MathsCBSE Class 11 · Linear Inequalities

Systems of Linear Inequalities, Modulus Inequalities and Word Problems (Class 11)

Updated 17 October 2026 · Fact-checked

A system of inequalities in one variable is solved by solving each one and taking the intersection of the solution sets. For modulus, |u| < c means −c < u < c, and |u| > c means u < −c or u > c (c > 0). For word problems, define the unknown, translate each condition (at least: ≥, at most: ≤, between: double inequality) and check that the answer fits the real-world domain.

Read |x−2| as distance from 2: at most 3 gives one block, at least 1 removes the middle, and the overlap leaves two pieces.

Illustration drawn for QuestionBuddy.

Understand Systems of Inequalities, Modulus Inequalities and Word Problems

Systems. To solve two or more inequalities in $x$ together, solve each separately and keep the values that satisfy all of them: the intersection of the intervals. On a number line, mark each solution set on its own line, and shade only the overlap. For example, $x \leq 6$ and $x > - 2$ together give $(- 2, 6]$ ; if the intersection is empty, the system has no solution.

Modulus inequalities. $∣ u ∣$ is the distance of $u$ from $0$ . For $c > 0$ :

$∣ u ∣ < c ⟺ - c < u < c$ (inside: and, an interval);
$∣ u ∣ \leq c ⟺ - c \leq u \leq c$ ;
$∣ u ∣ > c ⟺ u < - c or u > c$ (outside: or, two pieces);
$∣ u ∣ \geq c ⟺ u \leq - c or u \geq c$ .

Here $u$ may be an expression: $∣2 x - 3∣ \leq 5$ gives $- 5 \leq 2 x - 3 \leq 5$ , then $- 1 \leq x \leq 4$ . For $∣ x - a ∣ < r$ think "all points within $r$ of $a$ ". For a statement like $1 \leq ∣ x - 2∣ \leq 3$ , combine the two conditions by intersection: $x$ is at distance at least $1$ and at most $3$ from $2$ , giving $[- 1, 1] \cup [3, 5]$ . If $c < 0$ : $∣ u ∣ < c$ has no solution and $∣ u ∣ > c$ is true for every real $u$ .

Word problems. Follow the routine:
1. Define the variable with units and its domain (a length is positive; people are whole numbers).
2. Translate the words: "at least" is $\geq$ , "at most" or "not more than" is $\leq$ , "more than" is $>$ , "between $p$ and $q$ " is $p < \cdot < q$ .
3. Solve the inequality (or system).
4. Interpret: restrict to whole numbers if needed, state the answer in words with units, and check the end points.

Typical models.

*Cost versus revenue:* profit needs revenue $>$ cost.
*Comparing two plans:* set one cost less than the other and solve for the distance or quantity at which the cheaper plan changes.
*Mixtures:* amount of pure substance $=$ percentage $\times$ volume; the concentration of the mixture must satisfy the given limits, so write (substance)/(total volume) between the bounds.
*Averages and marks:* total required $=$ average $\times$ number of tests.

Warning. Do not square or multiply by quantities of unknown sign. Do not combine "or" results into a single interval unless they overlap.

Systems of Inequalities, Modulus Inequalities and Word Problems: formulas to remember

Modulus less than: $∣ u ∣ < c i s t h es am e a s - c < u < c$; c > 0; an interval.
Modulus greater than: $∣ u ∣ > c i s t h es am e a s u < - cor u > c$; Two separate pieces.
Within a distance: $∣ x - a ∣ < r : a - r < x < a + r$; Points within r of a.
System: $S o l u t i o n se t = in t er sec t i o n o f t h eso l u t i o n se t s$; Empty if no overlap.
Profit condition: $R e v e n u e > C os t$; Then solve for the number of items.

How to solve Systems of Inequalities, Modulus Inequalities and Word Problems questions

Use these steps for systems, modulus inequalities and word problems.

1For a system, solve each inequality on its own and write each solution as an interval.
2Draw each interval on a number line and shade only the common part; write it as the final answer.
3For modulus, rewrite $∣ u ∣ < c$ as a double inequality, or $∣ u ∣ > c$ as two inequalities joined by "or", and solve.
4For a word problem, define the unknown with its domain and units.
5Translate each phrase into an inequality sign and form one inequality or a system.
6Solve, restrict to the allowed values (whole numbers, positive), and state the answer in words with units.

Systems of Inequalities, Modulus Inequalities and Word Problems 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.

Modulus as tolerance

Suppose a factory makes bolts whose diameter must be 12 mm with a tolerance of 0.05 mm, so $∣ d - 12∣ \leq 0.05$ .

Between which values must the diameter lie?

Try it first, then show the solutionSolution

$∣ d - 12∣ \leq 0.05$ means $- 0.05 \leq d - 12 \leq 0.05$ .
Add 12: $11.95 \leq d \leq 12.05$ .

Answer: 11.95 mm to 12.05 mm.

Memory hook: Modulus less than c means "within c of the centre".

