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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 4 - Linear Inequations (in one variable)

## Selina Textbook Solutions Chapter 4 - Linear Inequations (in one variable)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 4 - Linear Inequations (in one variable).

All Selina textbook questions of Chapter 4 - Linear Inequations (in one variable) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 4 - Linear Inequations (in one variable) Page/Excercise 4(B)

Solution 1

Solution on number line is:

Solution on number line is:

Solution on number line is:

Solution on number line is:

Solution on number line is:

Solution on number line is:

Solution on number line is:

Solution 2

Solution 3

The solution set on the real number line is:

The solution set on the real number line is:

Solution 4

The solution on number line is as follows:

The solution on number line is as follows:

The solution on number line is as follows:

The solution on number line is as follows:

The solution on number line is:

The solution on number line is:

Solution 5

-1 < 3 - 2x 7

-1 < 3 - 2x and 3 - 2x 7

2x < 4 and -2x 4

x < 2 and x -2

Solution set = {-2 x < 2, x R}

Thus, the solution can be represented on a number line as:

Solution 6

-3 < x - 2 9 - 2x

-3 < x - 2 and x - 2 9 - 2x

-1 < x and 3x 11

-1 < x

Since, x N

Solution set = {1, 2, 3}

Solution 7

-3 x and x < 3

-3 x < 3

The required graph of the solution set is:

Solution 8

Thus, the solution set is {x N: -2 ≤ x ≤3.75}

Since x N, the values of x are 1, 2, 3

The solution on number line is given by

Solution 8(old)

Since, x N

Solution set = {1, 2, 3}

The required graph of the solution set is:

Solution 9

-5 2x - 3 < x + 2

-5 2x - 3 and 2x - 3 < x + 2

-2 2x and x < 5

-1 x and x < 5

Required range is -1 x < 5.

The required graph is:

Solution 10

5x - 3 5 + 3x 4x + 2

5x - 3 5 + 3x and 5 + 3x 4x + 2

2x 8 and -x -3

x 4 and x 3

Thus, 3 x 4.

Hence, a = 3 and b = 4.

Solution 11

2x - 3 < x + 2 3x + 5

2x - 3 < x + 2 and x + 2 3x + 5

x < 5 and -3 2x

x < 5 and -1.5 x

Solution set = {-1.5 x < 5}

The solution set can be graphed on the number line as:

Solution 12

(i) 2x - 9 < 7 and 3x + 9 25

2x < 16 and 3x 16

x < 8 and x 5

Solution set = { x 5, x R}

The required graph on number line is:

(ii) 2x - 9 7 and 3x + 9 > 25

2x 16 and 3x > 16

x 8 and x > 5

Solution set = {5 < x 8, x I} = {6, 7, 8}

The required graph on number line is:

(iii) x + 5 4(x - 1) and 3 - 2x < -7

9 3x and -2x < -10

3 x and x > 5

Solution set = Empty set

Solution 13

(i) 3x - 2 > 19 or 3 - 2x -7

3x > 21 or -2x -10

x > 7 or x 5

Graph of solution set of x > 7 or x 5 = Graph of points which belong to x > 7 or x 5 or both.

Thus, the graph of the solution set is:

(ii) 5 > p - 1 > 2 or 7 2p - 1 17

6 > p > 3 or 8 2p 18

6 > p > 3 or 4 p 9

Graph of solution set of 6 > p > 3 or 4 p 9

= Graph of points which belong to 6 > p > 3 or 4 p 9 or both

= Graph of points which belong to 3 < p 9

Thus, the graph of the solution set is:

Solution 14

(i) A = {x R: -2 x < 5}

B = {x R: -4 x < 3}

(ii) A B = {x R: -2 x < 5}

It can be represented on number line as:

B' = {x R: 3 < x -4}

A B' = {x R: 3 x < 5}

It can be represented on number line as:

Solution 15

(i) x > 3 and 0 < x < 6

Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:

x > 3

0 < x < 6

From both graphs, it is clear that their common range is

3 < x < 6

(ii) x < 0 and -3 x < 1

Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:

x < 0

-3 x < 1

From both graphs, it is clear that their common range is

-3 x < 0

(iii) -1 < x 6 and -2 x 3

Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:

-1 < x 6

-2 x 3

From both graphs, it is clear that their common range is

-1 < x 3

Solution 16

Graph of solution set of -3 x < 0 or x > 2

= Graph of points which belong to -3 x < 0 or x > 2 or both

Thus, the required graph is:

Solution 17

(i) A B = {x: -1 < x < 3, x R}

It can be represented on a number line as:

(ii) Numbers which belong to B but do not belong to A = B - A

A' B = {x: -4 x -1, x R}

It can be represented on a number line as:

(iii) A - B = {x: 3 x 5, x R}

It can be represented on a number line as:

Solution 18

and

(i)

(ii) P - Q = {x: 1 < x < 5, x R}

(iii) {x: 1 < x < 5, x R}

Solution 19

Solution 20

(i)If x W, range of values of x is {0, 1, 2, 3, 4, 5, 6}.

(ii) If x Z, range of values of x is {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}.

(iii)If x R, range of values of x is .

