MathsICSE Class 10 · Linear Inequations

Double (Combined) Inequations: Methods and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

A double inequation such as −3 < 2x + 1 ≤ 7 sets two limits at once. If x appears only in the middle, apply each step to all three parts. If x appears in more than one part, split it into two inequations, solve each, and take the values common to both (their intersection).

Do every step to all three parts, then graph: the open circle at −2 leaves it out, the closed circle at 3 keeps it in.

Illustration drawn for QuestionBuddy.

Understand Double (Combined) Inequations

A double (or combined) inequation puts two conditions on $x$ at once, for example $- 3 < 2 x + 1 \leq 7$ , which means both $- 3 < 2 x + 1$ and $2 x + 1 \leq 7$ . Its solution set is the set of values that satisfy both conditions, which is the intersection of the two separate solution sets. Double inequations are the most common inequation question in the ICSE paper: both the 2024 and 2025 papers had a 3-mark question of this type, and a 1-mark MCQ in 2025.

Method 1: x only in the middle. When the unknown appears only in the middle expression, work on all three parts together. Whatever you add, subtract, multiply or divide, do it to all three:

$- 3 < 2 x + 1 \leq 7 \Rightarrow - 4 < 2 x \leq 6 \Rightarrow - 2 < x \leq 3.$

If you multiply or divide by a negative number, both signs reverse. Then it is neatest to rewrite the result with the smaller number on the left: $4 \geq x > - 6$ is written $- 6 < x \leq 4$ .

Method 2: x in more than one part. For $2 x - \frac{5}{3} < \frac{3 x}{5} + 10 \leq \frac{4 x}{5} + 11$ (2025 Q4(i)), you cannot isolate $x$ in the middle. Split it:

Left part: $2 x - \frac{5}{3} < \frac{3 x}{5} + 10$ .
Right part: $\frac{3 x}{5} + 10 \leq \frac{4 x}{5} + 11$ .

Solve each separately, then combine. If the left gives $x < a$ and the right gives $x \geq b$ , the answer is $b \leq x < a$ (when $b < a$ ). If the two conditions do not overlap, the solution set is empty, $\emptyset$ .

Intersections of two separate inequations. A question may give two inequations $A$ and $B$ separately and ask for $A \cap B$ . Solve each in the given replacement set, list (or picture) the solution sets, and keep the common members. A number line makes this visual: the answer is where the two shaded parts overlap.

End-points. Keep track of which ends are included: $\leq$ and $\geq$ include the end ('closed'), $<$ and $>$ exclude it ('open'). In the 2025 MCQ $0 < - \frac{x}{3} < 2$ both ends were excluded, so neither $0$ nor $- 6$ belonged; the examiners reported candidates who included end values.

Double (Combined) Inequations: formulas to remember

Meaning of a double inequation: $a < f (x) \leq b ⟺ a < f (x) and f (x) \leq b$; The solution set is the intersection.
Operating on all three parts: $a < p x + q \leq b \Rightarrow \frac{a - q}{p} < x \leq \frac{b - q}{p}$; For $p > 0$ ; if $p < 0$ , both signs reverse.
Combining two conditions: $x \geq m and x < n \Rightarrow m \leq x < n$; Empty if $m \geq n$ .
Intersection of solution sets: $A \cap B = {x : x \in A and x \in B}$; Keep only the common values.

How to solve Double (Combined) Inequations questions

First decide whether x is only in the middle or in more than one part.

1Note the replacement set ( $N, W, Z$ or $R$ ).
2If $x$ is only in the middle, clear fractions by multiplying all three parts by the LCM, then isolate $x$ in the middle.
3If you divide by a negative number, reverse both signs and rewrite with the smaller number on the left.
4If $x$ appears in more than one part, split into left and right inequations and solve each.
5Combine: the answer is where both conditions hold. Watch the end-points (open or closed).
6Write the solution set: roster form for $N, W, Z$ ; set-builder form for $R$ .
7Check an end-point value in the original inequation.

Double (Combined) Inequations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Two limits on one quantity

Suppose a courier accepts parcels weighing from 2 kg to 10 kg (both included), and the packaging adds 0.5 kg to whatever you send.