A system for a hall

Suppose a hall is set out in rows of 8 seats. The organiser needs at least 200 seats, but the fire rules allow fewer than 260 seats.

How many rows $x$ can be set up?

Try it first, then show the solutionSolution

At least 200 seats: $8 x \geq 200$ , so $x \geq 25$ .
Fewer than 260 seats: $8 x < 260$ , so $x < 32.5$ .
Whole numbers satisfying both: $25 \leq x \leq 32$ .
Check: $8 \times 32 = 256 < 260$ , and $8 \times 33 = 264$ is too many.

Answer: 25 to 32 rows.

Memory hook: Both limits must hold: intersect.

Tricks to remember Systems of Inequalities, Modulus Inequalities and Word Problems

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

Inside is AND, outside is OR
|u| < c: one interval. |u| > c: two pieces.
Use it when: Modulus inequalities.
Distance picture
|x − a| < r reads: x is within r of a.
Use it when: Quick answers.
Intersect with a picture
Shade each solution on its own line and keep only the overlap.
Use it when: Systems.
Round down for capacity
If x must be a whole number and x < 32.5, the largest value is 32.
Use it when: Word problems with integer answers.

Systems of Inequalities, Modulus Inequalities and Word Problems: solved examples

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

Example 1: A system · 3 marks · Short answerBoard level

Solve the system $3 x + 2 \geq 5 x - 10$ and $2 (x - 3) > x - 8$ for real $x$ ; show the solution on the number line and list the integer solutions with their sum.

Try it first, then show the solutionSolution

First: $12 \geq 2 x$ , so $x \leq 6$ . Second: $2 x - 6 > x - 8$ , so $x > - 2$ . (1 mark)
Intersection: $- 2 < x \leq 6$ , i.e. $(- 2, 6]$ , drawn with an open circle at $- 2$ and a filled dot at $6$ . (1 mark)
Integers $- 1, 0, 1, \dots, 6$ : eight numbers with sum $20$ . (1 mark)

Answer: $(- 2, 6]$ ; integers $- 1$ to $6$ , sum 20

Example 2: Modulus, less than · 2 marks · Very short answerBoard level

Solve $∣2 x - 3∣ \leq 5$ .

Try it first, then show the solutionSolution

$- 5 \leq 2 x - 3 \leq 5$ . (1 mark)
Add 3 and divide by 2: $- 1 \leq x \leq 4$ , i.e. $[- 1, 4]$ . (1 mark)

Answer: $[- 1, 4]$

Example 3: Modulus, greater than · 3 marks · Short answerHarder board level

Solve $∣3 x - 2∣ > 7$ and write the solution as a union of intervals.

Try it first, then show the solutionSolution

$3 x - 2 > 7$ or $3 x - 2 < - 7$ . (1 mark)
First: $x > 3$ . Second: $3 x < - 5$ , so $x < - \frac{5}{3}$ . (1 mark)
Solution $(- \infty, - \frac{5}{3}) \cup (3, \infty)$ . (1 mark)

Answer: $(- \infty, - \frac{5}{3}) \cup (3, \infty)$

Example 4: A mixture · 4 marks · Short answerHarder board level

A shopkeeper has 50 litres of a 30% sugar syrup. How many litres of a 10% syrup must be added so that the sugar concentration of the mixture is more than 20% but less than 25%?

Try it first, then show the solutionSolution

Let $x$ litres be added. Sugar $= 0.3 \times 50 + 0.1 x = 15 + 0.1 x$ ; volume $= 50 + x$ . (1 mark)
Lower limit: $15 + 0.1 x > 0.2 (50 + x) = 10 + 0.2 x$ , so $5 > 0.1 x$ , $x < 50$ . (1 mark)
Upper limit: $15 + 0.1 x < 0.25 (50 + x) = 12.5 + 0.25 x$ , so $2.5 < 0.15 x$ , $x > \frac{50}{3}$ . (1 mark)
Together $\frac{50}{3} < x < 50$ : between about $16.7$ and $50$ litres. Check $x = 30$ : $\frac{18}{80} = 22.5%$ . (1 mark)

Answer: More than $\frac{50}{3} \approx 16.7$ litres and less than 50 litres

Example 5: Comparing two taxi plans · 5 marks · Long answerHOTS / competency

Plan A charges Rs 50 plus Rs 12 per km. Plan B charges Rs 80 plus Rs 9 per km. (i) For what distances is Plan B cheaper? (ii) With Rs 260, which plan lets you travel further, and by how much?

