Solving Absolute Value Equations

Absolute value measures distance from zero, so it is never negative. The equation |x − 4| = 3 asks which numbers are 3 units from 4: 1 and 7. In algebra, that distance idea becomes two cases. Once an absolute value expression is isolated, set its inside equal to the positive right side and then to the negative of that right side. Solve both equations and check the candidates.

Understand the two cases

If |A| = c and c is positive, then A can be c or −c because both numbers have absolute value c. Therefore, write A = c or A = −c. The word 'or' matters: either case can produce a valid solution. For |x − 4| = 3, solve x − 4 = 3 to get x = 7, and x − 4 = −3 to get x = 1. Both are exactly 3 units from 4.

The two-case rule applies to the entire expression inside the absolute value. If the equation is |2x + 1| = 9, the cases are 2x + 1 = 9 and 2x + 1 = −9. Their solutions are x = 4 and x = −5. Do not change just the x term's sign; set the whole inside expression equal to each value.

Isolate the absolute value first

When other operations appear outside the bars, undo them before creating the two cases. For 3|x − 2| + 4 = 19, subtract 4 from both sides: 3|x − 2| = 15. Divide by 3: |x − 2| = 5. Then solve x − 2 = 5 or x − 2 = −5, giving x = 7 or x = −3. Substituting either value into the original equation returns 19.

Use inverse operations in the usual order while preserving the equality. If a coefficient multiplies the absolute value, divide by it after undoing any addition or subtraction outside the bars. If an outside operation changes the right side to a negative value, pause: a nonnegative absolute value cannot equal a negative number, so the equation has no real solution.

The right side determines the number of solutions

For |A| = c, if c is negative there are no real solutions because absolute value cannot be less than zero. If c = 0, there is one case, A = 0, because a number has absolute value zero only when the number itself is zero. If c is positive, there are generally two cases, A = c and A = −c. They may simplify to two different values, but check for repeated or invalid candidates in the original equation.

For example, |2x + 6| = −4 has no real solution. The left side is at least zero, so it cannot equal −4. For |2x + 6| = 0, solve 2x + 6 = 0 to obtain the single solution x = −3. For |2x + 6| = 4, use both cases: 2x + 6 = 4 gives x = −1, and 2x + 6 = −4 gives x = −5.

Equations with expressions on both sides

If an absolute value is set equal to another expression containing a variable, do not immediately split into two cases unless you can establish that expression is nonnegative. For |x − 1| = x − 5, the right side must be at least zero, which requires x ≥ 5. But for x ≥ 5, x − 1 is nonnegative, so |x − 1| = x − 1. The equation becomes x − 1 = x − 5, which is impossible. Therefore there is no solution.

Alternatively, a careful case analysis can be used, but it must account for the sign of the right side and of the expression inside the bars. The shortcut |A| = B implies A = B or A = −B is only safe when B is known to be nonnegative. On a basic exam, the right side is often a constant, but recognizing this condition prevents invalid answers in more complex equations.

Check each answer in the original equation

Solving an equation with absolute values can introduce errors if the bars are mishandled or if a sign is changed only in part of an expression. Substitute each candidate into the original equation, not just the two case equations. For 2|x + 3| − 1 = 9, isolate: |x + 3| = 5. Cases produce x = 2 or x = −8. Check: 2|5| − 1 = 9 and 2|−5| − 1 = 9.

A number-line check is also intuitive. The equation |x − a| = r describes points exactly r units from a. If r is positive, there are two points, a − r and a + r. If r is zero, both points coincide at a. If r is negative, no distance can satisfy it. This geometric view confirms the algebra and helps detect when a result is implausible.

Common mistakes

A frequent error is writing only A = c and forgetting A = −c. Another is changing the sign of one term rather than the entire expression inside the bars. Students also sometimes split before isolating the absolute value, which can make the cases harder to track. Finally, some apply the two-case rule when the other side is negative or variable without checking its sign. Isolate first, determine the right-side condition, then solve and verify.

1. Use the fact that an absolute value is nonnegative to check for no solution or a zero case.
2. Undo operations outside the bars to isolate the absolute value.
3. For a positive constant c, set the entire inside expression equal to c and −c.
4. Solve each linear equation and combine distinct solutions.
5. Substitute every candidate into the original equation.

Exam takeaway

Absolute value represents distance. Isolate it, check whether the right side can be a distance, and solve the positive and negative cases when the right side is positive. A zero right side gives one case; a negative right side gives none. Verify the solutions in the original equation.

Interpret absolute value as distance

The equation |x − 4| = 3 asks which numbers are three units from 4. There are two: x = 1 and x = 7. Algebraically, set x − 4 = 3 or x − 4 = −3 and solve both linear equations. This distance meaning explains why a positive right-hand side usually creates two candidates.

Isolate the absolute value first

For 2|x + 1| − 5 = 9, add 5 and divide by 2 to obtain |x + 1| = 7. Then x + 1 = 7 or x + 1 = −7, giving x = 6 or x = −8. Do not split into cases while the expression is still surrounded by addition or multiplication; isolate the absolute value before creating the two equations.

Check zero and negative right-hand sides

An absolute value cannot be negative. Thus |3x − 2| = −1 has no solution. If isolation gives |3x − 2| = 0, solve just 3x − 2 = 0; both branches would be identical and yield one solution. These cases explain why an absolute-value equation can have zero, one, or two solutions.

Check both candidates in the original equation

After solving the branches, substitute each candidate into the original equation, including any operations that were outside the absolute-value bars. This catches arithmetic errors and invalid candidates in equations with fractions or multiple steps. For example, solving |x − 2| = x can produce a candidate from one branch that fails the original equation; direct substitution exposes it.

If variables appear on both sides after splitting, solve each resulting equation and then test the candidates. If a branch simplifies to a false statement such as 4 = 7, that branch contributes no solution. If it simplifies to a true statement such as 4 = 4, every value satisfying the branch’s conditions may work; do not automatically report two numbers without checking what the equation means.