Modulus • Topic 2 of 3

Modulus Equations

A modulus equation is solved by removing the bars correctly. For |ax + b| = c, first check c ≥ 0 (if c < 0 there is no solution), then split into ax + b = c and ax + b = −c and solve each. Always substitute solutions back, because squaring or combining moduli can create false roots. The richer CAT type is the sum-of-distances equation such as |x − a| + |x − b| = c. Here the critical points are a and b; they cut the number line into three regions, and in each region every modulus opens with a fixed sign, turning the equation into a simple linear one. A geometric shortcut: |x − a| + |x − b| is the total distance from x to a and to b, whose minimum value is |a − b|. So if c < |a − b| there is no solution; if c = |a − b| every point between a and b works; if c > |a − b| there are exactly two solutions, one on each side.

✅ Solved examples

1. Solve |2x − 3| = 7.
2x − 3 = 7 ⇒ x = 5; or 2x − 3 = −7 ⇒ 2x = −4 ⇒ x = −2. So x = 5 or x = −2.
2. Solve |x − 1| + |x − 5| = 8.
Minimum of the left side is |1 − 5| = 4, and 8 > 4 so two solutions lie outside [1,5]. For x > 5: (x−1)+(x−5) = 8 ⇒ 2x − 6 = 8 ⇒ x = 7. For x < 1: (1−x)+(5−x) = 8 ⇒ 6 − 2x = 8 ⇒ x = −1. So x = 7 or x = −1.
3. Solve |x + 2| = 3x − 4.
Need RHS ≥ 0, i.e. x ≥ 4/3. Case x ≥ −2: x + 2 = 3x − 4 ⇒ 6 = 2x ⇒ x = 3 (valid). Case x < −2: −(x+2) = 3x − 4 ⇒ −x − 2 = 3x − 4 ⇒ 2 = 4x ⇒ x = 0.5 (rejected, not < −2). So x = 3.
4. For how many real x does |x − 3| + |x − 6| = 2 hold?
Minimum of the left side is |3 − 6| = 3, which already exceeds 2. Since the sum can never be less than 3, there is no solution. Count = 0.

✏️ Practice — try these, take hints as needed

1. Solve |3x + 1| = 10.
Split into ±10.
3x + 1 = 10 or 3x + 1 = −10.
x = 3 or 3x = −11.
x = 3 or x = −11/3
2. Solve |x − 2| + |x − 8| = 10.
Critical points 2 and 8; min sum = 6.
For x > 8: 2x − 10 = 10.
For x < 2: 10 − 2x = 10.
x = 10 or x = 0
3. How many solutions does |x − 4| + |x − 9| = 5 have?
Min sum = |4 − 9| = 5.
c equals the minimum.
Every point in [4,9] works.
Infinitely many (all x with 4 ≤ x ≤ 9)
4. Solve |2x − 5| = x + 1.
Need x + 1 ≥ 0.
2x − 5 = x + 1 or 2x − 5 = −(x+1).
Check each root.
x = 6 or x = 4/3
5. Solve |x + 3| = |2x − 1|.
Equal moduli ⇒ equal or opposite.
x + 3 = 2x − 1 or x + 3 = −(2x − 1).
Solve both.
x = 4 or x = −2/3

📝 Topic test — 8 questions

Auto-graded with full solutions; saved to your dashboard. Use the calculator and formula sheet (top-right) any time.

Loading questions…