These questions use the skills from the inequalities and feasible regions module: reversing a sign, number line intervals, boundaries, feasible regions and point checks. They are original and ordered from easier to harder.
Write full working on paper before you open each answer. Mark the first line where your working and ours differ, because that line is where the learning sits.
Questions 1 to 5: solving inequalities
Q1. Solve 2x + 3 < 11.
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Subtract 3: 2x < 8. Divide by 2 (positive, so no flip): x < 4.
Check: x = 3 gives 9 < 11, true. x = 4 gives 11 < 11, false.
x < 4
Q2. Dina solves −2x > 6 and writes x > −3. Find the error and give the correct answer.
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She divided by −2 but did not reverse the sign. The correct step gives x < −3.
Check: x = −4 gives −2(−4) = 8, and 8 > 6 is true. x = −3 gives 6, and 6 > 6 is false. Her answer x > −3 would accept x = 0, and 0 > 6 is false.
x < −3
Q3. Solve 4 − 5x ≥ 19.
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Subtract 4: −5x ≥ 15. Divide by −5 and reverse the sign: x ≤ −3.
Check: x = −3 gives 4 + 15 = 19, and 19 ≥ 19 is true. x = 0 gives 4, and 4 ≥ 19 is false.
x ≤ −3
Q4. Solve −x/3 ≤ 2.
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Multiply both sides by −3 and reverse the sign: x ≥ −6.
Check: x = −6 gives 6/3 = 2, and 2 ≤ 2 is true. x = 0 gives 0 ≤ 2, true. x = −9 gives 3 ≤ 2, false.
x ≥ −6
Q5. Solve 3(x − 2) > 5x + 4.
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Expand: 3x − 6 > 5x + 4. Subtract 5x: −2x − 6 > 4. Add 6: −2x > 10. Divide by −2 and reverse the sign: x < −5.
Check: x = −6 gives 3(−8) = −24 on the left and −30 + 4 = −26 on the right. −24 > −26 is true. x = −5 gives −21 and −21, and −21 > −21 is false.
x < −5
Questions 6 to 8: number line work and integers
Q6. Solve 7 − 2x ≥ −3 and list the positive integers that satisfy it.
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Subtract 7: −2x ≥ −10. Divide by −2 and reverse the sign: x ≤ 5.
The positive integers up to and including 5 are 1, 2, 3, 4 and 5.
Check: x = 5 gives 7 − 10 = −3, and −3 ≥ −3 is true. x = 6 gives −5, and −5 ≥ −3 is false.
x ≤ 5; integers 1, 2, 3, 4, 5
Q7. List the integers that satisfy −2 ≤ x < 3, and say which circle goes at each end.
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−2 is included because of ≤, so draw a closed circle at −2. 3 is excluded because of <, so draw an open circle at 3.
The integers are −2, −1, 0, 1 and 2, which is 5 integers.
−2, −1, 0, 1, 2
Q8. Solve 1 < 3x − 2 ≤ 10 and list the integer solutions.
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Add 2 to every part: 3 < 3x ≤ 12. Divide every part by 3 (positive, so no flip): 1 < x ≤ 4.
Integers: 2, 3 and 4.
Check: x = 4 gives 12 − 2 = 10, which is ≤ 10. x = 1 gives 1, and 1 < 1 is false.
1 < x ≤ 4; integers 2, 3, 4
Questions 9 to 12: boundaries, regions and checks
Q9. For y < 3x − 6, state the line style, the intercepts of the boundary, and which side to shade.
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The symbol is <, so the line is dashed.
Boundary y = 3x − 6. When x = 0, y = −6. When y = 0, 3x = 6, so x = 2. The intercepts are (0, −6) and (2, 0).
Test (0, 0): 0 < −6 is false, so the origin side is not wanted. Shade the side below the line.
Check: (0, −7) gives −7 < −6, true, and it lies below the line.
Q10. Find the corners of the region x ≥ 0, y ≥ 0, x + y ≤ 5 and y ≤ 3, then count the integer points inside or on its boundary.
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Corners: (0, 0); (5, 0) where y = 0 meets x + y = 5; (2, 3) where y = 3 meets x + y = 5, since x = 5 − 3 = 2; and (0, 3).
Integer points by row: y = 0 has x from 0 to 5, which is 6. y = 1 has x from 0 to 4, which is 5. y = 2 has x from 0 to 3, which is 4. y = 3 has x from 0 to 2, which is 3.
Total: 6 + 5 + 4 + 3 = 18.
Corners (0, 0), (5, 0), (2, 3), (0, 3); 18 integer points
Q11. Constraints: x ≥ 1, y > 0 and 2x + y ≤ 12. Which of A (3, 5), B (5, 2), C (5, 3), D (1, 0) are feasible?
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A: 3 ≥ 1, 5 > 0 and 6 + 5 = 11 ≤ 12. All true, feasible.
B: 5 ≥ 1, 2 > 0 and 10 + 2 = 12 ≤ 12. All true, feasible.
C: 10 + 3 = 13 and 13 ≤ 12 is false. Not feasible.
D: y > 0 becomes 0 > 0, which is false. Not feasible.
Q12. A baker makes x small cakes and y large cakes. There are at most 12 cakes altogether, at least 3 small cakes, and no more than twice as many large cakes as small cakes. Write the constraints, then test (4, 7) and (3, 7).
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Constraints: x + y ≤ 12, x ≥ 3 and y ≤ 2x.
(4, 7): 11 ≤ 12, 4 ≥ 3 and 7 ≤ 8 are all true. Allowed.
(3, 7): 10 ≤ 12 and 3 ≥ 3 are true, but 7 ≤ 6 is false. Not allowed.
If you got these wrong
| What went wrong | Go to |
|---|---|
| Wrong direction after dividing by a negative (Q2 to Q5) | Reverse an inequality when multiplying by a negative |
| Wrong circle or wrong integer list (Q6 to Q8) | Represent an interval on a number line |
| Wrong line style, intercepts or shading side (Q9) | Draw a boundary for a linear inequality |
| Wrong corners or integer count (Q10) | Identify a feasible region from combined constraints |
| A point accepted that breaks a constraint (Q11, Q12) | Check whether a proposed point satisfies every constraint |
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