Secondary Mathematics · Worked Repair Guide 10
An equation asks for values that make two expressions equal. An inequality asks for a set of values that make one expression larger or smaller than another. This difference changes the shape of the answer: equations often produce isolated values, while inequalities usually produce intervals or regions.
This guide develops one central habit: treat an inequality as a statement about order, not as an equation with a different symbol. Every transformation must preserve which side is greater, and every final answer should be interpreted as a set.
All examples are original teaching material. Graphical and compound-inequality extensions should be matched to the learner’s actual Secondary Mathematics route and school programme.
1. Inequalities describe ranges of values
The statement x>4 means every real number greater than 4. The answer is not one number. It is an infinite set.
The symbols < and > exclude the boundary value. The symbols ≤ and ≥ include it.
Entry check: decide whether x=5 satisfies x>4, x=4 satisfies x≥4, and x=4 satisfies x>4. The answers are yes, yes and no.
This simple substitution habit becomes a powerful checking tool later. Test a value that should be inside the solution set and one that should be outside it.
2. Addition and subtraction preserve order
If a<b, adding the same number c to both sides gives a+c<b+c. Subtraction is addition of a negative, so it also preserves the inequality direction.
Worked example: x+7<12. Subtract 7 from both sides to obtain x<5.
Check with x=4: 4+7=11<12. Check a value outside, x=6: 13<12 is false.
The method resembles equation solving because the same operation is applied to both sides, but the final object remains a set of values.
3. Multiplication by a positive number preserves order
If a<b and c>0, then ac<bc. Positive scaling keeps the number line in the same orientation.
Worked example: 3x≤18 gives x≤6 after dividing by positive 3.
Multiplying or dividing by a positive quantity therefore behaves like the familiar equation operation with no change to the sign.
Do not memorise “division never changes the sign”. The sign depends on the sign of the quantity used to divide.
4. Multiplication by a negative number reverses order
If 2<5, multiplying both sides by −1 gives −2>−5. The order reverses because reflection through zero reverses left and right on the number line.
Therefore when an inequality is multiplied or divided by a negative number, reverse the inequality sign.
Worked example: −3x<12. Divide by −3 and reverse the sign: x>−4.
Check x=0: −3(0)=0<12, so 0 belongs to the solution set x>−4. If the sign had not been reversed, the incorrect set x<−4 would fail this check.
5. The reversal rule can be understood geometrically
On a number line, 3 lies to the right of 1. Multiplying both by −1 maps them to −3 and −1. Now −3 lies to the left of −1.
Negative multiplication is a reflection through zero combined with scaling. Reflection reverses orientation, which is why the inequality sign changes.
This explanation is more durable than a slogan because it reveals the mathematical reason.
It also helps when the negative factor is hidden inside an algebraic step. Ask whether the transformation effectively multiplies or divides both sides by a negative quantity.
6. Solve multi-step linear inequalities in controlled stages
Worked example: 5−2x≥11.
Subtract 5: −2x≥6. Divide by −2 and reverse the inequality: x≤−3.
Check the boundary x=−3: 5−2(−3)=11, so equality is allowed. Check x=−4: 13≥11 true. Check x=0: 5≥11 false.
The three checks confirm both direction and boundary inclusion.
7. Expand brackets before deciding the final direction
Consider 4(2−x)>3x+1. Expand first: 8−4x>3x+1. Then 7>7x, so 1>x or x<1.
The inequality can be written x<1 to place the variable first. Reordering the two sides also reverses the visual direction because “1>x” and “x<1” are the same statement.
Do not reverse the sign merely because a negative term appears. Reversal is required when both sides are multiplied or divided by a negative quantity.
Keep each transformation explicit enough that the sign decision is visible.
8. Compound inequalities describe intersections or unions
The compound inequality 2<x≤7 means x is greater than 2 and at most 7. Both conditions must hold. This is the intersection of two sets.
By contrast x<−1 or x≥4 describes two separated parts of the number line. This is a union.
Worked example: solve 3<2x+1≤9. Subtract 1 throughout: 2<2x≤8. Divide all parts by positive 2: 1<x≤4.
For a three-part inequality, apply the same legal operation to all three parts.
9. Number lines make boundary information visible
A strict inequality such as x<3 is usually drawn with an open circle at 3 and shading to the left. An inclusive inequality x≤3 uses a closed or filled circle at 3.
Different textbooks may use slightly different drawing conventions, but the mathematical distinction is always the same: is the boundary included?
For 1<x≤4, mark 1 open and 4 closed, shading only between them.
Use the visual representation to check whether your symbolic answer matches the intended set.
10. Interval notation is another representation of the same set
Where interval notation is used, x<3 can be written (−∞,3), while x≤3 can be written (−∞,3].
The parenthesis at 3 signals exclusion; the square bracket signals inclusion. Infinity is not a reachable endpoint, so it is written with a parenthesis.
The set 1<x≤4 becomes (1,4].
Interval notation may not be required in every school route, but understanding it reinforces the set interpretation of inequalities.
11. Inequalities can model real constraints
Suppose a lift has a stated load limit of 600 kg. If the current load is 420 kg and each identical box has mass 18 kg, the number n of additional boxes must satisfy 420+18n≤600.
Solving gives 18n≤180, so n≤10. Since n counts boxes, n must also be a non-negative integer. Thus at most 10 additional boxes fit the model.
The algebraic solution set contains many real numbers, but the context restricts the meaningful values further.
