Did you know? The vertical bars in |x − 4| < 3 describe something you can picture without doing algebra: a distance from four that must be less than three. Once a JC1 student sees that, the two inequality signs in the final answer stop feeling like a mysterious formula.
In Singapore’s H2 Mathematics, syllabus 9758 includes the concept of modulus and relationships such as |x − a| < b and |x − a| > b. For JC1 learners in Bukit Timah, this is one of the fastest opportunities to replace memorised procedures with genuine understanding. A wrong inequality direction, an excluded endpoint or a forgotten second branch can each cost a correct-looking answer.
This JC1 H2 Mathematics Tutor Bukit Timah guide focuses on a single practical skill: reading and solving absolute value inequalities as distance conditions. You will find worked examples, boundary cases, a short self-test and a teaching approach that reveals exactly where a student’s reasoning breaks down.
Quick answer: close means BETWEEN, far means OUTSIDE
When b is positive, |x − a| < b says that x is less than b units away from a. It means a − b < x < a + b. The answer lies between the two boundary points.
By contrast, |x − a| > b says that x is more than b units away from a. It means x < a − b or x > a + b. The answer lies outside the two boundary points. The word “or” matters: the two outer regions are separate.
- |x − a| < b, with b > 0: a − b < x < a + b.
- |x − a| ≤ b, with b ≥ 0: a − b ≤ x ≤ a + b.
- |x − a| > b, with b ≥ 0: x < a − b or x > a + b.
- |x − a| ≥ b, with b > 0: x ≤ a − b or x ≥ a + b.
- The strict signs < and > exclude their boundary points; ≤ and ≥ include them when the expression is defined.
An important qualification: these familiar interval rules need the stated conditions on b. If b is negative or zero, first use the fact that an absolute value can never be negative.
Why modulus is a distance, not a decoration
On a number line, the distance between x and four is |x − 4|. If x = 6, the distance is |6 − 4| = 2. If x = 2, the distance is |2 − 4| = 2. Points equally far on opposite sides of four have the same modulus.
This is why there are usually two boundary locations. Saying “within three of four” means starting at four and travelling three units left and three units right. The boundary points are one and seven. Nothing about the formula requires us to memorise an arrow first.
The same meaning appears in real-world tolerances. If a measurement x must be within 0.2 units of a target value five, we can express it as |x − 5| ≤ 0.2. The allowed interval is 4.8 ≤ x ≤ 5.2. Mathematics turns an ordinary tolerance sentence into an exact statement.
Worked example 1: |x − 4| < 3
The distance between x and four must be strictly less than three. Draw the number line, mark four, then mark points one and seven at distance three. The acceptable values sit between those two boundary points. Therefore 1 < x < 7.
Check x = 4: |4 − 4| = 0, which is less than three. Check x = 2: the distance is two, also allowed. Check x = 8: the distance is four, not allowed. Check x = 1: the distance is exactly three, so it is excluded by the strict inequality.
In symbols, we can write −3 < x − 4 < 3. Add four to all three parts: 1 < x < 7. The algebra and the distance picture agree. That agreement is useful because each method checks the other.
Worked example 2: |x − 4| ≥ 3
Now the distance from four must be at least three. This time the middle is forbidden; values must lie on either outer side. The boundary points remain one and seven, but equality is allowed.
Therefore x ≤ 1 or x ≥ 7. For example x = −2 has distance six from four and is allowed. The value x = 5 has distance one and is not allowed. The values one and seven both work because their distance is exactly three.
A common error is to write 1 ≤ x ≤ 7, simply because the sign contains equality. The equality determines endpoint inclusion; it does not determine whether we keep the inside or outside intervals. The direction < or > determines the region.
Worked example 3: |2x − 3| ≤ 7
The bars now surround an expression, not x − a in its simplest form. We still have a distance from zero: the quantity 2x − 3 must lie between −7 and 7. Write −7 ≤ 2x − 3 ≤ 7.
Add three to every part: −4 ≤ 2x ≤ 10. Divide every part by positive two: −2 ≤ x ≤ 5. That is the complete solution. Do not accidentally divide only the middle part of a compound inequality.
Check x = −2: |−4 − 3| = 7, so the lower boundary belongs. Check x = 5: |10 − 3| = 7, so the upper boundary belongs. Check x = 6: |12 − 3| = 9, which fails. These simple tests prevent a surprising number of careless marks being lost.
Could we divide by two earlier? Yes, since |2x − 3| = 2|x − 3/2|. We can then express the inequality as |x − 3/2| ≤ 7/2. Its centre is 1.5, its radius is 3.5, and its boundaries are again −2 and five. This version makes the geometric meaning explicit.
Worked example 4: |3 − 2x| ≥ 5
A large modulus means two possible cases. Either 3 − 2x ≥ 5, or 3 − 2x ≤ −5. Solve each case separately.
