Absolute Value Calculator & Equation Solver — |ax + b| = c (Free)

An absolute value calculator measures distance from 0, so |−7| = 7 and |7| = 7. To solve |2x − 3| = 7, split it into two cases: 2x − 3 = 7 gives x = 5, and 2x − 3 = −7 gives x = −2. Both answers are 7/2 units from 3/2, the point where the inside equals 0.

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How to use the absolute value equation solver

1. Pick a tab: Equation for |ax + b| = c (and two harder forms), Inequality for <, ≤, > or ≥, Evaluate to work out an expression such as |−7 + 3| − |2 − 9|, Distance for the gap between two numbers, or Graph for y = a|x − h| + k.
2. Type the numbers. Whole numbers, decimals and fractions such as 3/4 all work, and the answer is kept exact.
3. Read the answer, the solution set or interval, the center and distance, and the number line. Open Show steps to see both cases and a check of every answer.

What absolute value means

The absolute value of a number is its distance from 0 on the number line. Distance is never negative, so |x| is always 0 or positive: |−4| = 4, |4| = 4 and |0| = 0. As a rule, |x| = x when x ≥ 0 and |x| = −x when x < 0. The second part is what turns −4 into −(−4) = 4.

The same idea measures the distance between any two numbers: the distance from a to b is |a − b|. The distance between −3 and 5 is |−3 − 5| = |−8| = 8, and the order does not matter, because |5 − (−3)| is also 8.

How to solve absolute value equations

An equation |ax + b| = c asks which x-values put the inside exactly c units from 0. There are two such numbers, c and −c, so you get two cases.

1. Isolate the absolute value. In 3|2x + 1| + 4 = 25, subtract 4 and divide by 3 first: |2x + 1| = 7.
2. Look at c. If c is negative there is no solution. If c = 0 there is one solution (solve ax + b = 0). If c is positive, continue.
3. Split into two cases: ax + b = c and ax + b = −c. Solve each as an ordinary linear equation.
4. Check each answer in the original equation.

Why there are two cases

Both 7 and −7 are 7 units from 0, so |2x + 1| = 7 holds when 2x + 1 = 7 (x = 3) and when 2x + 1 = −7 (x = −4). On the number line, the center is where the inside is 0, here x = −1/2, and the two solutions sit 7/2 units on either side of it: −1/2 − 7/2 = −4 and −1/2 + 7/2 = 3. The calculator draws this picture for every equation.

Two absolute values, or x on both sides

For |ax + b| = |cx + d|, the insides are either equal or opposite, so solve ax + b = cx + d and ax + b = −(cx + d). For |ax + b| = cx + d, the right side may be negative for some x, so each candidate must make cx + d ≥ 0. A candidate that fails is an extraneous solution and is rejected.

Worked examples

Example 1: isolate first

Solve 3|2x + 1| + 4 = 25. Subtract 4: 3|2x + 1| = 21. Divide by 3: |2x + 1| = 7. Case 1: 2x + 1 = 7, so x = 3. Case 2: 2x + 1 = −7, so x = −4. Check: 3|7| + 4 = 25 and 3|−7| + 4 = 25.

Example 2: an extraneous solution

Solve |x − 2| = 2x − 7. Case 1: x − 2 = 2x − 7 gives x = 5. Case 2: x − 2 = −(2x − 7) gives 3x = 9, so x = 3. Check: at x = 5 both sides equal 3, so it works. At x = 3 the right side is 2(3) − 7 = −1, and an absolute value cannot equal −1, so x = 3 is rejected. The only solution is x = 5.

Example 3: an inequality with a sign flip

Solve −2|x + 3| + 5 ≥ −3. Subtract 5: −2|x + 3| ≥ −8. Divide by −2 and flip the sign: |x + 3| ≤ 4. “Less than or equal” gives one three-part inequality: −4 ≤ x + 3 ≤ 4, so −7 ≤ x ≤ 1, or [−7, 1] in interval notation.

Absolute value inequalities

After you isolate the absolute value, the sign tells you the shape of the answer. “Less than” keeps x close to the center, so it becomes one AND inequality (a single interval). “Greater than” pushes x away from the center, so it becomes two inequalities joined by OR (two rays).

Watch the special values of c. |u| < −1 has no solution, and |u| > −2 is true for every x, because an absolute value is never negative. |u| < 0 has no solution, |u| ≤ 0 means u = 0, and |u| > 0 is true everywhere except where u = 0.

Graphing y = a|x − h| + k

The graph of an absolute value function is a V. Its vertex is (h, k), its axis of symmetry is x = h, and the arms have slopes a and −a. If a is positive the V opens up and k is the minimum; if a is negative it opens down and k is the maximum. For y = 2|x − 1| − 4, the vertex is (1, −4), the y-intercept is 2|0 − 1| − 4 = −2, and the x-intercepts come from |x − 1| = 2: x = −1 and x = 3.

Common mistakes to avoid

• Splitting before isolating. In 3|2x + 1| + 4 = 25, writing 3(2x + 1) + 4 = ±25 gives wrong answers. Get the bars alone first.
• Dropping the negative case. |x| = 5 has two answers, 5 and −5.
• Solving |x| = −5. There is no solution; do not write x = ±5.
• Forgetting the flip. Dividing an inequality by a negative number reverses the sign.
• Mixing up AND and OR. |u| < c is a single “between” interval; |u| > c is two pieces going outward.
• Skipping the check when x appears outside the bars, which is where extraneous answers hide.

Where absolute value is used

Tolerances are absolute value inequalities. A part meant to be 70 mm long that may be off by at most 5 mm must satisfy |x − 70| ≤ 5, so 65 ≤ x ≤ 75. Differences in temperature, elevation or account balances are distances, so they use |a − b|: the change from −15 °F to 40 °F is |40 − (−15)| = 55 degrees.

Checking your answer by hand

Before you trust any answer, substitute it into the original equation, including any numbers outside the bars. For an inequality, test one value inside your answer and one outside: the inside value should make the inequality true and the outside value should make it false. The calculator’s steps show both tests, so you can compare them with your own working.

Absolute value practice quiz: 10 questions

Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.

Frequently asked questions

Why does |ax + b| = c have two solutions?

Because two numbers are c units from 0: c and −c. The inside ax + b can equal either one, and each case gives its own x. For |2x − 3| = 7 the solutions are x = 5 and x = −2.

What if c is negative?

Then there is no solution. An absolute value is a distance, and a distance is never negative, so |4x − 1| = −2 can never be true. Isolate the bars first, though: −2|x| + 10 = 4 becomes |x| = 3, which does have solutions.

What if c is zero?

Then there is exactly one solution, because only 0 has absolute value 0. Solve ax + b = 0: |x + 5| = 0 gives x = −5.

Can an absolute value be negative?

No. |x| is always 0 or positive. An expression like −|−5| is negative, but that is the opposite of an absolute value: −|−5| = −5.

How do you solve absolute value inequalities?

Isolate the absolute value, then rewrite. For |u| < c use −c < u < c; for |u| > c use u < −c or u > c. Solve the pieces, write the answer in interval notation and test one point inside and one outside.

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