Hard GCSE Linear inequalities Questions

Challenging, exam-style GCSE Linear inequalities questions with worked solutions. Stretch yourself on the hardest unknowns on both sides, collecting like terms, inclusive inequality, negative coefficient problems.

unknowns on both sidescollecting like termsinclusive inequalitynegative coefficientreversing the inequality signexpanding brackets
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Solve 2≤5−3x≤112 \le 5 - 3x \le 11 and list all the integer values of xx in the solution set.
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Worked solution

  1. Write down the double inequality

    2≤5−3x≤112 \le 5 - 3x \le 11

    The xx term is negative, so a flip is coming — but do the additions first.

  2. Subtract 5 from all three parts

    −3≤−3x≤6-3 \le -3x \le 6

    Subtracting leaves both signs pointing the same way: 2−5=−32 - 5 = -3 and 11−5=611 - 5 = 6.

  3. Divide all three parts by −3-3 and flip both signs

    1≥x≥−21 \ge x \ge -2

    Dividing by the negative number −3-3 reverses **both** inequality signs: −3÷−3=1-3 \div -3 = 1 and 6÷−3=−26 \div -3 = -2.

  4. Rewrite smallest-to-largest

    −2≤x≤1-2 \le x \le 1

    A double inequality is always written with the smaller number on the left, reading along the number line.

  5. Draw the solution set

    −2≤x≤1-2 \le x \le 1

    Both signs are inclusive, so **both** circles are closed and the segment between them is shaded.

  6. Check the lower boundary

    x=−2:  5−3(−2)=11,2≤11≤11  ✓x = -2: \; 5 - 3(-2) = 11, \quad 2 \le 11 \le 11 \; \checkmark

    At x=−2x = -2 the middle hits the **upper** limit exactly — the flip has swapped which end is which. Since the sign is ≤\le, it is included.

  7. Check the upper boundary

    x=1:  5−3(1)=2,2≤2≤11  ✓x = 1: \; 5 - 3(1) = 2, \quad 2 \le 2 \le 11 \; \checkmark

    At x=1x = 1 the middle hits the **lower** limit exactly, and ≤\le allows it.

  8. Check a value inside

    x=0:  5−3(0)=5,2≤5≤11  ✓x = 0: \; 5 - 3(0) = 5, \quad 2 \le 5 \le 11 \; \checkmark

    00 is comfortably inside the solution set.

  9. Check a value outside

    x=2:  5−3(2)=−1,2≤−1  ×x = 2: \; 5 - 3(2) = -1, \quad 2 \le -1 \; \times

    22 is just outside, and it fails the lower condition.

  10. Check the other side

    x=−3:  5−3(−3)=14,14≤11  ×x = -3: \; 5 - 3(-3) = 14, \quad 14 \le 11 \; \times

    −3-3 is just outside on the other end and fails the upper condition.

  11. List the integers

    x=−2,  −1,  0,  1x = -2, \; -1, \; 0, \; 1

    Both endpoints are included because both circles are closed, giving four integer solutions.

  12. Set notation

    {x:−2≤x≤1}\{x : -2 \le x \le 1\}

    A closed interval — both ends belong to the set.

  13. Why the order swapped

    dividing by −3 reverses order\text{dividing by } -3 \text{ reverses order}

    The lower limit 22 produced the **upper** bound x=1x = 1, and the upper limit 1111 produced the **lower** bound x=−2x = -2. That reversal is the flip in action.

  14. Watch the classic error

    −13≤x≤−2  is wrong-\frac{1}{3} \le x \le -2 \;\text{is wrong}

    Flipping only one of the two signs, or forgetting to reorder, produces a nonsensical inequality with the bigger number on the left.

  15. State the integer solutions

    x=−2,  −1,  0,  1x = -2, \; -1, \; 0, \; 1

    Four integers lie in the solution set −2≤x≤1-2 \le x \le 1.

Answer
x=−2,  −1,  0,  1x = -2, \; -1, \; 0, \; 1
Question 2
5 markschallenging
A rectangle has width ww cm and length 3w3w cm. Its perimeter is at least 2424 cm and less than 4040 cm. Find the range of possible values of ww.
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Worked solution

  1. Write the perimeter in terms of w

    P=2(w+3w)=2×4w=8wP = 2(w + 3w) = 2 \times 4w = 8w

    The perimeter of a rectangle is twice the sum of the width and the length, so P=8wP = 8w.

