Write down the inequality: x + 5 < 12 Get the x term on its own: x < 7 State the solution: x < 7 Answer: x < 7

Write down the inequality: 6x ≤ -12 Divide both sides by 6: x ≤ -2 State the solution: x ≤ -2 Answer: x ≤ -2

Write down the inequality: 8 - 2x ≤ 2 Get the x term on its own: -2x ≤ -6 Divide both sides by -2 — and flip the sign: x ≥ 3 State the solution: x ≥ 3 Show the solution set on a number line: x ≥ 3 Check a value inside the solution set (x = 4): 8 - 2 (4) ≤ 2 \;\; 0 ≤ 2 \; Answer: x ≥ 3

Solve 4(1 - 2x) ≤ 3(2 - x).

Write down the inequality: 4(1 - 2x) ≤ 3(2 - x) Expand both brackets: 4 - 8x ≤ 6 - 3x Collect the x terms on one side: -5x + 4 ≤ 6 Get the x term on its own: -5x ≤ 2 Divide both sides by -5 — and flip the sign: x ≥ -(2)/(5) State the solution: x ≥ -(2)/(5) Show the solution set on a number line: x ≥ -(2)/(5) Check a value inside the solution set (x = 0): 4 - 8 (0) ≤ 6 - 3 (0) \;\; 4 ≤ 6 \; Test the boundary value x = -(2)/(5): 4 - 8 (-2/5) ≤ 6 - 3 (-2/5) \;\; (36)/(5) ≤ (36)/(5) \; Final check: x ≥ -(2)/(5) Answer: x ≥ -(2)/(5)

Solve 2 ≤ 5 - 3x ≤ 11 and list all the integer values of x in the solution set.

Write down the double inequality: 2 ≤ 5 - 3x ≤ 11 Subtract 5 from all three parts: -3 ≤ -3x ≤ 6 Divide all three parts by -3 and flip both signs: 1 ≥ x ≥ -2 Rewrite smallest-to-largest: -2 ≤ x ≤ 1 Draw the solution set: -2 ≤ x ≤ 1 Check the lower boundary: x = -2: \; 5 - 3(-2) = 11, 2 ≤ 11 ≤ 11 \; Check the upper boundary: x = 1: \; 5 - 3(1) = 2, 2 ≤ 2 ≤ 11 \; Check a value inside: x = 0: \; 5 - 3(0) = 5, 2 ≤ 5 ≤ 11 \; Check a value outside: x = 2: \; 5 - 3(2) = -1, 2 ≤ -1 \; × Check the other side: x = -3: \; 5 - 3(-3) = 14, 14 ≤ 11 \; × List the integers: x = -2, \; -1, \; 0, \; 1 Set notation: \x : -2 ≤ x ≤ 1\ Why the order swapped: dividing by -3 reverses order Watch the classic error: -(1)/(3) ≤ x ≤ -2 \;is wrong State the integer solutions: x = -2, \; -1, \; 0, \; 1 Answer: x = -2, \; -1, \; 0, \; 1

GCSE Linear inequalities Practice Questions

Free GCSE Linear inequalities practice questions with full step-by-step worked solutions. Covers one-step inequality, inverse operations, inclusive inequality, dividing by a positive. Practise exam-style problems and check your method.

one-step inequalityinverse operationsinclusive inequalitydividing by a positivemultiplying both sidestwo-step inequality

GCSE Foundation70 questionsStep-by-step solutions

Question 1

1 markeasy

Solve

x + 5 < 12

Show worked solution

Worked solution

Write down the inequality
$x + 5 < 12$
Treat it exactly like an equation, but keep the inequality sign — the aim is to get $x$ on its own.
Get the $x$ term on its own
$x < 7$
Subtract $5$ from both sides — again the sign stays as it is.
State the solution
$x < 7$
This is the complete solution set: every value of $x$ that makes the original inequality true, and no others.

Answer

x < 7

Question 2

1 markeasy

Solve

6x \le -12

Show worked solution

Worked solution

Write down the inequality
$6x \le -12$
Treat it exactly like an equation, but keep the inequality sign — the aim is to get $x$ on its own.
Divide both sides by $6$
$x \le -2$
Dividing by the positive number $6$ leaves the inequality sign pointing the same way.
State the solution
$x \le -2$
This is the complete solution set: every value of $x$ that makes the original inequality true, and no others.

Answer

x \le -2

Question 3

2 marksintermediate

Solve

8 - 2x \le 2

Show worked solution

Worked solution

Write down the inequality
$8 - 2x \le 2$
Treat it exactly like an equation, but keep the inequality sign — the aim is to get $x$ on its own.
Get the $x$ term on its own
$-2x \le -6$
Subtract $8$ from both sides — again the sign stays as it is.
Divide both sides by $-2$ — and flip the sign
$x \ge 3$
Dividing by the negative number $-2$ reverses the inequality: $\le$ becomes $\ge$ . This is the one rule that separates inequalities from equations.
State the solution
$x \ge 3$
This is the complete solution set: every value of $x$ that makes the original inequality true, and no others.
Show the solution set on a number line
$x \ge 3$
Put a closed (filled-in) circle at $3$ — it is closed because $3$ itself does satisfy the inequality — then shade the line to the right, because every number greater than or equal to $3$ works.
Check a value inside the solution set ( $x = 4$ )
$8 - 2 (4) \le 2 \;\Rightarrow\; 0 \le 2 \; \checkmark$
$4$ satisfies $x \ge 3$ , and it makes the original inequality true — a good sign the direction is right.