Try it first, then show the solutionSolution

Let the distance be $d$ km. Plan B cheaper: $80 + 9 d < 50 + 12 d$ . (1 mark)
$30 < 3 d$ , so $d > 10$ km. (1 mark)
Plan A with Rs 260: $50 + 12 d \leq 260$ , so $d \leq 17.5$ km. (1 mark)
Plan B with Rs 260: $80 + 9 d \leq 260$ , so $d \leq 20$ km. (1 mark)
Plan B takes you 2.5 km further (consistent with $d > 10$ , where B is cheaper). (1 mark)

Answer: (i) $d > 10$ km (ii) Plan B: 20 km against 17.5 km, so 2.5 km further

Where students lose marks in Systems of Inequalities, Modulus Inequalities and Word Problems

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

Mistake: Writing "and" for a greater-than modulus
|u| > c gives two outside pieces, not one interval.
Wrong working
$∣3 x - 2∣ > 7 \Rightarrow - 7 < 3 x - 2 < 7$ .
Right working
$3 x - 2 > 7$ or $3 x - 2 < - 7$ .
Do this: Greater than means outside: "or".
Usually costs: The whole answer.
Mistake: Writing the union as one interval
The two pieces do not overlap.
Wrong working
$- \frac{5}{3} > x > 3$ .
Right working
$x < - \frac{5}{3}$ or $x > 3$ .
Do this: Use the union sign.
Usually costs: 1 mark.
Mistake: Taking the union instead of the intersection for a system
A system needs all conditions at once.
Wrong working
$x \leq 6$ and $x > - 2$ written as "all real $x$ ".
Right working
$(- 2, 6]$ .
Do this: Shade each condition on its own line and keep the overlap.
Usually costs: The answer.
Mistake: Ignoring the real-world domain
A number of items or rows must be a whole number.
Wrong working
"At least 160.01 items" or $x \leq 32.5$ rows.
Right working
At least 161 items; at most 32 rows.
Do this: Restrict to whole numbers in the last line.
Usually costs: The final mark.

Exam tips

In a system, give the final answer as one interval and show the overlap on the number line.
For modulus inequalities write the unfolded form (double inequality or "or") as a separate line.
Define the unknown with units at the start of a word problem; it earns a mark by itself.
Check end points: strict or non-strict depends on the wording ("more than" versus "at least").
Mixture problems: write the amount of pure substance and the total volume separately before forming the ratio.

Practice questions on Systems of Inequalities, Modulus Inequalities and Word Problems

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 $∣ x ∣ < 4$ is:
(a) $(- 4, 4)$
(b) $(- \infty, - 4) \cup (4, \infty)$
(c) $[- 4, 4]$
(d) $(0, 4)$
Hint
Less than means between.
Answer
(a). $- 4 < x < 4$ .
Q2 · 1 mark · MCQHarder board level
The number of integers satisfying $∣2 x + 1∣ < 3$ is:
(a) 1
(b) 2
(c) 3
(d) 4
Hint
Rewrite as a double inequality.
Answer
(b). $- 3 < 2 x + 1 < 3$ gives $- 2 < x < 1$ ; integers $- 1$ and $0$ .
Q3 · 3 marks · Short answerHarder board level
Solve $1 \leq ∣ x - 2∣ \leq 3$ .
Hint
Take the intersection of $∣ x - 2∣ \geq 1$ and $∣ x - 2∣ \leq 3$ .
Answer
$∣ x - 2∣ \geq 1$ : $x \leq 1$ or $x \geq 3$ . $∣ x - 2∣ \leq 3$ : $- 1 \leq x \leq 5$ . Intersection: $[- 1, 1] \cup [3, 5]$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The inequality $∣ x ∣ < - 2$ has no real solution.
Reason (R): The modulus of a real number is never negative.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Can a non-negative number be less than $- 2$ ?
Answer
(a). $∣ x ∣ \geq 0 > - 2$ , so $∣ x ∣ < - 2$ is impossible.
Q5 · 4 marks · Short answerHOTS / competency
Suppose a workshop sells an item for Rs 55 and spends Rs 4000 plus Rs 30 per item on making $x$ items. (i) How many items must be sold to make a profit? (ii) How many to make a profit of at least Rs 1500?
Hint
Profit $= 55 x - (4000 + 30 x)$ .
Answer
Profit $= 25 x - 4000$ . (i) $25 x - 4000 > 0$ gives $x > 160$ , so at least 161 items. (ii) $25 x - 4000 \geq 1500$ gives $x \geq 220$ .

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.What does |u| < c mean when c > 0?Show answer
−c < u < c.
2.What does |u| > c mean when c > 0?Show answer
u < −c or u > c.
3.What is the solution of a system of inequalities?Show answer
The intersection of the individual solution sets.
4.How is "at least" written?Show answer
With ≥.
5.When does Revenue > Cost give a profit?Show answer
When the number of items exceeds the break-even value.

Systems of Inequalities, Modulus Inequalities and Word Problems: frequently asked questions

How do I know whether to use "and" or "or" for a modulus inequality?

Less than (or less than or equal) gives "and", an interval between −c and c. Greater than gives "or", two pieces outside.

What if the system has no common values?

Then the solution set is empty. Say "no solution".

Why must some answers be whole numbers?

Quantities such as items, rows or people cannot be fractions, so after solving you keep only the whole numbers that satisfy the inequality.

Are two-variable inequalities in the exam?

The chapter page notes that graphical solution of inequalities in two variables is assessed only formatively in 2026-27.