Mathematical modelling therefore needs a return step from symbolic inequality to the actual quantity being counted or measured.
12. Bounds are inequalities created by rounding
If a length is recorded as 8.4 cm to the nearest 0.1 cm, then 8.35≤L<8.45.
This is a compound inequality describing all exact values that would round to 8.4 cm under the standard convention.
The lower boundary is included because 8.35 rounds up to 8.4. The upper boundary is excluded because 8.45 rounds to 8.5.
Bounds therefore connect measurement and inequality thinking. The Units, Scale and Measurement guide develops that connection further.
13. Two-variable inequalities describe regions
The equation y=2x+1 describes a line. The inequality y>2x+1 describes all points above that line.
If the boundary is strict, the boundary line is not included. If the inequality is y≥2x+1, the boundary line is included.
To test which side to shade, choose a point not on the line. The origin (0,0) is convenient when it does not lie on the boundary. For y>2x+1, substituting (0,0) gives 0>1, false, so the origin side is not included.
Graphical regions are set solutions in two dimensions, just as intervals are set solutions on a one-dimensional number line.
14. Several inequalities create a feasible region
Suppose x≥0, y≥0 and x+y≤6. The valid points lie in the first quadrant and on or below the line x+y=6.
The overlap of all three conditions is a triangular region with vertices (0,0), (6,0) and (0,6).
Each inequality removes part of the plane. The feasible region is the intersection of what remains.
This idea later supports optimisation and linear programming where those topics appear, but the underlying mathematics is already visible in basic region shading.
15. Quadratic inequalities require sign analysis
Where appropriate for the learner’s route, consider x²−5x+6>0. Factorise: (x−2)(x−3)>0.
The critical values are 2 and 3. Test the intervals x<2, 2<x<3 and x>3. The product is positive outside the roots and negative between them.
Therefore the solution is x<2 or x>3.
Do not solve each factor independently as though (x−2)>0 and (x−3)>0 were the only possibility. A product is also positive when both factors are negative.
16. Capstone: build, solve and interpret a constraint
An invented event budget is at most $840. A fixed venue charge is $300 and each participant pack costs $18. Let n be the number of packs. Build and solve the constraint.
The model is 300+18n≤840. Subtract 300: 18n≤540. Divide by 18: n≤30.
If n is a count of packs, then n is a non-negative integer, so n can be 0,1,2,…,30 within this simplified model.
Check n=30: cost=300+540=840, allowed. Check n=31: cost=858, not allowed. The boundary and direction are both confirmed.
17. Independent practice
- Solve x+5<12.
- Solve 4x≥28.
- Solve −3x≤15.
- Solve 7−2x>1.
- Solve 3(x−2)≤12.
- Solve 5−x<2x+11.
- Solve 2<x+1≤8.
- Solve −6≤2x<10.
- State whether x=4 belongs to x<4.
- State whether x=4 belongs to x≤4.
- Represent −2<x≤3 on a number line.
- Write x≥5 in interval notation if your course uses it.
- A value is rounded to 12.7 to the nearest 0.1. State its bounds.
- Describe the region y<x+2 relative to its boundary line.
- Does (0,0) satisfy y≥2x−1?
- For x≥0, y≥0, x+y≤4, list the three vertices of the feasible region.
- Solve x²−9>0 by factorisation and interval testing.
- A fixed fee is $120 and each item costs $8. The total must not exceed $400. Find the maximum whole number of items.
- Explain why dividing by −5 reverses an inequality.
- Check whether x=−2 satisfies 5−3x≥10.
18. Worked answers
1. x<7.
2. x≥7.
3. x≥−5. Divide by −3 and reverse the sign.
4. x<3. −2x>−6 then divide by −2.
5. x≤6.
6. x>−2. Rearrangement gives −3x<6, then divide by −3.
7. 1<x≤7.
8. −3≤x<5.
9. No. Strict inequality excludes 4.
10. Yes.
11. Open at −2, closed at 3, shaded between.
12. [5,∞).
13. 12.65≤x<12.75.
14. Below the line y=x+2, excluding the line.
15. Yes. 0≥−1 is true.
16. (0,0), (4,0), (0,4).
17. x<−3 or x>3. Factor (x−3)(x+3)>0.
18. 35 items. 120+8n≤400 gives n≤35.
19. Multiplication by a negative reflects the number line and reverses order.
20. Yes. 5−3(−2)=11≥10.
19. Diagnose the first inequality error
Common failures include forgetting sign reversal, reversing the sign when merely adding a negative term, shading the wrong side of a boundary, confusing “and” with “or”, or forgetting that a contextual variable may need to be a whole number.
A useful correction note names the order-preservation issue: “divided by negative, sign must reverse”, “strict boundary excluded”, or “both compound conditions must hold”.
Then test a boundary value, one interior value and one exterior value. Three small substitutions can reveal whether the set has the correct direction and endpoints.
20. Continue through the BTT Mathematics library
Return to the BTT Mathematics Hub. Use Equations, Balance and Checking for equality-preserving transformations, Graphs, Tables and Relationships for lines and coordinates, and Units, Scale and Measurement for bounds created by rounding.
The BTT Mathematical Lab is the diagnostic route when inequality direction or boundary errors recur.
21. Sources and scope
The algebraic, graphical and modelling examples in this guide are original teaching material. They are designed to expose structure, not to reproduce examination questions.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match interval notation, quadratic inequalities and region work to the learner’s actual subject level and school programme.