  2. Turn the words into inequality signs

    "at least"⇒≥,"less than"⇒<\text{"at least"} \Rightarrow \ge, \quad \text{"less than"} \Rightarrow <

    "At least 2424 cm" allows exactly 2424, so that end is inclusive; "less than 4040 cm" does not allow 4040, so that end is strict.

  3. Form the double inequality

    24≤8w<4024 \le 8w < 40

    Both conditions apply to the same perimeter 8w8w, so they can be written as one double inequality.

  4. Divide all three parts by 88

    3≤w<53 \le w < 5

    88 is positive, so both inequality signs stay as they are.

  5. State the solution

    3≤w<53 \le w < 5

    This is the complete solution set: every value of ww that makes the original inequality true, and no others.

  6. Show the solution set on a number line

    3≤w<53 \le w < 5

    A closed circle at 33 and an open circle at 55, with the line between them shaded. Open circles mark values that are excluded; closed circles mark values that are included.

  7. Check a value inside the solution set (w=4w = 4)

    24≤8(4)<40  ⇒  24≤32<40  ✓24 \le 8(4) < 40 \;\Rightarrow\; 24 \le 32 < 40 \; \checkmark

    44 lies between the two boundaries, and substituting it back into the original inequality makes both parts true.

  8. Test the lower boundary w=3w = 3

    24≤8(3)<40  ⇒  24≤24<40  ✓24 \le 8(3) < 40 \;\Rightarrow\; 24 \le 24 < 40 \; \checkmark

    At w=3w = 3 one part becomes an equality. The sign there is inclusive, so equality is allowed and 33 is part of the solution set — a closed circle.

  9. Test the upper boundary w=5w = 5

    24≤8(5)<40  ⇒  24≤40<40  ×24 \le 8(5) < 40 \;\Rightarrow\; 24 \le 40 < 40 \; \times

    At w=5w = 5 the other part becomes an equality. The sign there is strict, so 55 is excluded — an open circle.

  10. List the integers in the solution set

    w=3,  4w = 3, \; 4

    Only whole numbers strictly inside the shading count, plus any endpoint whose circle is closed. That gives 2 integer solutions.

  11. Check a value below the range (w=2w = 2)

    24≤8(2)<40  ⇒  24≤16<40  ×24 \le 8(2) < 40 \;\Rightarrow\; 24 \le 16 < 40 \; \times

    22 lies to the left of the shaded region and it breaks the inequality, exactly as it should.

  12. Check a value above the range (w=6w = 6)

    24≤8(6)<40  ⇒  24≤48<40  ×24 \le 8(6) < 40 \;\Rightarrow\; 24 \le 48 < 40 \; \times

    66 lies to the right of the shaded region and it also fails. The solution really is bounded on both sides.

  13. Write the solution set in set notation

    {w:3≤w<5}\{w : 3 \le w < 5\}

    The double inequality is shorthand for two conditions holding at the same time: w≥3w \ge 3 **and** w<5w < 5.

  14. Pick out the smallest and largest integers

    wmin=3,wmax=4w_{\text{min}} = 3, \quad w_{\text{max}} = 4

    A quick way to check the open/closed circles: if an endpoint is a whole number and its circle is open, that whole number must be missing from the list.

  15. Final check

    3≤w<53 \le w < 5

    Both boundaries behave exactly as the open/closed circles predict, the interior value works, and values on either side fail.

Answer
3≤w<53 \le w < 5
Question 3
5 markschallenging
Solve −1≤4−x<6-1 \le 4 - x < 6.
Show worked solution

Worked solution

  1. Write down the double inequality

    −1≤4−x<6-1 \le 4 - x < 6

    Whatever you do, do it to **all three** parts so that xx ends up alone in the middle.

  2. Subtract 44 from all three parts

    −5≤−x<2-5 \le -x < 2

    The same number is applied to every part, so the two inequality signs keep pointing the same way.

  3. Divide all three parts by −1-1 — both signs flip

    5≥x>−25 \ge x > -2

    Dividing by the negative number −1-1 reverses **both** inequality signs.

  4. Rewrite with the smaller number first

    −2<x≤5-2 < x \le 5

    A double inequality is always written smallest-to-largest, reading left to right along the number line.

  5. State the solution

    −2<x≤5-2 < x \le 5

    This is the complete solution set: every value of xx that makes the original inequality true, and no others.

  6. Show the solution set on a number line

    −2<x≤5-2 < x \le 5

    An open circle at −2-2 and a closed circle at 55, with the line between them shaded. Open circles mark values that are excluded; closed circles mark values that are included.

