ICSE Class 10 Maths Linear Inequations Selina Solutions
ICSE Class 10 Maths Linear Inequations Solutions: Chapter 4 Summary
ICSE Class 10 Maths linear inequations in one variable are solved like linear equations, but the answer is usually a range of values instead of one value. In Selina Concise Mathematics Class 10 Chapter 4, the main skills are to solve inequations, apply the replacement set such as N, W, Z, or R, and represent the solution on a number line.
This replacement page is written as a teacher’s solution note for Chapter 4, Linear Inequations (in one variable). It keeps the study focus on correct algebraic steps, interval notation, number-line interpretation, and common mistakes that students make in ICSE board-style answers.
Formula and Method Reference for Linear Inequations
A linear inequation in one variable has the variable only to the first power. The symbols used are \lt, \gt, \leq, and \geq. The table below gives the rules students must apply before writing the final answer.
| Operation | Effect on inequality sign | Example |
|---|---|---|
| Add the same number to both sides | Sign does not change | x-3 \lt 5 \Rightarrow x \lt 8 |
| Subtract the same number from both sides | Sign does not change | x+4 \geq 9 \Rightarrow x \geq 5 |
| Multiply both sides by a positive number | Sign does not change | \frac{x}{3} \leq 2 \Rightarrow x \leq 6 |
| Divide both sides by a positive number | Sign does not change | 5x \gt 20 \Rightarrow x \gt 4 |
| Multiply or divide by a negative number | Sign reverses | -2x \lt 8 \Rightarrow x \gt -4 |
Concept snapshot: think of the inequality sign as an order arrow
An inequation compares two sides in order. Adding the same quantity to both sides moves both sides together, so the order remains the same. Multiplying by a negative number turns the number line around; the larger side becomes the smaller side. That is why \lt becomes \gt, and \leq becomes \geq, when you multiply or divide by a negative number.
For more syllabus-aligned study support, students can also use the ICSE solutions, the ICSE syllabus, the ICSE sample papers, and the ICSE question papers sections. For official board information, refer to CISCE. NCERT algebra resources at NCERT may help with overlapping algebra basics, although ICSE students should follow the CISCE syllabus and prescribed textbook for exam preparation.
Worked Examples Before the Exercise
These examples are original practice models based on the Chapter 4 method. They show how to move from the inequation to the final replacement-set answer.
Worked Example 1: Solve 7-2x \leq 15, where x \in Z
Step 1: Start with the given inequation.
7-2x \leq 15
Step 2: Subtract 7 from both sides.
-2x \leq 8
Step 3: Divide by -2. Since the divisor is negative, reverse the inequality sign.
x \geq -4
Final answer: Since x \in Z, the solution set is \{x:x\in Z,\ x\geq -4\}.
Worked Example 2: Solve -3 \lt 2x+1 \leq 9, where x \in R
Step 1: Subtract 1 from all three parts.
-3-1 \lt 2x+1-1 \leq 9-1
-4 \lt 2x \leq 8
Step 2: Divide all three parts by 2. The divisor is positive, so the signs remain unchanged.
-2 \lt x \leq 4
Final answer: \{x:x\in R,\ -2 \lt x \leq 4\}.
Worked Example 3: Find the smallest integer satisfying 4x-3 \gt 10
Step 1: Add 3 to both sides.
4x \gt 13
Step 2: Divide by 4.
x \gt \frac{13}{4}
Step 3: Convert the boundary to understand the integer values.
\frac{13}{4}=3.25
Final answer: The smallest integer greater than 3.25 is 4.
Selina Exercise 4(A) Solutions for Linear Inequations
The following solutions use the visible Exercise 4(A) question pattern from the existing page and present the working in a clean school-answer format. Edition wording may vary slightly, but the algebraic method remains the same.
Question 1(a): If x\in W, solve -x \gt -7
Step 1: Start with the inequation.
-x \gt -7
Step 2: Multiply both sides by -1. Reverse the sign.
x \lt 7
Step 3: Select whole numbers less than 7.
Final answer: \{0,1,2,3,4,5,6\}.
Question 1(b): Solve 4(2x-5)\lt 2x+28, where x\in R
Step 1: Expand the left side.
4(2x-5)\lt 2x+28
8x-20\lt 2x+28
Step 2: Bring x-terms to one side and constants to the other side.
8x-2x\lt 28+20
6x\lt 48
Step 3: Divide by 6.
x\lt 8
Final answer: \{x:x\in R,\ x\lt 8\}.
Question 1(c): Solve -2x+7\leq 3, where x\in R
Step 1: Subtract 7 from both sides.
-2x+7\leq 3
-2x\leq -4
Step 2: Divide by -2. Reverse the sign.
x\geq 2
Final answer: \{x:x\in R,\ x\geq 2\}.
