Hi students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get * S Chand ICSE Maths Solutions for Class 10 Chapter 4 Linear Inequations in One Variable Exercise 4*. This is the fourth chapter ‘

*‘ included in*

**Linear Inequations in One Variables***.*

**ICSE 2020 Class 10 Maths syllabus**First, we understand some concepts of Linear Inequations in One Variable in this post.

Table of Contents

## ICSE Class 10 Maths Chapter 4 Linear Inequations Important Concepts

### What is an Inequality?

In mathematics, a statement that contains > (greater than), ≥ (greater than or equal to), < (less than), ≤ (less than or equal to), ≠ (not equal to) sign are called an inequality. The inequality defines the comparison between two numbers or other mathematical expressions.

For example:

(i) *a* < *b* means that *a* is less than *b*.

(ii) *a* > *b* means that *a* is greater than *b*.

(iii) *a* ≤ *b* or *a* ⩽ *b* means that *a* is less than or equal to *b. *

(iv) *a* ≥ *b* or *a* ⩾ *b* means that *a* is greater than or equal to *b.*

(v) *a* ≠ *b* means that *a* is not equal to *b.*

### What is an Inequations?

A mathematical statement that indicate the value of variables or an algebraic expression which is not equal to other value are called an inequation. These inequations are written using the inequalities signs, > (greater than), ≥ (greater than or equal to), < (less than), ≤ (less than or equal to), ≠ (not equal to).

For example:

(i) x < 5

(ii) y > 10

(iii) (3x + 6) ≤ 5

(iv) (2y – 9) ≥ 15

(v) (3k – 2) ≠ 5

### What is an Linear Inequations?

A mathematical statements of any of the forms ax + b > 0, ax + b ≥ 0, ax + b < 0, ax + b ≤ 0 are called as linear inequations in variable x, where a, b are real numbers and a ≠ 0.

For example:

(i) x < 15

(ii) y > 12

(iii) (3x – 8) ≤ 15

(iv) (2y + 9) ≥ 15

(v) (3k + 2) ≠ 5

### What are Replacement Set and Solution Set?

For an inequation, the set of elements or values from which the values of the variable are taken is called the replacement set or also known as domain of variable.

Now, the solution of an inequation depends upon the replacement set. So, an inequation may have one, many or no solution, depending upon the replacement set.

The solutions of inequation make a set of values of variable, and it is known as solution set of the inequation.

**Every solution set is a subset of replacement set**.

For example: we need to solve x + 2 > 7.

Depending the replacement set, the solutions set will be different.

If replacement set A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then solution of given inequation, means solution is S = {6, 7, 8, 9, 10}.

If replacement set B = {0, 1, 2, 3, 4, 5, 6, 7}, then solution of given inequation, means solution is S = {6, 7}.

If replacement set C = {0, 1, 2, 3, 4}, then solution of given inequation, means solution is S = φ (No Solution).

### What are Inequality Rules or Postulates?

There are following inequality rules:

(i) *An inequation is NOT changed if the same number is added to both sides of inequation. *

For example:

x – 5 > 12

⇒ x – 5 + 5 > 12 + 5 (**Adding 5 to both sides**)

⇒ x > 17

(ii) *An inequation is NOT changed if the same number is subtracted from both sides of inequation. *

For example:

y + 12 < 15

⇒ y + 12 – 12 < 15 – 12 (**Subtracting 12 from both sides**)

⇒ y < 3

(iii) *An inequation is NOT changed if the same positive number is multiplied to both sides of inequation. *

For example:

x/5 < 2

⇒ x/5 * 5 < 2 * 5 (**Multiplying 5 to both sides**)

⇒ x < 10

(iv) *An inequation is changed if the same negative number is multiplied to both sides of inequation. And, the inequality sign is reversed.*

For example:

-x/3 < 4

⇒ -x/3 * (-3) > 4 * (-3) (**Multiplying -3 to both sides**)

⇒ x > -12

(v) *An inequation is NOT changed if the same positive number divides both sides of inequation. *

For example:

3x < 15

⇒ 3x / 3 < 15 / 3 (**Both sides are divided by 3**)

⇒ x < 5

(vi) *An inequation is changed if the same negative number divides both sides of inequation. And, the inequality sign is reversed.*

For example:

-3x < 21

⇒ -3x / -3 > 21 / -3 (**Both sides are divided by -3**)

⇒ x > -7

(vii) *Any term of an inequation may be taken to other side with its sign changed without affecting the sign of inequality.*

