Utilizing OP Malhotra Class 10 Solutions Chapter 4 Linear Inequations in One Variable Ex 4 as a study aid can enhance exam preparation.
S Chand Class 10 ICSE Maths Solutions Chapter 4 Linear Inequations in One Variable Ex 4
Question 1.
On a bargain counter, the shopkeeper put labels on various goods showing their prices, Rs. P, where P e {real numbers}. Write a mathematical sentence 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
Solution:
(a) more than Rs. 7.50 : > Rs. 7.50
(b) not less than Rs. 10 : ≮ Rs. 10
(c) not more than Rs. 22 : ≯ Rs. 22
(d) less than Rs. 11 : < Rs. 11
Question 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 = {A : A ≤ -3} = {…..}
(b) B = {A : A ≤ 1} = {…..}
Solution:
Given numbers are ; -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7
(a) A = {x : x ≥ -3}
= {-3, -2.6, 0.4, 1.2, 4.7, 5.1}
(b) B = {x : x ≤ 1} = {-3.1, -3, -2.6, 0.4}
Question 3.
If the replacement set is {-2, -1, +1, +2, +4, +5, +9}, what is the solution set of each of the following mathematical sentences?
(a) x + \(\frac { 3 }{ 2 }\) > \(\frac { 5 }{ 2 }\)
(b) x – 4 = – 3
(c) 2x – 5 ≥ 10
(d) 3y ÷ 2 ≤ \(\frac { 5 }{ 2 }\)
(e) 4x² = 16
Solution:
Replacement set = {-2, -1, +1, +2, +4, +5, +9}
(a) x + \(\frac { 3 }{ 2 }\) > \(\frac { 5 }{ 2 }\)
⇒ x > \(\frac { 5 }{ 2 }\) – \(\frac { 3 }{ 2 }\) ⇒ x > \(\frac { 5 – 3}{ 2 }\)
⇒ x > \(\frac { 2 }{ 2 }\) ⇒ x > 1
∴ Solution set = {+2, +4, +5, +9}
(b) x – 4 = – 3
⇒ x = – 3 + 4
⇒ x = 1
Solution set = {+1}
(c) 2x – 5 ≥ 10
⇒ 2x ≥ 10 + 5 ⇒ 2x ≥ 15
⇒ x ≥ y
⇒ A ≥ 7.5
∴ Solution set = {+9}
(d) 3y ÷ 2 ≤ \(\frac { 5 }{ 2 }\)
⇒ 3y ≤ \(\frac { 5 }{ 2 }\) x 2 ⇒ 3y ≤ 5
⇒ y ≤ \(\frac { 5 }{ 3 }\)
⇒ y ≤ 1\(\frac { 2 }{ 3 }\)
∴ Solution set = {+1, -1, -2} = {-2. -1, +1}
(e) 4x² = 16
⇒ x² = \(\frac { 16 }{ 4 }\)
⇒ x² = 4 ⇒ x² = (±2)²
∴ x = ±2,
∴ Solution set = {-2, +2}
Question 4.
List the solution set of 30 – 4 (2x – 1) < 30, given that x is a positive integer.
Solution:
30 – 4 (2x – 1) < 30
⇒ 30 – 8x + 4 < 30
⇒ 34 – 8x < 30
⇒ 34 – 30 < 8x ⇒ 4 < 8x ⇒ 8A > 4 ⇒ x > \(\frac { 4 }{ 8 }\)
⇒ x > \(\frac { 1 }{ 2 }\)
∴ A is a positive integer,
∴ Solution set = {1, 2, 3, 4, }
Question 5.
If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, what is the solution set of the following mathematical sentences?
