| < a and |x| > a?

|x| < a describes the numbers within a units of zero, a single interval from −a to a. |x| > a describes the numbers farther than a from zero, which form two rays: x < −a or x > a.

How do I write the answer in interval notation?

Use parentheses for endpoints that are not included (strict < or >) and square brackets for endpoints that are included (≤ or ≥). Two separate pieces are joined with the union symbol ∪, and infinity always takes a parenthesis.

Can the calculator handle fractions and decimals?

Yes. Enter values such as 1/2, -3/4 or 2.5 for a, b and c. The solution is computed exactly with fractions, so the endpoints are shown as reduced fractions rather than rounded decimals.

What happens when the right-hand side is zero?

|u| = 0 and |u| ≤ 0 have exactly one solution, where the inside equals zero. |u| < 0 has no solution, and |u| > 0 is true for every number except the one that makes the inside zero.

Solve your absolute value problem

Enter a, b, the relation and c above to get both cases, the exact solution set and the interval notation.

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