Find the range of weights $w$ of the contents that the courier will accept.

Try it first, then show the solutionSolution

$2 \leq w + 0.5 \leq 10$ .
Subtract 0.5 from all three parts: $1.5 \leq w \leq 9.5$ .

Answer: Contents from 1.5 kg to 9.5 kg.

Memory hook: Whatever you do to the middle, do to both ends.

A double inequation with whole-number answers

Imagine a theme-park ride needs riders to be at least 120 cm tall and under 190 cm. A child is 112 cm tall now and grows about 4 cm a year.

After how many whole years $t$ (from now) will the child meet the height rule, assuming the growth stays the same?

Try it first, then show the solutionSolution

$120 \leq 112 + 4 t < 190$ .
Subtract 112: $8 \leq 4 t < 78$ ; divide by 4: $2 \leq t < 19.5$ .
Whole years: $t \in {2, 3, \dots, 19}$ .

Answer: From 2 years from now up to 19 years from now.

Memory hook: At least … but less than: a closed start and an open finish.

Tricks to remember Double (Combined) Inequations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sandwich rule
A double inequation is a sandwich: whatever you do to the filling (middle), do to both slices (ends).
Use it when: Method 1 when $x$ is only in the middle.
Negative flips both
Dividing a sandwich by a negative flips it over: both signs reverse, then turn it round so the smaller number is on the left.
Use it when: Questions like $0 < - \frac{x}{3} < 2$ .
Split and overlap
x in two places? Split into two, solve each, shade both on one number line, keep the overlap.
Use it when: Method 2.
Filled for equal, hollow for not
≤ or ≥: the end-point is in (filled dot). < or >: it is out (hollow dot).
Use it when: End-points in the solution set.

Double (Combined) Inequations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: x only in the middle, integers · 1 mark · MCQBoard level

The solution set of $- 3 < 2 x + 1 \leq 7$ , $x \in Z$ , is:
(a) ${- 1, 0, 1, 2, 3}$
(b) ${- 2, - 1, 0, 1, 2, 3}$
(c) ${- 1, 0, 1, 2}$
(d) ${- 2, - 1, 0, 1, 2}$

Try it first, then show the solutionSolution

Subtract 1: $- 4 < 2 x \leq 6$ ; divide by 2: $- 2 < x \leq 3$ .
$- 2$ is excluded and $3$ included: ${- 1, 0, 1, 2, 3}$ .

Answer: (a)

Example 2: A fraction in the middle · 3 marks · Short answerBoard level

Solve $- 2 \leq \frac{x - 1}{3} < 2$ , $x \in Z$ , and write the solution set.

Try it first, then show the solutionSolution

Multiply all parts by 3: $- 6 \leq x - 1 < 6$ .
Add 1: $- 5 \leq x < 7$ .
Solution set ${- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6}$ .

Answer: ${- 5, - 4, \dots, 5, 6}$

Example 3: Splitting into two parts · 3 marks · Short answerHarder board level

Solve $2 x - 3 < x + 2 \leq 3 x + 5$ , $x \in R$ .

Try it first, then show the solutionSolution

Left part: $2 x - 3 < x + 2 \Rightarrow x < 5$ .
Right part: $x + 2 \leq 3 x + 5 \Rightarrow - 3 \leq 2 x \Rightarrow x \geq - 1.5$ .
Both: $- 1.5 \leq x < 5$ ; solution set ${x : - 1.5 \leq x < 5, x \in R}$ .

Answer: ${x : - 1.5 \leq x < 5, x \in R}$

Example 4: Dividing by a negative number · 3 marks · Short answerHarder board level

Solve $- 1 \leq \frac{3 - 2 x}{5} < 3$ , $x \in R$ .

Try it first, then show the solutionSolution

Multiply by 5: $- 5 \leq 3 - 2 x < 15$ .
Subtract 3: $- 8 \leq - 2 x < 12$ .
Divide by $- 2$ and reverse both signs: $4 \geq x > - 6$ .
Rewrite: ${x : - 6 < x \leq 4, x \in R}$ .