(a) x + \(\frac { 4 }{ 3 }\) = \(\frac { 7 }{ 3 }\)
(b) 2x + 1 < 3 (c) x – 6 > 10 – 6
(d) x + 5 = 20
(e) 2x + 3 ≥ 17
Solution:
Replacement set = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(a) x + \(\frac { 4 }{ 3 }\) = \(\frac { 7 }{ 3 }\) ⇒ x = \(\frac { 7 }{ 3 }\) – \(\frac { 4 }{ 3 }\)
⇒ x = \(\frac { 7 – 4 }{ 3 }\) = \(\frac { 3 }{ 3 }\) ⇒ x = 1
∴ Solution set = 1
(b) 2A + 1 – 1 < 3 ⇒ 2x < 3 – 1
⇒ 2A < 2 ⇒ x < \(\frac { 2 }{ 2 }\)
⇒ x < 1
∴ Solution set = {0}
(c) x – 6 > 10 – 6
⇒ x > 10 – 6 + 6
⇒ x > 10 – 12 ⇒ x > – 2
∴ Solution set = Φ {∵ – 2 ∉ to the given set}
(d) x + 5 = 20 ⇒ x = 20 – 5
⇒ x = 15
∴ Solution set = Φ {∵ 15 ∉ to the given set}
(e) 2A + 3 ≥ 17 ⇒ 2x ≥ 17 – 3
⇒ 2A ≥ 14 ⇒ x ≥ \(\frac { 14 }{ 2 }\)
⇒ A ≥ 7
∴ Solution set = {7, 8, 9}
Question 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 factory 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), (6, 5)}.
Solution:
(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}
∵ A is a factor of y and x < y ∴ Ordered pairs will be {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)} (b) x > y + 2 where
(A, y) ∈ {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6), (6, 5)}
If (A, y) = (1, 2) then 1 > 2 + 2 ⇒ 1 > 4 which is not true
If (A, y) = (2, 3) then 2 > 3 + 2 ⇒ 2 > 5 which is not true
If (A, y) = (5, 1) then 5 > 1 + 2 ⇒ 5 > 3 which is true
∴ (5, 1) is its solution
If (x, y) = (7, 3), then 7 > 3 + 2 ⇒ 7 > 5 which is true
∴ (7, 3) is its solution
If (x, y) = (5, 6). then 5 > 6 + 2 ⇒ 5 > 8 which is not true
If (x, y) = (6, 5). then 6 > 5 + 2 ⇒ 6 > 7 which is not true
∴ Solution or true set is {(5, 1), (7, 3)}
Question 7.
Find out the truth sets of the following open sentences; replacement sets are given against them.
(i) \(\frac { 5 }{ x }\) > 7 ; {1, 2}
(ii) \(\frac { 5 }{ x }\) > 2; {1, 2, 3, 4, 5, 6}
(iii) x² = 9; {-3, -2, -1, 1, 2, 3}
(tv) x + \(\frac { 1 }{ x }\) = 2 ; {0, 1,2, 3}
(v) 3x² < 2x ; {-4, -3, -2, -1, 0, 1, 2, 3, 4}
(vi) 2 (x – 3) < 1 ; {1, 2, 3, 4, …………… 10}. Solution: (i) \(\frac { 5 }{ x }\) > 7; {1, 2}
If x = 1, then \(\frac { 5 }{ 1 }\) > 7 ⇒ 5 > 7 which is not true
If x = 2, then \(\frac { 5 }{ 2 }\) > 7 which is not true
∴ x has no solution
Solution set = Φ
(ii) \(\frac { 5 }{ x }\) > 2; {1,2, 3,4, 5,6}
If x = 1, then \(\frac { 5 }{ 1 }\) > 2 which is true 5
If x = 2, then \(\frac { 5 }{ 1 }\) > 2 which is true
If x = 3, then \(\frac { 5 }{ 2 }\) > 2 which is not true
If x = 4, then \(\frac { 5 }{ 4 }\) > 2 which is not true
If x = 5, then \(\frac { 5 }{ 5 }\) > 2 ⇒ 1 > 2 which is not true