Question 1(d): Solve 7-3x\lt x-5
Step 1: Add 3x to both sides.
7-3x\lt x-5
7\lt 4x-5
Step 2: Add 5 to both sides.
12\lt 4x
Step 3: Divide by 4.
3\lt x
Final answer: x\gt 3.
Question 1(e): Solve x(8-x)\gt 0, where x\in N
Step 1: For the product x(8-x) to be positive, both factors must have the same sign.
Step 2: Since x\in N, x\gt 0. So test the other factor.
8-x\gt 0\Rightarrow x\lt 8
Step 3: Combine the two conditions.
0\lt x\lt 8
Final answer: Since x\in N, the solution set is \{1,2,3,4,5,6,7\}. In interval form, 0\lt x\lt 8.
Question 2: State true or false
(i) If x\lt -y, then multiplying by -1 reverses the sign.
-x\gt y
Answer: True.
(ii) From -5x\geq 15, divide by -5 and reverse the sign.
x\leq -3
Answer: False, because the given statement says x\geq -3.
(iii) Dividing 2x\leq -7 by -4 reverses the sign.
\frac{2x}{-4}\geq \frac{-7}{-4}
Answer: True.
(iv) Since 7\gt 5, their positive reciprocals reverse order.
\frac{1}{7}\lt \frac{1}{5}
Answer: True.
Question 3: Check the properties of inequalities
(i) If a\lt b, subtracting c from both sides gives a-c\lt b-c. True.
(ii) If a\gt b, adding c to both sides gives a+c\gt b+c. True.
(iii) If a\lt b, then ac\gt bc is not always true because the result depends on the sign of c. False.
(iv) If a\gt b, then \frac{a}{c}\lt \frac{b}{c} is not always true because c may be positive or negative. False.
(v) From a-c\gt b-d, add c+d to both sides.
a-c+c+d\gt b-d+c+d
a+d\gt b+c
Answer: True.
(vi) If a\lt b and c\gt 0, subtracting c gives a-c\lt b-c, not a-c\gt b-c. False.
Question 4: Solve 3-2x\geq x-12, where x\in N
Step 1: Add 2x to both sides.
3-2x\geq x-12
3\geq 3x-12
Step 2: Add 12 to both sides.
15\geq 3x
Step 3: Divide by 3.
5\geq x
Final answer: Since x\in N, the solution set is \{1,2,3,4,5\}.
Question 5: Solve 25-4x\leq 16 and find the smallest value of x
Step 1: Subtract 25 from both sides.
25-4x\leq 16
-4x\leq -9
Step 2: Divide by -4. Reverse the sign.
x\geq \frac{9}{4}
Final answer: If x\in R, the smallest value is \frac{9}{4}. If x\in Z, the smallest value is 3.
Question 6: Solve -4x\geq -16 and 8-3x\leq 20, where x\in R
(i) Step 1: Divide -4x\geq -16 by -4 and reverse the sign.
x\leq 4
Answer for (i): \{x:x\in R,\ x\leq 4\}.
(ii) Step 1: Subtract 8 from both sides.
8-3x\leq 20\Rightarrow -3x\leq 12
Step 2: Divide by -3 and reverse the sign.
x\geq -4
Answer for (ii): \{x:x\in R,\ x\geq -4\}.
Question 7: Find the smallest integer for 5-2x\lt 5\frac{1}{2}-\frac{5}{3}x
Step 1: Convert the mixed number.
5\frac{1}{2}=\frac{11}{2}
Step 2: Write the inequation.
5-2x\lt \frac{11}{2}-\frac{5}{3}x
Step 3: Move constants to the left and x-terms to the right.
5-\frac{11}{2}\lt -\frac{5}{3}x+2x
-\frac{1}{2}\lt \frac{x}{3}
Step 4: Multiply by 3.
-\frac{3}{2}\lt x
Final answer: x\gt -\frac{3}{2}. The smallest integer value is -1.
Question 8: Find the largest whole-number value of x for 2(x-1)\leq 9-x
Step 1: Expand and simplify.
2(x-1)\leq 9-x
2x-2\leq 9-x
Step 2: Add x and add 2 to both sides.
3x\leq 11
Step 3: Divide by 3.
x\leq \frac{11}{3}
Final answer: Since x\in W, the largest value is 3.
Selina Exercise 4(B) Solutions for Number Line Inequations
Exercise 4(B) focuses on compound inequations and graphing. The answer must show whether each endpoint is included or excluded. A closed endpoint means \leq or \geq. An open endpoint means \lt or \gt.
Question 1(a): Read the number line -2\lt x\leq 4
Step 1: The left endpoint -2 is not included, so use x\gt -2.
Step 2: The right endpoint 4 is included, so use x\leq 4.
Final answer: \{x:x\in R,\ -2\lt x\leq 4\}.
