Class 10 SELINA Solutions Maths Chapter 4 - Linear Inequations (in one variable)
Linear Equations in One Variable Exercise Ex. 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}
Linear Equations in One Variable Exercise Ex. 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 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 - 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 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:
Solution 37
As, 