For example:

x + 10 < 51

⇒ x < 51 – 10 (**10 is moved to RHS and it becomes negative without affecting inequality sign**)

⇒ x < 41

## S Chand ICSE Maths Solutions for Class 10 Chapter 4 Exercise 4

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 1**

On a bargain counter, the shopkeeper puts labels on various goods showing their prices, Rs. P, where P is real numbers. Write a mathematical sentences for each of the following labels

(a) more than Rs. 7.50

(b) not less than Rs. 10

(c) not more than Rs. 22

(d) less than Rs. 11

**S Chand ICSE Maths Solutions: **

Since the price on label is Rs P, then

(a) more than Rs. 7.50 ⇒ P > 7.50

(b) not less than Rs. 10 ⇒ P ≮ 10 or P ≥ 10

(c) not more than Rs. 22 ⇒ P ≯ 22 or P ≤ 10

(d) less than Rs. 11 ⇒ P < 11

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 2**

You are given the following numbers : -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7

Fill in the blanks

(a) A = {x : x ≥ -3 } = {……..}

(b) B = {x : x ≤ 1 } = {……..}

**S Chand ICSE Maths Solutions:**

You are given the following numbers : -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7

It mean, you are given a replacement set as U = {-3.1, -3, -2.6, 0.4, 1.2, 4.7, 5.1}.

(a) Solution of A = {x : x ≥ -3 } = {**-3, -2.6, 0.4, 1.2, 4.7, 5.1**}

(b) Solution of B = {x : x ≤ 1 } = {**-3.1, -3, -2.6, 0.4**}

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 3**

If the replacement set is {-2, -1, +1, +2, +4, +5, +9}, what is the solution set of each of the following mathematical senetences.

(a) x + 3/2 > 5/2

(b) x – 4 = -3

(c) 2x – 5 ≥ 10

(d) 3y/2 ≤ 5/2

(e) 4x^{2} = 16.

**S Chand ICSE Maths Solutions:**

Since the replacement set is {-2, -1, +1, +2, +4, +5, +9}, then

(a) Solution set of x + 3/2 > 5/2 ⇒ x > 5/2 – 3/2 ⇒ x > 1 is {+2, +4, +5, +9}.

(b) Solution set of x – 4 = -3 ⇒ x = -3 + 4 ⇒ x = 1 is {+1}.

(c) Solution set of 2x – 5 ≥ 10 ⇒ 2x ≥ 15 ⇒ x ≥ 7.5 is {+9}.

(d) Solution set of 3y/2 ≤ 5/2 ⇒ 3y ≤ 5 ⇒ y ≤ 5/3 ⇒ y ≤ 1.66 is {-2, -1, +1}.

(e) Solution set of 4x^{2} = 16 ⇒ x^{2} = 4 ⇒ x = +2 or -2 is {-2, +2}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 4**

List the solution set of 30 – 4(2x – 1) < 30, given that x is a positive integer.

**S Chand ICSE Maths Solutions:**

Since the replacement set is a positive integer, then

Solution set of

30 – 4(2x – 1) < 30

⇒ -4(2x – 1) < 30 – 30

⇒ -4(2x – 1) < 0

⇒ (2x – 1) > 0

⇒ x > 1/2

⇒ x > 0.5

is x = {1, 2, 3, 4, 5,…}

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 5**

If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, what is the solution set of each of the following mathematical senetences.

(a) x + 4/3 = 7/3

(b) 2x + 1 < 3

(c) x – 6 > 10 – 6

(d) x + 5 = 20

(e) 2x + 3 ≥ 17

**S Chand ICSE Maths Solutions:**

Since the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then

(a) Solution set of x + 4/3 = 7/3 ⇒ x = 3/3 ⇒ x = 1 is {1}.

(b) Solution set of 2x + 1 < 3 ⇒ 2x < 2 ⇒ x < 1 is {0}.

(c) Solution set of x – 6 > 10 – 6 ⇒ x > 10 is φ.

(d) Solution set of x + 5 = 20 ⇒ x = 15 is φ.

(e) Solution set of 2x + 3 ≥ 17 ⇒ 2x ≥ 14 ⇒ x ≥ 7 is {7, 8, 9}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 6**

(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}. Form all ordered pairs (x, y) such that x is a factor of y and x < y.

(b) Find the truth set of the inequality x > y + 2, where (x, y) ∈ {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6)}

**S Chand ICSE Maths Solutions:**

(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}.