If x = 6, then \(\frac { 5 }{ 6 }\) > 2 which is not true o
∴ Solution set = {1, 2}
(iii) x² = 9; {-3, -2, -1, 1, 2, 3}
If x = -3, then (-3)² = 9 ⇒ 9 = 9 which is true
If x = -2, then (-2)² = 9 ⇒ 4 = 9 which is not true
If x = -1, then (-1)² = 9 ⇒ 1 = 9 which is not true
If x = 1, then (1 )² – 9 ⇒ 1 =9 which is not true
If x = 2, then (2)² – 9 ⇒ 4 = 9 which is not ture
If x = 3, then (3)² = 9 ⇒ 9 which is true
∴ Solution set = {-3, 3}
(iv) x + y = 2 ; {0, 1, 2, 3}
If x = 0, then 0 + \(\frac { 1 }{ 0 }\) = 2 which is not true
If x = 1, then 1 + \(\frac { 1 }{ 1 }\) = 2 ⇒ 1 + 1 = 2 ⇒ 2 = 2 which is true
If x = 2, then 2 + \(\frac { 1 }{ 2 }\) = 2 which not true 1
If x = 3, then 3 + \(\frac { 1 }{ 3 }\) = 2 which is not true
∴ Solution set = {1}
(v) 3x² < 2x ; {- 4, – 3, – 2, – 1, 0, 1, 2, 3, 4} If x = – 4, then 3 (- 4)² < 2 x (- 4)
⇒ 3 x 16 < – 8 ⇒ 48 < – 8 which is not true If x = – 3, then 3 (- 3)² < 2 x (- 3)
⇒ 3 x 9 < – 6 ⇒ 27 < – 6 which is not true If x = – 2, then 3 (-2)² < 2 x (-2)
⇒ 3 x 4 < – 4 ⇒ 12 < – 4 which is not true If x = – 1, then 3 (-1)² < 2 x (-1)
3 x 1 < – 2 ⇒ 3 < – 2 which is not true
If x = 0. then 3 (0)² < 2 (0) ⇒ 3 x 0 < 2 x 0 ⇒ 0 < 0 which is not true
If x = 1, then 3 (1)² <2xl⇒ 3xl<2 ⇒ 3 < 2 which is not true
If x = 2, then 3 (2)² <2×2⇒ 3×4<4 ⇒ 12 < 4 which is not true
If x = 3, then 3 (3)² <2×3⇒ 3*9<6 ⇒ 27 < 6 which is not true
If x = 4, then 3 (4)² <2×4⇒ 3xl6<8 ⇒ 48 < 8 which is not true
∴ Solution set = Φ
(vi) 2 (x – 3) < 1 ; {2, 3, 4, 10}
If x = 1. then 2 (1 – 3) < 1 ⇒ 2 x (-2) < 1
⇒ – 4 < 1 which is true
If x = 2, then 2 (2 – 3) < 1 ⇒ 2 (-1) < 1
⇒ – 2 < 1 which is true
If x = 3, then 2(3 – 3) < 1
⇒ 2 x 0 < 1
If x = 4, then 2(4 – 3) < 1 ⇒ 2 x 1
⇒ 2 < 1 which is not true
If x = 5, then 2 (5 – 3) < 1 ⇒ 2 x 2 < 1
⇒ 4 < 1 which is not true
If x = 10, then 2 (10 – 3) < 1 ⇒ 2 x 7 < 1
⇒ 14 < 1 which is not true
∴ Solution set = {1, 2, 3}
Question 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.
Solution:
It is given that in a triangle,
Sum of any two sides is greater than the third side
Let x, x + 1 and x + 2 be the lengths of the three sides of a triangle,
(i) then the three inequations can be (a) x > 1 (b) x > 2 (c) x > 3
(ii) When x > 1, then the sides will be (2, 3, 4) When x > 2, then the sides will be (3, 4, 5) and when x > 3, then the side will be (4, 5,6)
Set of solution = {2, 3, 4, }
P.Q. P is the solution set of 8x – 1 > 5x + 2 and Q is the solution set of 7x – 2 ≥ 3 (x + 6), where x ∈ N. Find the set P ∩ Q.
Solution:
P is the solution set of 8x – 1 > 5x + 2
Q is the solution set of 7x – 2 ≥ 3 (x + 6)
Where x ∈ N.