Answer

x \ge 3

Question 4

4 markshard

Solve

4(1 - 2x) \le 3(2 - x)

Show worked solution

Worked solution

Write down the inequality
$4(1 - 2x) \le 3(2 - x)$
Treat it exactly like an equation, but keep the inequality sign — the aim is to get $x$ on its own.
Expand both brackets
$4 - 8x \le 6 - 3x$
Multiply each bracket out. Watch the signs: $4 \times (-2x) = -8x$ and $3 \times (-x) = -3x$ .
Collect the $x$ terms on one side
$-5x + 4 \le 6$
Add $3x$ to both sides. Adding or subtracting the same thing on both sides never changes the direction of the inequality.
Get the $x$ term on its own
$-5x \le 2$
Subtract $4$ from both sides — again the sign stays as it is.
Divide both sides by $-5$ — and flip the sign
$x \ge -\frac{2}{5}$
Dividing by the negative number $-5$ reverses the inequality: $\le$ becomes $\ge$ . This is the one rule that separates inequalities from equations.
State the solution
$x \ge -\frac{2}{5}$
This is the complete solution set: every value of $x$ that makes the original inequality true, and no others.
Show the solution set on a number line
$x \ge -\frac{2}{5}$
Put a closed (filled-in) circle at $-\frac{2}{5}$ — it is closed because $-\frac{2}{5}$ itself does satisfy the inequality — then shade the line to the right, because every number greater than or equal to $-\frac{2}{5}$ works.
Check a value inside the solution set ( $x = 0$ )
$4 - 8 (0) \le 6 - 3 (0) \;\Rightarrow\; 4 \le 6 \; \checkmark$
$0$ satisfies $x \ge -\frac{2}{5}$ , and it makes the original inequality true — a good sign the direction is right.
Test the boundary value $x = -\frac{2}{5}$
$4 - 8 (-2/5) \le 6 - 3 (-2/5) \;\Rightarrow\; \frac{36}{5} \le \frac{36}{5} \; \checkmark$
At the boundary both sides are equal ( $\frac{36}{5} = \frac{36}{5}$ ). The sign $\ge$ allows equality, so $-\frac{2}{5}$ is part of the solution set — hence the closed circle.
Final check
$x \ge -\frac{2}{5}$
The boundary is right, the direction is right, and both test values agree with the number line.

Answer

x \ge -\frac{2}{5}

Question 5

6 markschallenging

Solve

2 \le 5 - 3x \le 11

and list all the integer values of

x

in the solution set.

Show worked solution

Worked solution

Write down the double inequality
$2 \le 5 - 3x \le 11$
The $x$ term is negative, so a flip is coming — but do the additions first.
Subtract $5$ from all three parts
$-3 \le -3x \le 6$
Subtracting leaves both signs pointing the same way: $2 - 5 = -3$ and $11 - 5 = 6$ .
Divide all three parts by $-3$ and flip both signs
$1 \ge x \ge -2$
Dividing by the negative number $-3$ reverses both inequality signs: $-3 \div -3 = 1$ and $6 \div -3 = -2$ .
Rewrite smallest-to-largest
$-2 \le x \le 1$
A double inequality is always written with the smaller number on the left, reading along the number line.
Draw the solution set
$-2 \le x \le 1$
Both signs are inclusive, so both circles are closed and the segment between them is shaded.
Check the lower boundary
$x = -2: \; 5 - 3(-2) = 11, \quad 2 \le 11 \le 11 \; \checkmark$
At $x = -2$ the middle hits the upper limit exactly — the flip has swapped which end is which. Since the sign is $\le$ , it is included.
Check the upper boundary
$x = 1: \; 5 - 3(1) = 2, \quad 2 \le 2 \le 11 \; \checkmark$
At $x = 1$ the middle hits the lower limit exactly, and $\le$ allows it.
Check a value inside
$x = 0: \; 5 - 3(0) = 5, \quad 2 \le 5 \le 11 \; \checkmark$
$0$ is comfortably inside the solution set.
Check a value outside
$x = 2: \; 5 - 3(2) = -1, \quad 2 \le -1 \; \times$
$2$ is just outside, and it fails the lower condition.
Check the other side
$x = -3: \; 5 - 3(-3) = 14, \quad 14 \le 11 \; \times$
$-3$ is just outside on the other end and fails the upper condition.
List the integers
$x = -2, \; -1, \; 0, \; 1$
Both endpoints are included because both circles are closed, giving four integer solutions.
Set notation
$\{x : -2 \le x \le 1\}$
A closed interval — both ends belong to the set.
Why the order swapped
$\text{dividing by } -3 \text{ reverses order}$
The lower limit $2$ produced the upper bound $x = 1$ , and the upper limit $11$ produced the lower bound $x = -2$ . That reversal is the flip in action.
Watch the classic error
$-\frac{1}{3} \le x \le -2 \;\text{is wrong}$
Flipping only one of the two signs, or forgetting to reorder, produces a nonsensical inequality with the bigger number on the left.
State the integer solutions
$x = -2, \; -1, \; 0, \; 1$
Four integers lie in the solution set $-2 \le x \le 1$ .

Answer

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

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