(x, y) = x is a factor of y and x < y.

2 is a factor of 4, 6, 18, 54 and 2 < 4, 2 < 6, 2 < 18, 2 < 54, then (x, y) forms the ordered pairs as (2, 4), (2, 6), (2, 18), (2, 54).

4 is a factor of 4 but 4 = 4, then (4, 4) is not in (x, y).

6 is a factor of 18, 54 and 6 < 18, 6 < 54, then (x, y) forms the ordered pairs as (6, 18), (6, 54).

9 is a factor of 18, 27, 54 and 9 < 18, 9 < 27, 9 < 54 then (x, y) forms the ordered pairs as (9, 18), (9, 27), (9, 54).

Thus, the solution set of (x, y) is {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)}.

(b) Given inequality is x > y + 2. It means x is greater than y + 2.

In {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6)}, we see that (5, 1) and (7, 3) are two pairs which satisfy the given inequality.

Thus, the solution set of (x, y) is {(5, 1), (7, 3)}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 7**

Find out the truth set of the following open sentences replacement sets are given

(i) 5/x > 7 ; {1, 2}

(ii) 5/x > 2 ; {1, 2, 3, 4, 5, 6}

(iii) x^{2} = 9 ; {-3, -2, -1, 1, 2, 3}

(iv) x + 1/x = 2 ; {0, 1, 2, 3}

(v) 3x^{2} < 2x ; {-4, -3, -2, -1, 0, 1, 2, 3, 4}

(vi) 2(x – 3) < 1 ; {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

**S Chand ICSE Maths Solutions:**

(i) Given replacement set is {1, 2}.

Now given inequality is 5/x > 7 ⇒ x < 5/7 ⇒ x < 0.71

Thus, the solution set of the given inequality is x ∈ φ.

(ii) Given replacement set is {1, 2, 3, 4, 5, 6}.

Now given inequality is 5/x > 2 ⇒ x < 5/2 ⇒ x < 2.5

Thus, the solution set of the given inequality is x ∈ {1, 2}.

(iii) Given replacement set is {-3, -2, -1, 1, 2, 3}.

Now given equation is x^{2} = 9 ⇒ x = 3 or -3.

Thus, the solution set of the given equation is x ∈ {-3, 3}.

(iv) Given replacement set is {0, 1, 2, 3}.

Now given equation is x + 1/x = 2 ⇒ (x^{2} + 1) = 2x ⇒ (x^{2} -2x + 1) = 0 ⇒ (x – 1)^{2} = 0 ⇒ (x – 1) = 0 ⇒ x = 1.

Thus, the solution set of the given equation is x ∈ {1}.

(v) Given replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.

Now given inequality is 3x^{2} < 2x ⇒ 3x^{2} – 2x < 0 ⇒ x(3x – 2) < 0 ⇒ 0 < x < 2/3 ⇒ 0 < x < 0.66

Thus, the solution set of the given inequality is x ∈ φ.

(vi) Given replacement set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Now given inequality is 2(x – 3) < 1 ⇒ x – 3 < 0.5 ⇒ x < 3.5.

Thus, the solution set of the given inequality is x ∈ {1, 2, 3}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 8**

**Statement**: The sum of the lengths of any two sides of a triangle is always greater than the length of its third side.

Let x, x+1, x+2 be the lengths of the three sides of a triangle.

(i) Write down the three inequations in x, each of which represents the given statement.

(ii) List the set of possible values of x which satisfy all the three inequations obtained in your answer to part (i) above, given that x is an integer.

**S Chand ICSE Maths Solutions:**

The lengths of the three sides of a triangle are x, x+1, x+2.

(i)

Since the sum of the lengths of any two sides of a triangle is always greater than the length of its third side, we have three inequality as below.

x + (x + 1) > (x + 2) ⇒ 2x + 1 > x + 2 ⇒ x > 1 …(1)

x + (x + 2) > (x + 1) ⇒ 2x + 2 > x + 1 ⇒ x > -1 …(2)

(x + 1) + (x + 2) > x ⇒ 2x + 3 > x ⇒ x > -3 …(3)

On combining (1), (2), (3), we get x > 1.

(ii) Since replacement set of x is an integer, means x ∈ {…, -3, -2, -1, 0, 1, 2, 3, …}.