Now 8x – 1 > 5x + 2
⇒ 8x – 5x > 2 + 1 ⇒ 3x > 3
⇒ x > \(\frac { 3 }{ 3 }\) ⇒ x > 1
∴ P = {2, 3, 4, 5, 6, 7, 8, }
and 7x – 2 ≥ 3 (x + 6)
⇒ 7x – 2 ≥ 3x + 18
⇒ 7x – 3x ≥ 18 + 2 ⇒ 4x ≥ 20
⇒ x ≥ \(\frac { 20 }{ 4 }\) ⇒ x ≥ 5
∴ Q = {5, 6, 7, 8, }
∴ P ∩ Q = {5, 6, 7, 8, }
Question 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 \(\frac { 1 }{ a }\) < \(\frac { 1 }{ a }\) (f) 3 ∈ {x : 3x – 2 > 5}.
Solution:
(a) True
∵ x + 10 = y + 14
⇒ x = y + 14 – 10
⇒ x = y + 4
∴ x > y
(b) False
∵ | – 4 | – 4 = 8
⇒ 4 – 4 = 8 and 0 = 8
Which is not possible
(c) True
∵ 10 – x > 3
∴ 10 – 3 > x ⇒ 7 > x
⇒ x < 7 (d) True ∵ p = q+ 2 ∴ p > q
(e) False
∵ a, b are two negative integers and a < b then \(\frac { 1 }{ a }\) > \(\frac { 1 }{ b }\)
(f) True
∵ 3 ∈ {x : 3x – 2 > 5}
3x – 2 ≥ 5 ⇒ 3x ≥ 5 + 2
⇒ 3x ≥ 7 ⇒ x ≥ \(\frac { 7 }{ 3 }\)
∴ Solution set = {3, 4, 5, 6, …………. } (∵ x ∈ N)
Question 10.
Find the solution of the inequation 2 ≤ 2p – 3 ≤ 5.
Hence graph the solution set on the number line given below.
Solution:
2 ≤ 2p – 3 ≤ 5
(i) 2 ≤ 2p – 3
⇒ 2 + 3 < 2p
⇒ 5 ≤ 2p
⇒ \(\frac { 5 }{ 2 }\) ≤ 2p
(ii) 2p – 3 ≤ 5
⇒ 2p ≤ 5 + 3
⇒ 2p ≤ 8
⇒ p ≤ \(\frac { 8 }{ 2 }\)
⇒ p ≤ 4
From (i) and (ii)
\(\frac { 5 }{ 2 }\) ≤ P ≤ 4 ⇒ 2\(\frac { 1 }{ 2 }\) ≤ p ≤ 4
∴ Solution set = {2\(\frac { 1 }{ 2 }\) ≤ P ≤ 4, p ∈ R}
The solution set is given below on the number line.
Question 11.
If x is a negative integer, find the solution set of \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) (x + 1) > 0.
Solution:
x is a negative integer
\(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) (x + 1) > 0.
⇒ \(\frac { 2 }{ 3 }\) + \(\frac { x }{ 3 }\) + \(\frac { 1 }{ 3 }\) > 0.
⇒ \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { x }{ 3 }\) > 0.
⇒ \(\frac { 2+1 }{ 3 }\) + \(\frac { x }{ 3 }\) > 0 ⇒ 1 + \(\frac { x }{ 3 }\) > 0
⇒ \(\frac { x }{ 3 }\) > – 1 ⇒ x > – 3
∴ Solution set = {-2, -1}
Question 12.
Write open mathematical sentences, using x for the variable whose graphs would be
Solution:
(i) Here solution set = {x : x ≤ – 2, x ∈ R}
(ii) Here solution set = {x : x ≤ 4, x ∈ R}
(iii) Here solution set = {x : 4 ≤ x ≤ 5, x ∈ N}
(iv) Here solution set = {x : 1 ≤ x ≤ 5, x ∈ N, x is odd}
(v) Here solution set = (x : x > – 2, x ∈ R}
Question 13.
Answer true or false
(i) If 2 – x < 0, then x > 2
(ii) The graph of the inequation y ≤ 2x includes the origin.
Solution:
(i) True
∵ 2 – x < 0 ⇒ 2 < x ⇒ x > 2
(ii) True
∵ y ≤ 2x and origin is (0, 0)
∴ 0 ≤ 2 x 0 ⇒ 0 ≤ 0
Question 14.
