To solve an equation with an absolute value (modulus), split it into two cases: the expression inside equals the right side, or it equals the negative of the right side. Then substitute each answer back to confirm it works.
This is part of algebraic equations and inequalities and pairs naturally with solving a quadratic inequality using sign intervals, since both rely on thinking about where an expression is positive or negative.
What does the modulus actually do?
|a| means the distance of a from zero, so it always returns a value that is zero or positive. |5| = 5 and |−5| = 5.
Because the bars hide the sign, an equation |A| = B can only be true if A = B or A = −B, and only when B ≥ 0. That single sentence is the whole method.
Method, step by step
- Isolate the modulus on one side, if it is not already alone.
- Check the right side: if it is a number, it must be positive or zero. If it contains x, remember it must end up non-negative.
- Write two equations: inside = right side, and inside = −(right side).
- Solve both as ordinary linear equations.
- Substitute each answer into the original equation and reject any that fail.
Worked example
Solve |3x − 1| = 2x + 4.
Case 1: 3x − 1 = 2x + 4, so x = 5.
Case 2: 3x − 1 = −(2x + 4) = −2x − 4. Then 5x = −3, so x = −3/5.
Check x = 5: left side |15 − 1| = 14. Right side 10 + 4 = 14. Matches.
Check x = −3/5: left side |−9/5 − 1| = |−14/5| = 14/5. Right side −6/5 + 4 = 14/5. Matches.
Answer: x = 5 or x = −3/5.
Both cases passed, but the check was still essential. The right side 2x + 4 is negative for x < −2, and any answer below −2 would have had to be rejected.
The mistake to watch for
The usual slip is to solve only the positive case, or to apply the negative to only part of the right side.
Mistaken working: 3x − 1 = −2x + 4
The student negated the 2x but left the 4 alone, which changes the equation.
The negative applies to the whole right side, so it must be written with brackets: 3x − 1 = −(2x + 4). Expanding gives −2x − 4, and the solution x = −3/5 follows. Writing the brackets every time removes this error.
Check yourself
Try these, then open each answer.
1. Solve |x + 3| = 8.
Show answer
Case 1: x + 3 = 8, so x = 5. Case 2: x + 3 = −8, so x = −11. Check: |8| = 8 and |−8| = 8.
x = 5 or x = −11
2. Solve |4x − 5| = 3x.
Show answer
Case 1: 4x − 5 = 3x, so x = 5. Case 2: 4x − 5 = −3x, so 7x = 5 and x = 5/7.
Check x = 5: |15| = 15 and 3(5) = 15. Check x = 5/7: |20/7 − 35/7| = 15/7 and 3(5/7) = 15/7.
x = 5 or x = 5/7
3. Explain why |2x + 1| = −3 has no solution.
Show answer
An absolute value is a distance, so it cannot equal a negative number. No value of x can make |2x + 1| equal −3.
No solution
Where this leads next
The same habit of testing answers returns in rejecting an extraneous root after squaring. When you are ready, try the mixed practice set, and use the non-calculator working trainer to practise the exact-fraction checks.
Modulus questions reward careful case-writing more than speed. If you would like someone to watch how you set out cases and catch slips as they happen, our teachers do this in online one-to-one Additional Mathematics tuition.