Math · College algebra · Concept
Absolute value equations and inequalities
To solve an absolute value equation, isolate the absolute value, then split |A| = c into two equations, A = c or A = −c, and solve both. For an inequality with c > 0, |A| < c becomes the single interval −c < A < c, and |A| > c becomes two pieces, A < −c or A > c.
Absolute value is distance
|x| is the distance from x to 0 on the number line, so it is never negative: |5| = 5 and |−5| = 5. More generally, |x − a| is the distance between x and a. Reading an equation as a distance often shows the answer at once: |x − 2| = 3 asks for the numbers 3 units from 2, which are −1 and 5.
Isolate, then split into two cases
First get the absolute value alone on one side, just as you would isolate x. Only then split: |A| = c means A is c units from zero, so A = c or A = −c. Solve both equations; each usually gives one solution.
When there is no solution, or only one
A distance cannot be negative, so |A| = c has no solution when c < 0. Stop as soon as the isolated form shows it: |x + 2| + 5 = 2 gives |x + 2| = −3, which no number satisfies. When c = 0 the two cases coincide, and A = 0 gives one solution.
Less than: one interval
|A| < c says A is less than c units from zero, so A lies between −c and c. Write it as one three-part inequality and solve all three parts together. With ≤, the endpoints are included.
Greater than: two pieces
|A| > c says A is more than c units from zero, on either side, so A < −c or A > c. The answer is two rays joined by “or”, never a single interval between them. With ≥, the endpoints are included.
See it on a graph
Graph y = |A| and the horizontal line y = c. The solutions of |A| = c are where they meet, |A| < c is where the V lies below the line, and |A| > c is where it lies above. The graph also shows why a negative c gives no solution: the V never dips below the x-axis.
Common mistakes
- Splitting before isolating: in 2|x − 1| + 3 = 11, first get |x − 1| = 4, then split.
- Solving only A = c and missing the case A = −c.
- Writing |A| > c as one interval, −c > A > c, which no number satisfies: it is two pieces joined by “or”.
- Treating |x + 3| as |x| + 3: absolute value does not distribute over a sum.
- Solving |A| = c when c is negative instead of stopping: a distance cannot be negative.
Key terms
- Absolute value
- A number’s distance from zero, so it is never negative: |−3| = 3 and |3| = 3. In general, |x| = x when x ≥ 0 and |x| = −x when x < 0.
- Absolute value equation
- An equation with the variable inside absolute value bars. Once the absolute value is isolated, |A| = c with c > 0 splits into A = c or A = −c; |A| = 0 gives A = 0, and |A| = c with c < 0 has no solution.
- Absolute value inequality
- An inequality with the variable inside absolute value bars. For c > 0, |A| < c means −c < A < c, one interval; |A| > c means A < −c or A > c, two pieces.
- Compound inequality
- Two inequalities joined by “and” or “or”. “And” keeps the values that satisfy both, an intersection such as −2 ≤ x < 3; “or” keeps the values that satisfy either, a union such as x < −3 or x > 3.
- Solution set
- All the allowed values that make an equation or inequality true, such as {−2, 3}. Equivalent equations have the same solution set.
- Interval
- An unbroken stretch of the number line, such as all x from 2 to 5. A bracket [ ] includes an endpoint and a parenthesis ( ) leaves it out.
Work through an example
Solve |2x − 3| + 4 = 11 and check both solutions.
Solve an absolute value equation →Sources and scope
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Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
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