The solution set of x > 1 is x ∈ { 2, 3, 4, …}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 9**

**Answer True or False**

(a) If x + 10 = y + 14, then x > y

(b) |-4| – 4 = 8

(c) If 10 – x > 3, then x < 7

(d) If p = q + 2, then p > q

(e) If a and b are two negative integers such that a < b, then 1/a < 1/b

(f) 3 ∈ {x : 3x – 2 ≥ 5}

**S Chand ICSE Maths Solutions:**

(a) x + 10 = y + 14 ⇒ x – y = 14 – 10 ⇒ x – y = 4 ⇒ x – y > 0 ⇒ x > y.

Thus, the given statement “If x + 10 = y + 14, then x > y” is **TRUE**.

(b) |-4| – 4 = 4 – 4 = 0

Thus, the given statement “|-4| – 4 = 8” is **FALSE**.

(c) 10 – x > 3 ⇒ 10 – 3 > x ⇒ 7 > x ⇒ x > 7

Thus, the given statement “If 10 – x > 3, then x < 7” is **TRUE**.

(d) p = q + 2 ⇒ p – q = 2 ⇒ p – q > 0 ⇒ p > q

Thus, the given statement “If p = q + 2, then p > q” is **TRUE**.

(e) a < b ⇒ 1/a > 1/b

Thus, the given statement “If a and b are two negative integers such that a < b, then 1/a < 1/b” is **FALSE**.

(f) 3x – 2 ≥ 5 ⇒ 3x ≥ 7 ⇒ x ≥ 7/3 ⇒ x ≥ 2.3

Thus, the given statement “3 ∈ {x : 3x – 2 ≥ 5}” is **TRUE**.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 10**

Find the solution of the inequation 2 ≤ 2p – 3 ≤ 5. Hence graph the solution set on the number line.

**S Chand ICSE Maths Solutions:**

2 ≤ 2p – 3 ≤ 5 ⇒ 5 ≤ 2p ≤ 8 ⇒ 5/2 ≤ p ≤ 4 ⇒ 2.5 ≤ p ≤ 4

The solution on the number line is as below.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 11**

If x is a negative integer, find the solution set of 2/3 + (x + 1)/3 > 0.

**S Chand ICSE Maths Solutions:**

2/3 + (x + 1)/3 > 0 ⇒ (x + 1)/3 > -2/3 ⇒ x + 1 > -2 ⇒ x > -3 …(1)

Since the replacement set of x is negative integer, it means x ∈ {…, -5, -4, -3, -2, -1}.

From (1), x ∈ {-2, -1}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 12**

Write open mathematical sentences, using x for the variable whose graphs would be

**S Chand ICSE Maths Solutions:**

(i)

From the graph, we see that the shaded arrow is left of -2 (included) and all the real numbers are taken up to -2.

Thus, required mathematical sentence is {x : x ≤ -2 and x ∈ R}.

(ii)

From the graph, we see that the shaded arrow is left of 4 (included) and all the real numbers are taken up to 4.

Thus, required mathematical sentence is {x : x ≤ 4 and x ∈ R}.

(iii)

From the graph, we see that the solid circles are at 4 and 5 and only natural numbers are taken.

Thus, required mathematical sentence is {x : 4 ≤ x ≤ 5 and x ∈ N}.

(iv)

From the graph, we see that the solid circles are at 1, 3 and 5 and only natural numbers are taken.

Thus, required mathematical sentence is {x : 1 ≤ x ≤ 5 and x ∈ N}.

(v)

From the graph, we see that the shaded arrow is right of -2 (excluded) and all the real numbers are ahead of -2.

Thus, required mathematical sentence is {x : x > -2 and x ∈ R}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 13**

Answer **TRUE** or **FALSE**:

(i) If 2 – x < 0, then x > 2.

(ii) The graph of the inequations y ≤ 2x includes the origin.

**S Chand ICSE Maths Solutions:**

(i) 2 – x < 0 ⇒ 2 < x.

Thus, the given statement is **TRUE**.

(ii) On putting x = 0 and y = 0 in y ≤ 2x, we get that 0 = 0, which is true.

Thus, the given statement is **TRUE**.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 14**

If 25 – 4x ≤ 16, then find

(i) the smallest value of x when x is real number.

(ii) the smallest values of x when x is an integer.

**S Chand ICSE Maths Solutions:**

25 – 4x ≤ 16 ⇒ 25 – 16 ≤ 4x ⇒ 9 ≤ 4x ⇒ x ≥ 9/4.

(i) Thus, the smallest value of x is 9/4 = 2 1/4.

