Class 10 SELINA Solutions Maths Chapter 4 - Linear Inequations (in one variable)
Linear Inequations (in one variable) Exercise Ex. 4(A)
Solution 1(a)
Correct Option: (i) {–8, –6, –5, –4, 6}
The elements in the set {–8, –6, –5, –4, 6} do not satisfy the replacement set.
Solution 1(b)
Correct Option: (ii) x < 8
4(2x – 5) < 2x + 28
8x – 20 < 2x + 28
8x – 2x < 28 + 20
6x < 48
x < 8
Solution 1(c)
Correct Option: (iv) {x : x ∊ R, x ≥ 2}
–2x + 7 ≤ 3
–2x ≤ 3 – 7
–2x ≤ –4
–x ≤ –2
x ≥ 2
Therefore, the solution set is {x : x ∊ R, x ≥ 2}.
Solution 1(d)
Correct Option: (i) x > 3
7 – 3x < x – 5
7 + 5 < x + 3x
12 < 4x
3 < x
i.e. x > 3
Solution 1(e)
Correct Option: (iii) 0 < x < 8
x(8 – x) > 0
8 – x > 0
8 > x
Since x ∊ N, 0 < x < 8.
Solution 2
Solution 3
(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 4
(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 5
(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}
Solution 6
3 - 2x 
x - 12
-2x - x -12 - 3
-3x -15
x 
5
Since, x 
N, therefore,
Solution set = {1, 2, 3, 4, 5}
Solution 7
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 8
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 
}
Solution 9
Thus, the required smallest value of x is -1.
Solution 10
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.
Linear Inequations (in one variable) Exercise Ex. 4(B)
Solution 1(a)
Correct Option: (iii) {x : x ∊ R and –2 < x ≤ 4}
The solution set is {x : x ∊ R and –2 < x ≤ 4}.
Solution 1(b)
Correct Option: (ii) {x : x ∊ Z and –3 ≤ x}
The solution set is {x : x ∊ Z and –3 ≤ x}.
Solution 1(c)
Correct Option: (ii) {(x < 10) ⋃ (x > 10)}
The solution set is {(x < 10) ⋃ (x > 10)}.
Solution 1(d)
Correct Option: (iv) {x : x ∊ R, x ≤ –2 or x ≥ 3}
The solution set is {x : x ∊ R, x ≤ –2 or x ≥ 3}.
Solution 1(e)
Correct Option: (ii)
The required number line is
Solution 2
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 3
Solution 4
The solution set on the real number line is:
The solution set on the real number line is:
Solution 5
The solution on number line is as follows:
The solution on number line is as follows:
The solution on number line is as follows:
(iv) 
2x + 5 > 9x -3
-7x > -14
x < 2
The solution on number line is:
Solution 6
-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 7
-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 8
-3 
x and x < 3
-3 x < 3
The required graph of the solution set is:
Solution 9
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 10
-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 11
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 12
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 13
(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 14
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:
Solution 15
(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 16
(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 17
(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 18
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 19
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 20
Solution set = {5, 6}
It can be graphed on number line as:
Linear Inequations (in one variable) Exercise TEST YOURSELF
Solution 1(a)
Correct Option: (ii) 4
4x ≤ 12 + x
4x – x ≤ 12
3x ≤ 12
x ≤ 4
Solution 1(b)
Correct Option: (i) 2
5x – 4 ≥ 18 – 6x
5x + 6x ≥ 18 + 4
11x ≥ 22
x ≥ 2
Solution 1(c)
Correct Option: (i) x > 1
3x + 15 < 5x + 13
15 – 13 < 5x – 3x
2 < 2x
1 < x
i.e., x > 1
Solution 1(d)
Correct Option: (i)
The solution for a number line A is –3 < x ≤ 1.
And, for a number line B, the solution is –4 ≤ x < 0.
Then, A ⋂ B = –3 < x < 0.
Hence, the required number line is
Solution 1(e)
Correct Option: (i)
The solution for a number line A is –3 < x ≤ 1.
And, for a number line B, the solution is –4 ≤ x < 0.
Then, A ⋃ B = –4 ≤ x ≤ 1.
Hence, the required number line is
Solution 2
Solution set = {x: x 
R and x 
6}
Solution 3
Since, x 
{whole numbers}
Solution set = {0, 1, 2, 3, 4}
Solution 4
7x + 3
3x - 5
4x -8
x -2
Since, x 
N
Solution set = {1, 2, 3, 4, 5}
Solution 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}
Solution 6
Since, x 
W
Solution set = {0, 1, 2}
The solution set can be represented on number line as:
Solution 7
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 8
Consider the given inequation:
⇒ -4 ≤ x < 5; where x ∊ R
The solution set can be represented on a number line as follows:
Solution 9
Solution 10
As, 