If 25 – 4x ≤ 16, find :
(i) the smallest value of x when x is a real number
(ii) the smallest value of x when x is an integer.
Solution:
25 – 4x ≤ 16
⇒ 25 – 16 ≤ 4x ⇒ 9 ≤ 4x
⇒ 4x ≥ 9 ⇒ x ≥ \(\frac { 9 }{ 4 }\)
∴ Smallest value of x when x is a real number
= \(\frac { 9 }{ 4 }\) = 2\(\frac { 1 }{ 4 }\) (∵ \(\frac { 9 }{ 4 }\) is a real number)
(ii) Smallest value of x when x is a real number = 3 (∵ 3 > \(\frac { 9 }{ 4 }\))
Question 15.
Given : x ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}, find the values of x for which – 3 < 2x – 1 < x + 4.
Solution:
– 3 < 2x – 1 < x + 4
(i) – 3 < 2x – 1 ⇒ – 3 + 1 < 2x
⇒ – 2 < 2x ⇒ \(\frac { -2 }{ 2 }\) < x ⇒ – 1 < x
(ii) 2x – 1 < x + 4 ⇒ 2x – x < 4 + 1 ⇒ x < 5
From (i) and (ii)
– 1 < x < 5
∴ Solution set will be {1, 2, 3, 4}
Question 16.
Solve the inequality : 2x – 10 < 3x – 15.
Solution:
2x – 10 < 3x – 15
⇒ – 10 + 15 < 3x – 2x
⇒ 5 < x ⇒ x > 5
∴ Solution set = {x : x > 5}
Question 17.
Solve the inequation : 3 – 2x ≥ x – 12, given that x ∈ N.
Solution:
3 – 2x ≥ x – 12
⇒ 3 + 12 ≥ x + 2x ⇒ 15 ≥ 3x
⇒ 3x ≤ 15 ⇒ x ≤ \(\frac { 15 }{ 2 }\)
⇒ x ≤ 5
∴ Solution set = {1, 2, 3, 4, 5} (∵ x ∈ N)
Question 18.
x ∈ {real numbers} and – 1 < 3 – 2x ≤ 7, evaluate x and represent it on a number line.
Solution:
x ∈ {Real numbers}
-1 < 3 – 2x ≤ 7
(i) – 1 < 3 – 2x ⇒ – 1 – 3 < – 2x
⇒ – 4 < – 2x ⇒ 4 > 2x
⇒ \(\frac { 4 }{ 2 }\) x ⇒ 2 > x ⇒ x < 2
(ii) and 3 – 2x < 7 ⇒ – 2x ≤ 7 – 3
⇒ – 2x ≤ 4 ⇒ – x ≤ \(\frac { 4 }{ 2 }\) ⇒ – x ≤ 2
⇒ – 2 ≤ x
From (i) and (ii)
The solution is – 2 ≤ x < 2
Number line
Question 19.
Find the range of values of x which satisfies
– 2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < 3\(\frac { 1}{ 3 }\), x ∈ R
Graph these values of x on the number line.
Solution:
– 2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < 3\(\frac { 1}{ 3 }\)
⇒ \(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < \(\frac { 10}{ 3 }\)
(i) \(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) ⇒ \(\frac { -8 }{ 3 }\) – \(\frac { 1}{ 3 }\) ≤ x
⇒ \(\frac { -9 }{ 3 }\) ≤ x ⇒ – 3 ≤ x
(ii) x + \(\frac { 1}{ 3 }\) < \(\frac { 10 }{ 3 }\) ⇒ \(\frac { 10 }{ 3 }\) – \(\frac { 1 }{ 3 }\)
⇒ x < \(\frac { 9 }{ 3 }\) ⇒ x < 3
From (i) and (ii)
Solution is – 3 < x < 3
Number line
Question 20.