(ii) The smallest integer greater than x ≥ 9/4 is 3.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 15**

Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1< x + 4

**S Chand ICSE Maths Solutions:**

Given replacement set is x ∈ {1, 2, 3, 4, 5, 6, 7, 9}.

Given inequation is -3 < 2x – 1 < x + 4.

On solving -3 < 2x – 1 ⇒ 2x > -3 + 1 ⇒ 2x > -2 ⇒ x > -1 … (1)

On solving 2x – 1 < x + 4 ⇒ 2x – x < 1 + 4 ⇒ x < 5 …(2)

From (1) and (2), we get -1 < x < 5.

Thus, the solution set of the given inequation is x ∈ {1, 2, 3, 4}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 16**

Solve the inequality 2x – 10 < 3x – 15

**S Chand ICSE Maths Solutions:**

Given inequality is 2x – 10 < 3x – 15

⇒ 2x – 3x < 10 – 15

⇒ -x < -5

⇒ x > 5

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 17**

Solve the inequation 3 – 2x ≥ x – 12, given that x ∈ N.

**S Chand ICSE Maths Solutions:**

Given inequation is 3 – 2x ≥ x – 12

⇒ -x – 2x ≥ -3 – 12

⇒ -3x ≥ -15

⇒ x ≤ 5

Since the replacement of x is natural number, then the solution set of the given inequation is

x ∈ {1, 2, 3, 4, 5}.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 18**

x ∈ {real numbers} and -1 < 3 – 2x ≤ 7, evaluate x and represent it on number line.

**S Chand ICSE Maths Solutions:**

Given inequation is -1 < 3 – 2x ≤ 7

From -1 < 3 – 2x,

⇒ -1 < 3 – 2x ⇒ -1 – 3 < – 2x ⇒ -4 < -2x ⇒ 2x < 4 ⇒ x < 2 …(1)

and 3 – 2x ≤ 7

⇒ 3 – 7 ≤ 2x ⇒ -4 ≤ 2x ⇒ -2 ≤ x …(2)

On combining (1) and (2), we get -2 ≤ x < 2.

The graph of this inequation is

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 19**

Find the range of values of x which satisfies and graph these values of x on number line.

**S Chand ICSE Maths Solutions:**

Given inequation is

On simplifying, we get

Now the graph of these values of x is as below.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 20**

Solution and graph the solution set of

(a)

(b)

**S Chand ICSE Maths Solutions:**

(a) Given inequation is

On solving first inequation, we get 6 ≥ 2 – x ⇒ x ≥ 2 – 6 ⇒ x ≥ -4 … (1)

On solving second inequation, we get x/3 + 2 < 3 ⇒ x/3 < 3 – 2 ⇒ x/3 < 1 ⇒ x < 3 … (2)

On combining (1) and (2), we get -4 ≤ x < 3.

The graph of this inequation is

(b) Given inequation is

On solving first inequation, we get

x/2 < (6 – x)/4 ⇒ 2x < 6 – x ⇒ 3x < 6 ⇒ x < 2… (1)

On solving second inequation, we get

(2 – x)/6 < (7 – x)/9 ⇒ 3(2 – x) < 2(7 – x) ⇒ 6 – 3x < 14 – 2x ⇒ x > -8 … (2)

On combining (1) and (2), we get -8 < x < 2.

The graph of this inequation is

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 21**

Find the range of values of x which satisfy

Graph these values of x on real number line.

**S Chand ICSE Maths Solutions:**

Given inequation is

On simplifying,we get

Graph of these values of x on number line is as below.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 22**

Write down the range of real values of x for which the inequation x > 3 and -2 ≤ x < 5 are both true.

**S Chand ICSE Maths Solutions:**

Graphing the given inequations x > 3 and -2 ≤ x < 5 on same number line, we get

Thus, we get 3 < x < 5.

**S Chand ICSE Solutions for Class 10 Maths Exercise 4: Ques No 23**

Solve and graph the solution set of 3x – 4 > 11 or 5 – 2x ≥ 7; x ∈ R.

**S Chand ICSE Maths Solutions:**

On solving the first inequation, we get 3x – 4 > 11 ⇒ 3x > 15 ⇒ x > 5 …(1)

On solving the second inequation, we get 5 – 2x ≥ 7 ⇒ – 2x ≥ 2 ⇒ x ≤ 1 …(2)

On combining (1) and (2), we get x ≤ 1 or x > 5.

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