Question 1(b): Read the integer number line starting from -3
Step 1: The values are integers, so use x\in Z.
Step 2: The point -3 is included and the arrow moves to the right.
Final answer: \{x:x\in Z,\ x\geq -3\}.
Question 1(c): Represent all real numbers except 10
Step 1: All values less than 10 are allowed.
Step 2: All values greater than 10 are allowed.
Step 3: The value 10 itself is excluded.
Final answer: \{x:x\in R,\ x\lt 10\}\cup\{x:x\in R,\ x\gt 10\}.
Question 1(d): Represent x\leq -2 or x\geq 3
Step 1: The left ray includes values up to -2.
Step 2: The right ray includes values from 3 onward.
Final answer: \{x:x\in R,\ x\leq -2\ \text{or}\ x\geq 3\}.
Question 1(e): Draw the number line for x\gt 5 and x\lt 10, where x\in R
Step 1: The word “and” means the common part of the two conditions.
Step 2: Combine the conditions.
5\lt x\lt 10
Final answer: \{x:x\in R,\ 5\lt x\lt 10\}. On the number line, use open endpoints at 5 and 10, and shade the part between them.
Question 2: Write inequations from the given graphs
(i) A shaded ray to the left of -1, including -1, represents x\leq -1, where x\in R.
(ii) A shaded ray to the right of 2, including 2, represents x\geq 2, where x\in R.
(iii) A shaded interval from -4 included to 3 excluded represents -4\leq x\lt 3, where x\in R.
(iv) A shaded interval from -1 excluded to 5 included represents -1\lt x\leq 5, where x\in R.
Final answers: (i) x\leq -1, (ii) x\geq 2, (iii) -4\leq x\lt 3, (iv) -1\lt x\leq 5.
Question 3(i): Solve and graph -4\leq 3x-1\lt 8
Step 1: Add 1 to all three parts.
-4+1\leq 3x-1+1\lt 8+1
-3\leq 3x\lt 9
Step 2: Divide all three parts by 3.
-1\leq x\lt 3
Final answer: \{x:x\in R,\ -1\leq x\lt 3\}. The point -1 is closed and the point 3 is open.
Question 3(ii): Solve and graph -1\lt 3-2x\leq 7
Step 1: Subtract 3 from all three parts.
-1-3\lt 3-2x-3\leq 7-3
-4\lt -2x\leq 4
Step 2: Divide all three parts by -2. Since the divisor is negative, reverse both inequality signs.
2\gt x\geq -2
Step 3: Rewrite in increasing order.
-2\leq x\lt 2
Final answer: \{x:x\in R,\ -2\leq x\lt 2\}. The point -2 is closed and 2 is open.
Question 3(iii): Solve and graph x-1\lt 3-x\leq 5
Step 1: Solve the left part.
x-1\lt 3-x
2x\lt 4\Rightarrow x\lt 2
Step 2: Solve the right part.
3-x\leq 5
-x\leq 2\Rightarrow x\geq -2
Step 3: Combine both conditions.
-2\leq x\lt 2
Final answer: \{x:x\in R,\ -2\leq x\lt 2\}.
Question 4: List the elements of -3\lt x-2\leq 9-2x, where x\in N
Step 1: Solve the left part.
-3\lt x-2\Rightarrow -1\lt x
Step 2: Solve the right part.
x-2\leq 9-2x
3x\leq 11\Rightarrow x\leq \frac{11}{3}
Step 3: Combine the range.
-1\lt x\leq \frac{11}{3}
Final answer: Since x\in N, the solution set is \{1,2,3\}.
Question 5: Solve -2\frac{2}{3}\leq x+\frac{1}{3}\lt 3\frac{1}{3}, where x\in R
Step 1: Convert mixed numbers into improper fractions.
-2\frac{2}{3}=-\frac{8}{3},\qquad 3\frac{1}{3}=\frac{10}{3}
Step 2: Write the inequation with fractions.
-\frac{8}{3}\leq x+\frac{1}{3}\lt \frac{10}{3}
Step 3: Subtract \frac{1}{3} from all three parts.
-\frac{8}{3}-\frac{1}{3}\leq x\lt \frac{10}{3}-\frac{1}{3}
-3\leq x\lt 3
Final answer: \{x:x\in R,\ -3\leq x\lt 3\}. On the number line, -3 is closed and 3 is open.
