Absolute value
Absolute value equations
Absolute value equations can be solved by distance or by cases. The number of solutions depends on what the absolute value is compared with.
Equation |x| = a
Because |x| is distance from zero, comparing it with a number a gives three common situations.
a < 0
No solutions, because distance cannot be negative.
a = 0
One solution: x = 0.
a > 0
Two solutions: x = a or x = -a.
|x - p| = a
For a > 0, the solutions lie a units away from p.
Equation of the form |x| = a
This is the basic model: we look for numbers whose distance from zero is a. You can immediately tell whether there are no solutions, one solution, or two solutions.
Method: check the sign of a first, then write the solutions.
No solutions|x| = -3 has no solutions because |x| >= 0
One solution|x| = 0 gives x = 0
Two solutions|x| = 4 gives x = -4 or x = 4
Result formfor a > 0: x ∈ {-a, a}
Common mistakes
- - Ignoring a < 0 and writing two solutions despite a negative right side.
- - Writing only x = a when a > 0.
- - Treating |x| = 0 as if it produced two different solutions.
Equation of the form |x - a| = b
The expression |x - a| means the distance from x to the point a. If b is positive, the solutions lie b units left and right of a.
Method: read the equation as distance, or write x - a = b and x - a = -b.
Example|x - 3| = 5
Interpretationx is 5 units from 3
Two directionsx = 3 - 5 or x = 3 + 5
Resultx ∈ {-2, 8}
Common mistakes
- - Confusing |x - 3| with distance from -3.
- - Solving only x - a = b and forgetting x - a = -b.
- - Forgetting to check whether b is negative.
Equation that requires cases
When the right side also contains x, distance intuition is usually not enough. Split into cases according to the sign of the expression inside the absolute value.
Method: find the zero of the expression inside the absolute value, solve on each interval, and check the case condition.
Equation|x - 1| = 2x + 3
Split pointx - 1 = 0, so x = 1
For x < 1-(x - 1) = 2x + 3, so x = -2/3
For x >= 1x - 1 = 2x + 3, so x = -4, reject
Resultx = -2/3
Common mistakes
- - Rejecting a result without checking which interval it belongs to.
- - Changing the sign only next to x instead of changing the whole expression inside the absolute value.
- - Forgetting that the right side of an absolute value equation must be nonnegative at a solution.
Equation with many absolute values
Each absolute value can change sign at a different point. First mark every split point, then solve the equation separately on each interval.
Method: find the zeros of all expressions inside absolute values, order them on the number line, and remove absolute values according to signs on each interval.
Equation|x - 2| + |x + 1| = 5
Split pointsx = -1 and x = 2
Intervals(-∞, -1), [-1, 2), [2, ∞)
After solvingwe get x = -2 or x = 3
Check|-4| + |-1| = 5 and |1| + |4| = 5
Common mistakes
- - Creating cases for only one absolute value.
- - Not ordering the split points on the number line.
- - Keeping a result that does not belong to the interval where it was computed.
More advanced exam-style example
Exam tasks often place absolute values on both sides or hide a parameter inside a linear expression. The same rule still works: split points, intervals, checking.
Method: solve by intervals and substitute each result into both its interval condition and the original equation.
Equation|2x - 4| - |x + 1| = 3
Split points2x - 4 = 0 gives x = 2, and x + 1 = 0 gives x = -1
For x < -1-(2x - 4) - (-(x + 1)) = 3, so -x + 5 = 3 and x = 2, reject
For -1 <= x < 2-(2x - 4) - (x + 1) = 3, so -3x + 3 = 3 and x = 0
For x >= 22x - 4 - (x + 1) = 3, so x - 5 = 3 and x = 8
Resultx ∈ {0, 8}
Common mistakes
- - Squaring both sides too quickly, which can introduce extraneous solutions.
- - Skipping one of the split points.
- - Not substituting results back into the original equation.
When is the geometric interpretation useful?
The geometric interpretation is best when the equation directly describes distance on the number line.
|x - a| = b
Use the number line when you are looking for points b units away from a.
Sum of distances
For equations such as |x - 2| + |x + 1| = 5, geometry helps explain what happens between split points.
Quick result check
If the result is supposed to be a distance, you can immediately notice when the right side is negative.
When to be careful
If the right side contains x, geometry is often only support. Splitting into cases is safer.
How do you check the solution?
Checking is short, but it prevents most mistakes in absolute value equations.
Substitute into the equation
Substitute every result into the original equation, not into a transformed version from one case.
Check the interval
If a result was computed for x < 1, make sure it actually satisfies that condition.
Inspect the right side
In an equation |expression| = right side, the right side must be nonnegative after substitution.
Write the complete set
Combine solutions from all cases and remove any that do not satisfy the equation.
Absolute value equation checklist
- - Check whether the equation can be read as distance.
- - For |x| = a or |x - p| = a, inspect the sign of a first.
- - Find the zeros of every expression inside an absolute value.
- - Split the number line into intervals and remove absolute values according to signs.
- - Solve the equation on every interval separately.
- - Check whether each result belongs to its interval.
- - Substitute every result into the original equation.
- - Write the solution set without duplicates.
Next step
Absolute value inequalities
After equations, move on to inequalities. You will see how the same distance interpretation leads to solution intervals.
Go to absolute value inequalities