  7. Check a value inside the solution set (x=2x = 2)

    −1≤4−(2)<6  ⇒  −1≤2<6  ✓-1 \le 4 - (2) < 6 \;\Rightarrow\; -1 \le 2 < 6 \; \checkmark

    22 lies between the two boundaries, and substituting it back into the original inequality makes both parts true.

  8. Test the lower boundary x=−2x = -2

    −1≤4−(−2)<6  ⇒  −1≤6<6  ×-1 \le 4 - (-2) < 6 \;\Rightarrow\; -1 \le 6 < 6 \; \times

    At x=−2x = -2 one part becomes an equality. The sign there is strict, so equality is not allowed and −2-2 is excluded — an open circle.

  9. Test the upper boundary x=5x = 5

    −1≤4−(5)<6  ⇒  −1≤−1<6  ✓-1 \le 4 - (5) < 6 \;\Rightarrow\; -1 \le -1 < 6 \; \checkmark

    At x=5x = 5 the other part becomes an equality. The sign there is inclusive, so 55 is included — a closed circle.

  10. List the integers in the solution set

    x=−1,  0,  1,  2,  3,  4,  5x = -1, \; 0, \; 1, \; 2, \; 3, \; 4, \; 5

    Only whole numbers strictly inside the shading count, plus any endpoint whose circle is closed. That gives 7 integer solutions.

  11. Check a value below the range (x=−3x = -3)

    −1≤4−(−3)<6  ⇒  −1≤7<6  ×-1 \le 4 - (-3) < 6 \;\Rightarrow\; -1 \le 7 < 6 \; \times

    −3-3 lies to the left of the shaded region and it breaks the inequality, exactly as it should.

  12. Check a value above the range (x=6x = 6)

    −1≤4−(6)<6  ⇒  −1≤−2<6  ×-1 \le 4 - (6) < 6 \;\Rightarrow\; -1 \le -2 < 6 \; \times

    66 lies to the right of the shaded region and it also fails. The solution really is bounded on both sides.

  13. Write the solution set in set notation

    {x:−2<x≤5}\{x : -2 < x \le 5\}

    The double inequality is shorthand for two conditions holding at the same time: x>−2x > -2 **and** x≤5x \le 5.

  14. Pick out the smallest and largest integers

    xmin=−1,xmax=5x_{\text{min}} = -1, \quad x_{\text{max}} = 5

    A quick way to check the open/closed circles: if an endpoint is a whole number and its circle is open, that whole number must be missing from the list.

  15. Final check

    −2<x≤5-2 < x \le 5

    Both boundaries behave exactly as the open/closed circles predict, the interior value works, and values on either side fail.

Answer
−2<x≤5-2 < x \le 5
Question 4
5 markschallenging
Solve 3(2x+1)−2≤6x+53(2x + 1) - 2 \le 6x + 5. Which statement about the solution set is correct?
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Worked solution

  1. Write down the inequality

    3(2x+1)−2≤6x+53(2x + 1) - 2 \le 6x + 5

    Expand the bracket on the left, then compare the two sides.

  2. Expand the bracket

    6x+3−2≤6x+56x + 3 - 2 \le 6x + 5

    3×2x=6x3 \times 2x = 6x and 3×1=33 \times 1 = 3.

  3. Simplify the left-hand side

    6x+1≤6x+56x + 1 \le 6x + 5

    3−2=13 - 2 = 1.

  4. Compare the two sides

    6x  appears on both sides6x \;\text{appears on both sides}

    The coefficient of xx is 66 on the left and 66 on the right, so the xx terms will cancel.

  5. Subtract 6x from both sides

    1≤51 \le 5

    Nothing involving xx survives — what is left is a statement about numbers only.

  6. Judge the statement that is left

    1≤5  is true1 \le 5 \;\text{is true}

    11 really is less than or equal to 55, and this is true no matter what xx was.

  7. Interpret the true statement

    true for all values of x\text{true for all values of } x

    The original inequality reduces to something always true, so **every** value of xx satisfies it. The solution set is the whole number line.

  8. Test x=0x = 0

    3(1)−2=1,6(0)+5=5,1≤5  ✓3(1) - 2 = 1, \quad 6(0) + 5 = 5, \quad 1 \le 5 \; \checkmark

    As predicted, 00 works.

  9. Test x=100x = 100

    6(100)+1=601,6(100)+5=605,601≤605  ✓6(100) + 1 = 601, \quad 6(100) + 5 = 605, \quad 601 \le 605 \; \checkmark

    A large value works.

