# SELINA Solutions for Class 10 Maths Chapter 4 - Linear Inequations (in one variable)

## Chapter 4 - Linear Equations in One Variable Exercise Ex. 4(A)

(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.

(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}.

(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, …}

3 - 2x x - 12

-2x - x -12 - 3

-3x -15

x 5

Since, x N, therefore,

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

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.

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}

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

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 set = {x: x R and x 6}

Since, x {integers}

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

Since, x {whole numbers}

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

## Chapter 4 - Linear Equations in One Variable Exercise Ex. 4(B)

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:

The solution set on the real number line is:

** **

The solution set on the real number line is:

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:

-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:

-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}

-3 x and x < 3

-3 x < 3

The required graph of the solution set is:

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

-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:

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.

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:

(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

(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:

(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:

(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

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:

(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:

and

(i)

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

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

(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 .

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}

2x - 5 ≤ 5x + 4 and 5x +4 < 11

2x - 9 ≤ 5x and 5x < 11 - 4

-9 ≤ 3x and 5x < 7

x - 3 and x <

x - 3 and x <

Since x I, the solution set is

And the number line representation is

Solution set = {5, 6}

It can be graphed on number line as:

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:

7x + 3 3x - 5

4x -8

x -2

Since, x N

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

(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}

Since, x W

Solution set = {0, 1, 2}

The solution set can be represented on number line as:

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.

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:

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:

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

Consider the given inequation:

⇒ -4 ≤ x < 5; where x ∊ R

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

The solution set is represented on number line as follows:

As,

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