Solve the inequality and give the answer in simplest form: -3x ≤ 12.
(2)
2
Solve 4(x - 3) > 8. Show your steps and give the solution as an inequality.
(2)
3
Solve the inequality 2x + 5 ≤ 2x + 9. State any special conclusion.
(2)
4
Solve the compound inequality and give the solution in inequality form: 1 ≤ 2x + 1 < 7.
(2)
5
Solve the inequality 7 - 2x > 1 and write the solution in inequality form.
(2)
6
Stretch: Solve the inequality and give the answer in inequality form: 2(3x - 1) ≤ 5x + 7. Show working.
(3)
7
Write down the inequality shown on the number line. An open circle at -1 and arrow to the left.
(2)
8
On the number line, a filled dot at 0 and arrow to the right. Write the inequality and give three example values that satisfy it.
(2)
9
Write the inequality represented by the shaded section between -2 and 4, including both endpoints.
(2)
10
Write the inequality shown: shading from 1 (open circle) to the right but not including 5 (open circle).
(2)
11
Solve the inequality 3x + 2 < 11. Show your working and give the final answer in inequality form.
(3)
12
Solve and represent on a number line: 2x - 7 > 1. Give the final answer and a short description of the number line representation.
(3)
13
Write down the inequality shown on the number line: a filled dot at 5 with an arrow pointing right.
(1)
14
Solve the inequality 7 - x > 2.
(2)
15
Solve the inequality -2x < 8.
(2)
16
Solve the inequality 3x + 5 ≤ x + 13.
(2)
17
Find all integer values of x that satisfy -3 ≤ x < 2. List them.
(2)
18
Solve the inequality 4(x - 1) > 8.
(3)
19
Solve the compound inequality and give the solution in inequality form: 3 ≤ 2x - 1 < 9.
(3)
20
A theme park ride requires riders to be at least 120 cm tall but under 200 cm tall. Using h for height in cm, write this rule as an inequality, then state whether a rider of height 118 cm is allowed on the ride.
(3)
21
Solve the inequality 3(2x - 1) ≤ 4x + 9. Then describe how the solution would be shown on a number line (state the type of dot used and the direction of the arrow).
(4)
22
A courier's box already weighs 4 kg. Extra identical items are added, each weighing 1.5 kg. The total weight must stay under 13 kg to qualify for standard postage. Using n for the number of extra items, form an inequality and solve it to find the greatest whole number of extra items that can be added.
(4)
23
Write down the inequality shown on the number line: an open circle at -2 with an arrow pointing left.
(1)
24
State whether x = 4 satisfies the inequality x > 3.
(1)
25
State whether x = -1 satisfies the inequality x ≤ -1.
(1)
26
Solve the inequality x - 2 < 6.
(1)
27
Solve the inequality 3x > 12.
(1)
28
Solve the inequality 2x + 3 < 11.
(2)
29
Solve the inequality 5x - 4 ≥ 16.
(2)
Mark scheme · 2.18 Inequalities and Number Lines
Question 1
M1 divides both sides by -3 and reverses inequality sign: x ≥ -4
A1 x ≥ -4 cao
Answer: x ≥ -4
Question 2
M1 expand or divide: 4x - 12 > 8 then 4x > 20
A1 x > 5 cao
Answer: x > 5
Question 3
M1 subtracts 2x from both sides and simplifies: 5 ≤ 9
A1 5 ≤ 9 is true so all real numbers satisfy the inequality, answer: all real x or (-infinity, infinity) cao
Answer: All real numbers
Question 4
M1 subtracts 1 across: 0 ≤ 2x < 6 then divides by 2 correctly
A1 0 ≤ x < 3 cao
Answer: 0 ≤ x < 3
Question 5
M1 subtracts 7: -2x > -6 then divides by -2 and reverses inequality sign
A1 x < 3 cao
Answer: x < 3
Question 6
M1 expand left side or apply distributive law: 6x - 2 ≤ 5x + 7 or equivalent rearrangement
M1 collect x terms and simplify: 6x - 5x ≤ 7 + 2 leading to x ≤ 9
A1 x ≤ 9 cao
Answer: x ≤ 9
Question 7
M1 identifies < relation and correct variable form such as x < -1
A1 x < -1 cao
Answer: x < -1
Question 8
M1 states inequality x ≥ 0 or similar
A1 gives three correct examples (eg 0, 1, 5) cao
Answer: x ≥ 0; examples: 0, 1, 5
Question 9
M1 recognises inclusive endpoints and writes compound inequality
A1 -2 ≤ x ≤ 4 cao
Answer: -2 ≤ x ≤ 4
Question 10
M1 formulates strict inequality with upper bound not included
A1 1 < x < 5 cao
Answer: 1 < x < 5
Question 11
M1 subtracts 2 from both sides or equivalent method: 3x < 9
M1 divides by 3 or equivalent: x < 3
A1 x < 3 cao
Answer: x < 3
Question 12
M1 adds 7 to both sides and divides by 2: 2x > 8 then x > 4
M1 identifies open circle at 4 on number line
A1 x > 4; number line: open circle at 4 with arrow to the right cao
Answer: x > 4
Question 13
B1 x ≥ 5 cao
Answer: x ≥ 5
Question 14
M1 correct rearrangement, e.g. -x > -5
A1 x < 5 cao (inequality sign flipped correctly)
Answer: x < 5
Question 15
M1 divides both sides by -2 and identifies the sign must flip
A1 x > -4 cao
Answer: x > -4
Question 16
M1 collects x terms on one side, e.g. 2x + 5 ≤ 13
A1 x ≤ 4 cao
Answer: x ≤ 4
Question 17
M1 identifies the correct range of integers, including -3 but excluding 2
A1 -3, -2, -1, 0, 1 cao
Answer: -3, -2, -1, 0, 1
Question 18
M1 expands the bracket correctly, e.g. 4x - 4 > 8
M1 correct rearrangement, e.g. 4x > 12
A1 x > 3 cao
Answer: x > 3
Question 19
M1 adds 1 to all three parts, e.g. 4 ≤ 2x < 10
M1 divides all three parts by 2 correctly
A1 2 ≤ x < 5 cao
Answer: 2 ≤ x < 5
Question 20
B1 correct inequality: 120 ≤ h < 200
M1 compares 118 to the lower bound 120 correctly
A1 not allowed, because 118 < 120
Answer: 120 ≤ h < 200; a rider of 118 cm is not allowed
Question 21
M1 expands the bracket correctly, e.g. 6x - 3 ≤ 4x + 9
M1 collects x terms correctly, e.g. 2x ≤ 12
A1 x ≤ 6 cao
B1 correctly describes a filled (closed) dot at 6 with an arrow pointing left
Answer: x ≤ 6; filled dot at 6, arrow pointing left
Question 22
M1 correct inequality formed: 4 + 1.5n < 13
M1 correct rearrangement, e.g. 1.5n < 9
A1 n < 6 cao
A1 greatest whole number of items is 5
Answer: n < 6; the greatest whole number of extra items is 5
Question 23
B1 x < -2 cao
Answer: x < -2
Question 24
B1 Yes, because 4 > 3
Answer: Yes
Question 25
B1 Yes, because -1 is equal to -1, which satisfies ≤