  10. Test x=−100x = -100

    6(−100)+1=−599,6(−100)+5=−595,−599≤−595  ✓6(-100) + 1 = -599, \quad 6(-100) + 5 = -595, \quad -599 \le -595 \; \checkmark

    A large negative value works too.

  11. See the structural reason

    RHS−LHS=4\text{RHS} - \text{LHS} = 4

    The right-hand side is always exactly 44 more than the left, whatever xx is, so the left can never exceed it.

  12. Draw the solution set

    the whole line is shaded\text{the whole line is shaded}

    Every point on the number line is a solution, so there are no circles and no unshaded region.

  13. Contrast with an empty solution set

    6x+1≥6x+5  has no solutions6x + 1 \ge 6x + 5 \;\text{has no solutions}

    Reverse the sign and the same cancellation leaves 1≥51 \ge 5, which is false — no solutions at all. The direction of the sign decides which of the two extremes you get.

  14. Watch the classic error

    do not write x≤4\text{do not write } x \le 4

    Students often "solve" 1≤51 \le 5 as though it were an equation in xx. There is no xx in it, so there is no boundary.

  15. State the conclusion

    true for all values of x\text{true for all values of } x

    Every real number satisfies the inequality.

Answer
true for all values of x\text{true for all values of } x
Question 5
6 markschallenging
Solve 5(x−2)−3(x+1)>2(x−4)−15(x - 2) - 3(x + 1) > 2(x - 4) - 1. Which statement about the solution set is correct?
Show worked solution

Worked solution

  1. Write down the inequality

    5(x−2)−3(x+1)>2(x−4)−15(x - 2) - 3(x + 1) > 2(x - 4) - 1

    Expand every bracket first, then collect like terms on each side.

  2. Expand the left-hand side

    5x−10−3x−3>2(x−4)−15x - 10 - 3x - 3 > 2(x - 4) - 1

    Note −3×(+1)=−3-3 \times (+1) = -3, so both terms from the second bracket are negative.

  3. Simplify the left-hand side

    2x−13>2(x−4)−12x - 13 > 2(x - 4) - 1

    5x−3x=2x5x - 3x = 2x and −10−3=−13-10 - 3 = -13.

  4. Expand the right-hand side

    2x−13>2x−8−12x - 13 > 2x - 8 - 1

    2×x=2x2 \times x = 2x and 2×(−4)=−82 \times (-4) = -8.

  5. Simplify the right-hand side

    2x−13>2x−92x - 13 > 2x - 9

    −8−1=−9-8 - 1 = -9.

  6. Subtract 2x from both sides

    −13>−9-13 > -9

    The xx terms are identical on both sides, so they vanish completely — there is no xx left to solve for.

  7. Judge the statement that is left

    −13>−9  is false-13 > -9 \;\text{is false}

    On the number line −13-13 lies to the **left** of −9-9, so −13-13 is smaller, not bigger. The statement is false.

  8. Interpret the false statement

    no solutions\text{no solutions}

    The original inequality reduces to something that is never true, so **no** value of xx can satisfy it. The solution set is empty.

  9. Test x=0x = 0

    5(−2)−3(1)=−13,2(−4)−1=−9,−13>−9  ×5(-2) - 3(1) = -13, \quad 2(-4) - 1 = -9, \quad -13 > -9 \; \times

    As predicted, 00 fails.

  10. Test x=100x = 100

    2(100)−13=187,2(100)−9=191,187>191  ×2(100) - 13 = 187, \quad 2(100) - 9 = 191, \quad 187 > 191 \; \times

    A huge value fails as well — the left side is always 44 less than the right.

  11. Test x=−100x = -100

    2(−100)−13=−213,2(−100)−9=−209,−213>−209  ×2(-100) - 13 = -213, \quad 2(-100) - 9 = -209, \quad -213 > -209 \; \times

    A large negative value fails too.

  12. See the structural reason

    LHS−RHS=−4\text{LHS} - \text{RHS} = -4

    The left-hand side minus the right-hand side is always −4-4, whatever xx is. It can never be positive, so the left side is never greater.

  13. Draw the solution set

    nothing is shaded\text{nothing is shaded}

    An empty solution set means the number line carries no shading and no circles at all.

  14. Watch the classic error

    do not invent a boundary\text{do not invent a boundary}

    Students often divide by the "zero" coefficient of xx or simply guess a boundary value. If the xx terms cancel, there is no boundary to find.

  15. State the conclusion

    no solutions\text{no solutions}

    The inequality is satisfied by no value of xx.

Answer
no solutions\text{no solutions}

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