Solution 21

A = {x: -8 < 5x + 2 17, x I}

= {x: -10 < 5x 15, x I}

= {x: -2 < x 3, x I}

It can be represented on number line as follows:

B = {x: -2 7 + 3x < 17, x R}

= {x: -9 3x < 10, x R}

= {x: -3 x < 3.33, x R}

It can be represented on number line as follows:

A B = {-1, 0, 1, 2, 3}

Solution 22

2x - 5 ≤ 5x +4 and 5x +4 < 11

2x - 5x ≤ 4 - 5 and 5x < 11 - 4

3x ≤ - 1 and 5x < 7

- 1 and x <

- 1 and x <

Since x I, the solution set is

And the number line representation is

Solution 23

Solution set = {5, 6}

It can be graphed on number line as:

Solution 24

A = {x: 11x - 5 > 7x + 3, x R}

= {x: 4x > 8, x R}

= {x: x > 2, x R}

B = {x: 18x - 9 15 + 12x, x R}

= {x: 6x 24, x R}

= {x: x 4, x R}

Range of A B = {x: x 4, x R}

It can be represented on number line as:

Solution 25

7x + 3 3x - 5

4x -8

x -2

Since, x N

Solution set = {1, 2, 3, 4, 5}

Solution 26

(i)

Since, x is a positive odd integer

Solution set = {1, 3, 5}

(ii)

Since, x is a positive even integer

Solution set = {2, 4, 6, 8, 10, 12, 14}

Solution 27

Since, x W

Solution set = {0, 1, 2}

The solution set can be represented on number line as:

Solution 28

Let the required integers be x, x + 1 and x + 2.

According to the given statement,

Thus, the largest value of the positive integer x is 24.

Hence, the required integers are 24, 25 and 26.

Solution 29

2y - 3 < y + 1 4y + 7, y R

2y - 3 - y < y + 1 - y 4y + 7 - y

y - 3 < 1 3y + 7

y - 3 < 1 and 1 3y + 7

y < 4 and 3y - 6 y - 2

- 2 y < 4

The graph of the given equation can be represented on a number line as:

Solution 30

3z - 5 z + 3 < 5z - 9

3z - 5 z + 3 and z + 3 < 5z - 9

2z 8 and 12 < 4z

z 4 and 3 < z

Since, z R

Solution set = {3 < z 4, Z R }

It can be represented on a number line as:

Solution 31

The solution set can be represented on a number line as:

Solution 32

Consider the given inequation:

-4 ≤ x < 5; where x  R

The solution set can be represented on a number line as follows:

Solution 33

Solution 34

Solution 35

Solution 36

The solution set is represented on number line as follows:

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 4 - Linear Inequations (in one variable) Page/Excercise 4(A)

Solution 1

Solution 2

(i) a < b a - c < b - c

The given statement is true.

(ii) If a > b a + c > b + c

The given statement is true.

(iii) If a < b ac < bc

The given statement is false.

(iv) If a > b

The given statement is false.

(v) If a - c > b - d a + d > b + c

The given statement is true.

(vi) If a < b a - c < b - c (Since, c > 0)

The given statement is false.

Solution 3

(i) 5x + 3 2x + 18

5x - 2x 18 - 3

3x 15

x 5

Since, x N, therefore solution set is {1, 2, 3, 4, 5}.

(ii) 3x - 2 < 19 - 4x

3x + 4x < 19 + 2

7x < 21

x < 3

Since, x N, therefore solution set is {1, 2}.

Solution 4

(i) x + 7 11

x 11 - 7

x 4

Since, the replacement set = W (set of whole numbers)

Solution set = {0, 1, 2, 3, 4}

(ii) 3x - 1 > 8

3x > 8 + 1

x > 3

Since, the replacement set = W (set of whole numbers)

Solution set = {4, 5, 6, …}

(iii) 8 - x > 5

- x > 5 - 8

- x > -3

x < 3

Since, the replacement set = W (set of whole numbers)

Solution set = {0, 1, 2}

(iv) 7 - 3x

-3x - 7

-3x

x

Since, the replacement set = W (set of whole numbers)

Solution set = {0, 1, 2}

(v)

Since, the replacement set = W (set of whole numbers)

Solution set = {0, 1}

(vi) 18 3x - 2

18 + 2 3x

20 3x

Since, the replacement set = W (set of whole numbers)

Solution set = {7, 8, 9, …}

Solution 5

3 - 2x x - 12

-2x - x -12 - 3

-3x -15

x 5

Since, x N, therefore,

Solution set = {1, 2, 3, 4, 5}

Solution 6

25 - 4x 16

-4x 16 - 25

-4x -9

x

x

(i) The smallest value of x, when x is a real number, is 2.25.

(ii) The smallest value of x, when x is an integer, is 3.

Solution 7

Since, the replacement set of real numbers.

Solution set = {x: x R and }

Since, the replacement set of real numbers.

Solution set = { x: x R and }

Since, the replacement set of real numbers.

Solution set = { x: x R and x > 80}

Since, the replacement set of real numbers.

Solution set = { x: x R and x > 13}

Solution 8

Thus, the required smallest value of x is -1.

Solution 9

2(x - 1) 9 - x

2x - 2 9 - x

2x + x 9 + 2

3x 11

Since, x W, thus the required largest value of x is 3.

Solution 10

Solution set = {x: x R and x 6}

Solution 11

Since, x {integers}

Solution set = {-1, 0, 1, 2, 3, 4}

Solution 12

Since, x {whole numbers}

Solution set = {0, 1, 2, 3, 4}

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# Text Book Solutions

ICSE X - Mathematics

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