Answer: ${x : - 6 < x \leq 4, x \in R}$

Example 5: Intersection of two solution sets · 4 marks · Short answerHOTS / competency

Given $A = {x : 3 x - 1 > 2 x + 1, x \in N}$ and $B = {x : 4 - x \geq - 3, x \in N}$ , find $A \cap B$ .

Try it first, then show the solutionSolution

$A$ : $3 x - 1 > 2 x + 1 \Rightarrow x > 2$ , so $A = {3, 4, 5, \dots}$ .
$B$ : $4 - x \geq - 3 \Rightarrow - x \geq - 7 \Rightarrow x \leq 7$ , so $B = {1, 2, \dots, 7}$ .
$A \cap B = {3, 4, 5, 6, 7}$ .

Answer: ${3, 4, 5, 6, 7}$

Example 6: Case study: a storage temperature · 4 marks · Case-basedHOTS / competency

A vaccine must be kept between 15 °C and 25 °C, both included. A thermometer reads in degrees Fahrenheit, where $C = \frac{5 ( F - 32 )}{9}$ .
(a) Write a double inequation in $F$ .
(b) Solve it.
(c) The thermometer reads 68 °F. Is the vaccine safe?

Try it first, then show the solutionSolution

(a) $15 \leq \frac{5 ( F - 32 )}{9} \leq 25$ .
(b) Multiply by $\frac{9}{5}$ : $27 \leq F - 32 \leq 45$ , so $59 \leq F \leq 77$ .
(c) $59 \leq 68 \leq 77$ , so yes (68 °F is 20 °C).

Answer: (a) $15 \leq \frac{5 ( F - 32 )}{9} \leq 25$ (b) $59 \leq F \leq 77$ (c) Yes, it is safe.

Where students lose marks in Double (Combined) Inequations

The slips examiners see most often, and what to write instead.

Mistake: Including the end values when the signs are strict
< and > exclude the end-points.
Wrong working
$0 < - \frac{x}{3} < 2$ : ${- 6, \dots, - 1, 0}$ .
Right working
$- 6 < x < 0$ : ${- 5, - 4, - 3, - 2, - 1}$ .
Do this: Mark each end 'in' or 'out' before listing.
Usually costs: The MCQ in 2025 Q1(xiv).
Mistake: Reversing only one sign
Dividing all three parts by a negative reverses both signs.
Wrong working
$- 8 \leq - 2 x < 12 \Rightarrow 4 \leq x > - 6$ .
Right working
$4 \geq x > - 6$ , that is $- 6 < x \leq 4$ .
Do this: Flip both signs together, then rewrite smallest first.
Usually costs: The answer mark and the number line.
Mistake: Writing an integer roster when x is real
For $x \in R$ , every real number in the interval is included.
Wrong working
$- 5 \leq x < 8 \frac{1}{3}$ , $x \in R$ : ${- 5, - 4, \dots, 8}$ .
Right working
${x : - 5 \leq x < 8 \frac{1}{3}, x \in R}$ .
Do this: Check the replacement set before writing the answer.
Usually costs: Reported for 2025 Q4(i).
Mistake: Including a value outside the strict upper limit
$x < 4$ excludes 4.
Wrong working
$- 2 \leq x < 4$ , $x \in I$ : ${- 2, \dots, 4}$ .
Right working
${- 2, - 1, 0, 1, 2, 3}$ .
Do this: Test the end values in the original.
Usually costs: Reported for 2024 Q7(ii).

Exam tips

2024 Q7(ii) ( $x \in I$ ) and 2025 Q4(i) ( $x \in R$ ) both asked: solve, write the solution set, and show it on a number line. Do all three parts.
In 2025 Q4(i), the expected points were $x < \frac{25}{3}$ from one part and $x \geq - 5$ from the other; write each part's result on its own line.
Keep improper fractions such as $\frac{25}{3}$ exact, and also show them as $8 \frac{1}{3}$ on the number line.
If the two parts do not overlap, write 'solution set = ∅'.