From 3 − 2x ≥ 5, subtract three: −2x ≥ 2. Divide by negative two and reverse the inequality: x ≤ −1. For the other case, 3 − 2x ≤ −5 gives −2x ≤ −8. Divide by negative two and reverse again: x ≥ 4.
The final solution is x ≤ −1 or x ≥ 4. Check x = −1 to see |3 + 2| = 5. Check x = 4 to see |3 − 8| = 5. A middle value such as x = 1 gives |1| = 1 and correctly fails.
Notice the precise point at which a student might lose the question: dividing by a negative number. This is an inequality-order error, not a misunderstanding of modulus. The remedy should target that first incorrect step.
What if the number on the right is zero or negative?
The familiar two-boundary picture assumes a positive radius. What does |x + 2| < 0 mean? It asks for a non-negative distance to be negative. There are no solutions. What does |x + 2| ≤ 0 mean? Only distance zero is allowed, so x = −2.
What about |x + 2| ≥ 0? Every real x satisfies it, because all absolute values are non-negative. And |x + 2| > −3 is also true for every real x. A learner who blindly substitutes b = −3 into the usual interval formula may invent a nonsensical answer.
Build the habit of checking the right-hand side before applying a remembered rule. Is the required distance positive, zero or negative? That small decision determines whether the usual interval pattern applies.
A 30-second diagnostic for JC1 learners
Give a student these two statements without allowing algebra: “within four units of six” and “more than four units from six.” Ask them to sketch the accepted points. The first is 2 < x < 10 if “within” is used strictly; the second is x < 2 or x > 10.
Then change the wording from “less than” to “at most,” and ask what happens only to the boundary dots. If the student switches inside to outside, they are confusing the direction of comparison with endpoint inclusion.
Next give |2x − 3| ≤ 7 and ask for a check at a proposed boundary. This isolates the algebra. The teaching sequence moves from meaning to symbols, then to verification; it avoids using many worksheets to conceal a single conceptual gap.
Five practice questions with worked answers
1. Solve |x − 5| < 2. Distance from five is less than two, so 3 < x < 7.
2. Solve |x + 1| ≥ 4. Write x + 1 ≤ −4 or x + 1 ≥ 4. Therefore x ≤ −5 or x ≥ 3.
3. Solve |3x + 6| ≤ 9. Write −9 ≤ 3x + 6 ≤ 9. Subtract six and divide by three: −5 ≤ x ≤ 1.
4. Solve |x − 1| > 0. Any non-zero distance from one is allowed. Thus x < 1 or x > 1, equivalently x ≠ 1.
5. Solve |x − 8| ≤ −1. There are no solutions because an absolute value is always at least zero.
After checking the answers, ask the learner to explain one boundary using substitution. A correct interval that cannot be explained may be a remembered pattern rather than transferable understanding.
Why this matters for A-Level H2 Mathematics
The 2027 official Singapore-Cambridge H2 Mathematics syllabus 9758 explicitly introduces the modulus inequalities |x − a| < b and |x − a| > b within functions, graphs and equations. Students may also need to read an inequality from a graph or connect it with a function’s domain.
The connection to the previous guide on rational inequalities and sign charts is useful but not identical. In a simple modulus inequality, distance can identify the intervals immediately. In a rational inequality, the expression’s numerator and denominator signs and excluded values need separate attention. Recognising which structure the question has is part of mathematical problem solving.
Questions parents frequently ask
Does modulus mean an answer can never be negative?
No. A modulus output cannot be negative, but the solutions for x can include negative numbers. For example |x| < 3 allows all x between −3 and 3.
Why do some answers use AND and others use OR?
A point within a fixed distance must satisfy both boundary inequalities at once: an AND condition. A point farther than the distance can lie on either the far left or far right: an OR condition.
What should I do if the right-hand side is negative?
Use the non-negativity of absolute value first. Depending on the comparison, the solution may be empty or all real numbers; do not apply a formula that assumes a positive radius.
Is this only relevant in JC1?
The concept is foundational and may be taught in JC1, but it belongs to the H2 Mathematics examination syllabus as a whole. School schemes of work can distribute topics differently.
Where to go next
Try the new guides on when a derivative equals zero but the curve does not turn, recurrence sequences and fixed points, and vector proofs of collinearity. For a wider H2 Maths topic route, see the Bukit Timah H2 Mathematics overview.
The immutable eduKateSG small-group Mathematics teaching reference documents a Secondary 1 approach to diagnose, explain, practise and check independence. It does not imply that the same class details automatically apply to JC1. Check current arrangements separately on the Bukit Timah JC1 tuition page.
The idea to remember is wonderfully ordinary: modulus measures distance. Decide whether a point must stay near a centre or far from it, and the signs will begin to make sense.