How to Represent Linear Inequations on a Number Line
In Concise Mathematics Selina Solutions Class 10 ICSE Chapter 4 Linear Inequations (In one variable), number-line answers are not decorative. They show whether the endpoint is part of the answer and whether the solution continues in one direction or stays between two points.
| Symbol | Endpoint | Direction or interval |
|---|---|---|
| x\lt a | Open at a | Shade left of a |
| x\leq a | Closed at a | Shade left of a, including a |
| x\gt a | Open at a | Shade right of a |
| x\geq a | Closed at a | Shade right of a, including a |
| a\lt x\leq b | Open at a, closed at b | Shade between a and b |
Examiner’s Mindset for Linear Inequations
In an ICSE-style algebra answer, credit is usually earned through the method, not only the final set. A clear solution should show the given inequation, the simplification step, the sign reversal when dividing by a negative number, and the replacement-set interpretation. For a number-line question, the endpoint type matters: a closed point matches \leq or \geq, while an open point matches \lt or \gt.
A common marking loss happens when a student solves x\leq \frac{11}{3} correctly but writes the final answer as x\leq 3 even when x\in R. That is only acceptable after applying an integer or whole-number replacement set. Always state the set condition before listing values.
Common Mistakes Students Make in Chapter 4
- Forgetting to reverse the sign: From -4x\leq -9, the correct result is x\geq \frac{9}{4}, not x\leq \frac{9}{4}.
- Mixing up N and W: In the usual school convention, W=\{0,1,2,3,\ldots\}, while N=\{1,2,3,\ldots\}. So 0 is included for whole numbers but not for natural numbers.
- Using an open endpoint for \leq or \geq: If the answer is x\geq -2, the point -2 must be included.
- Listing integers for a real-number answer: If x\in R, the answer -2\leq x\lt 2 includes all real numbers in that interval, not only -2,-1,0,1.
- Solving only one side of a compound inequation: In -4\leq 3x-1\lt 8, both boundaries must be solved before the final interval is written.
Quick Answer Index
Use this table only after reading the worked solution above. It is meant for checking the final answer, not for skipping the method.
| Exercise | Question | Answer |
|---|---|---|
| 4(A) | 1(a) | \{0,1,2,3,4,5,6\} |
| 4(A) | 1(b) | x\lt 8 |
| 4(A) | 1(c) | x\geq 2 |
| 4(A) | 1(d) | x\gt 3 |
| 4(A) | 1(e) | \{1,2,3,4,5,6,7\} |
| 4(A) | 2 | True, False, True, True |
| 4(A) | 3 | True, True, False, False, True, False |
| 4(A) | 4 | \{1,2,3,4,5\} |
| 4(A) | 5 | Smallest real value =\frac{9}{4}; smallest integer value =3 |
| 4(A) | 6(i) | x\leq 4 |
| 4(A) | 6(ii) | x\geq -4 |
| 4(A) | 7 | Smallest integer value =-1 |
| 4(A) | 8 | Largest whole-number value =3 |
| 4(B) | 1(a) | -2\lt x\leq 4,\ x\in R |
| 4(B) | 1(b) | x\geq -3,\ x\in Z |
| 4(B) | 1(c) | x\lt 10 or x\gt 10 |
| 4(B) | 1(d) | x\leq -2 or x\geq 3 |
| 4(B) | 1(e) | 5\lt x\lt 10,\ x\in R |
| 4(B) | 2 | x\leq -1, x\geq 2, -4\leq x\lt 3, -1\lt x\leq 5 |
| 4(B) | 3(i) | -1\leq x\lt 3 |
| 4(B) | 3(ii) | -2\leq x\lt 2 |
| 4(B) | 3(iii) | -2\leq x\lt 2 |
| 4(B) | 4 | \{1,2,3\} |
| 4(B) | 5 | -3\leq x\lt 3 |
Frequently Asked Questions
What is the main rule in ICSE Class 10 Maths linear inequations?
The main rule is that adding or subtracting the same quantity keeps the inequality sign unchanged, but multiplying or dividing by a negative number reverses the sign.
How do I decide the solution set when x belongs to natural numbers?
First solve the inequation over real numbers, then choose only the positive integers from that interval. For example, if the result is x\lt 5 and x\in N, the solution set is \{1,2,3,4\}.
How is a real number line answer different from an integer solution set?
A real number line answer represents every real value in an interval, while an integer solution set lists only integer points that satisfy the same condition.
Why does the inequality sign reverse when divided by a negative number?
The sign reverses because multiplication by a negative changes the order of numbers. For example, 2\gt 1, but after multiplying by -1, -2\lt -1.
What should I write in a Selina Chapter 4 solution to get full method credit?
Write the given inequation, perform one valid operation per line, reverse the sign when using a negative divisor or multiplier, state the replacement set, and give the final interval or listed set.
![For the inequations A and B [as given above in part (d)], A ∪ B is : Linear Inequations, Concise Mathematics Solutions ICSE Class 10.](https://cdn1.knowledgeboat.com/img/cm10/q1-e-test-linear-inequations-maths-concise-icse-class-10-solutions-1126x1099.png "ICSE Class 10 Maths Linear Inequations Selina Solutions 1")