Practice questions on Double (Combined) Inequations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The solution set of $1 \leq x + 2 < 4$ , $x \in W$ , is:
(a) ${- 1, 0, 1}$
(b) ${0, 1}$
(c) ${0, 1, 2}$
(d) ${1, 2}$
Hint
$- 1 \leq x < 2$ , but $x$ is a whole number.
Answer
(b) ${0, 1}$
Q2 · 3 marks · Short answerBoard level
Solve $- 4 < 3 x - 1 \leq 8$ , $x \in Z$ .
Hint
Add 1 to all parts, then divide by 3.
Answer
$- 3 < 3 x \leq 9 \Rightarrow - 1 < x \leq 3$ : ${0, 1, 2, 3}$ .
Q3 · 3 marks · Short answerHarder board level
Solve $\frac{x}{2} + 1 \leq x - 1 < 9 - x$ , $x \in R$ .
Hint
Split into two parts.
Answer
Left: $x \geq 4$ . Right: $2 x < 10 \Rightarrow x < 5$ . So ${x : 4 \leq x < 5, x \in R}$ .
Q4 · 1 mark · MCQHarder board level
The solution set of $0 < - \frac{x}{2} < 3$ , $x \in Z$ , is:
(a) ${- 6, - 5, - 4, - 3, - 2, - 1}$
(b) ${- 5, - 4, - 3, - 2, - 1}$
(c) ${- 5, - 4, - 3, - 2, - 1, 0}$
(d) ${1, 2, 3, 4, 5}$
Hint
Multiply by $- 2$ and reverse both signs.
Answer
(b). $0 > x > - 6$ , so $- 6 < x < 0$ .
Q5 · 3 marks · Short answerHarder board level
Solve $- 3 \leq \frac{1 - 2 x}{3} < 2$ , $x \in R$ .
Hint
Multiply by 3, subtract 1, divide by $- 2$ .
Answer
$- 9 \leq 1 - 2 x < 6 \Rightarrow - 10 \leq - 2 x < 5 \Rightarrow 5 \geq x > - 2.5$ : ${x : - 2.5 < x \leq 5, x \in R}$ .
Q6 · 4 marks · Short answerHOTS / competency
Find all integers $x$ such that $x + 7$ is at least 3 more than twice $x$ , and $x + 7$ is less than four times $x$ .
Hint
Write $2 x + 3 \leq x + 7 < 4 x$ .
Answer
$2 x + 3 \leq x + 7 \Rightarrow x \leq 4$ ; $x + 7 < 4 x \Rightarrow x > \frac{7}{3}$ . Integers: ${3, 4}$ .
Q7 · 3 marks · Short answerHarder board level
Show that the inequations $2 x + 3 > 5 x$ and $3 x - 2 > x + 6$ have no common real solution.
Hint
Solve each and compare.
Answer
First: $3 > 3 x \Rightarrow x < 1$ . Second: $2 x > 8 \Rightarrow x > 4$ . No real number is both less than 1 and greater than 4, so the intersection is $\emptyset$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Solve $2 < x + 3 \leq 6$ , $x \in Z$ .Show answer
$- 1 < x \leq 3$ : ${0, 1, 2, 3}$ .
2.What happens to both signs when you divide all parts by $- 1$ ?Show answer
Both reverse.
3.Rewrite $5 \geq x > - 2$ with the smaller number on the left.Show answer
$- 2 < x \leq 5$ .
4. $x > 3$ and $x \leq 7$ , $x \in N$ . Solution set?Show answer
${4, 5, 6, 7}$ .
5. $x < 1$ and $x > 4$ . Solution set?Show answer
$\emptyset$ (empty).

Double (Combined) Inequations: frequently asked questions

How do I solve a double inequation?

If x is only in the middle, do every step to all three parts. If x is in more than one part, split it into two inequations, solve each, and take the values that satisfy both.

What if a double inequation has no solution?

If the two conditions do not overlap, for example x < 1 and x > 4, the solution set is the empty set, written ∅ or { }.

Do I reverse both signs when dividing a double inequation by a negative number?

Yes. Both inequality signs reverse. Then rewrite the result so that the smaller number is on the left.