Solve and graph the solution set of :
(a) 6 ≥ 2 – x, \(\frac { x }{ 3 }\) + 2 < 3; x ∈ R
(b) \(\frac { x }{ 2 }\) < \(\frac { 6-x }{ 4 }\), \(\frac { 2-x }{ 6 }\) < \(\frac { 7-x }{ 9 }\); x ∈ R
Solution:
(a) 6 ≥ 2 – x ⇒ 6 – 2 ≥ – x
⇒ 4 ≥ – x ⇒ x ≥ – 4 ⇒ – 4 ≤ x … (i)
and \(\frac { x }{ 3 }\) + 2 < 3 ⇒ x + 6 < 9 (Multiplying by 3)
⇒ x < 9 – 6 ⇒ x < 3 … (ii),
From (i) and (ii)
∴ Solution = – 4 ≤ x < 3
Number line
(b) \(\frac { x }{ 2 }\) < \(\frac { 6-x }{ 4 }\) ⇒ 2x < 6 – x (Multiplying by 4)
⇒ 2x + x < 6 ⇒ 3x < 6
⇒ x < \(\frac { 6 }{ 3 }\) ⇒ x < 2 … (i)
and \(\frac { 2-x }{ 6 }\) < \(\frac { 7-x }{ 9 }\)
⇒ 3 (2 – x) 2 (7 – x)
(Multiplying by L.C.M. of 6, 9 = 18)
⇒ 6 – 3x < 14 – 2x
⇒ 6 – 14 < – 2x + 3x ⇒ – 8 < x … (ii)
From (i) and (ii)
– 8 < x < 2
Number line
Question 21.
Find the range of values of x which satisfy :
\(\frac { -1 }{ 3 }\) ≤ \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) < \(\frac { 1 }{ 6 }\); x ∈ R
Graph these values of x on the real number line.
Solution:
\(\frac { -1 }{ 3 }\) ≤ \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) < \(\frac { 1 }{ 6 }\)
(i) \(\frac { -1 }{ 3 }\) ≤ \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) ⇒ \(\frac { -1 }{ 6 }\)
Multiplying by the L.C.M. of 3, 2 = 6
– 2 < 3x – 8 ⇒ – 2 + 8 ≤ 3x
⇒ 6 ≤ 3x ⇒ \(\frac { 6 }{ 3 }\) ≤ x
⇒ 2 ≤ x …. (i)
and \(\frac { x }{ 2 }\) – 1\(\frac { 1 }{ 3 }\) < \(\frac { 1 }{ 6 }\) ⇒ \(\frac { x }{ 2 }\) – \(\frac { 4 }{ 3 }\) < \(\frac { 1 }{ 6 }\)
Multiplying by the L.C.M. of 2, 3,6 = 6
3x – 8 < 1 ⇒ 3x < 1 + 8
⇒ 3x < 9 ⇒ x < \(\frac { 9 }{ 3 }\)
⇒ x < 3 … (ii)
from (i) and (ii)
2 ≤ x ≤ 3
∴ Number line
Question 22.
Write down the range of real values of x for which the inequation x > 3 and – 2 < x < 5 are both true.
Solution:
x > 3
3 < x … (i)
and -2 < x < 5
∴ Range = {-2, -1,0, 1, 2, 3, 4} … (ii)
∴ From (i) and (ii)
Range = {3 < x < 5; x ∈ R}
Question 23.
Solve and graph the solution set of 3x – 4 > 11 or 5 – 2x < 7; x ∈ R}
Solution:
3x – 4> 11 ⇒ 3x > 11 + 4
⇒ 3x > 15 ⇒ x > \(\frac { 15 }{ 3 }\)
⇒ x > 5 … (i)
and 5 – 2x ≥ 7 ⇒ 5 – 7 > 2x
⇒ – 2 ≥ 2x ⇒ \(\frac { -2 }{ 2 }\) > x
⇒ -1 > x
⇒ x ≤ – 1, x ∈ R … (ii)
From (i) and (ii)
x > 5 or x ≤ – 1; x ∈ R
Graph of solution
Self Evaluation And Revision (LATEST ICSE QUESTIONS)
Question 1.
Solve the following inequation and graph the solution set on the number line 2x – 3 < x + 2 ≤ 3x + 5, x ∈ R.
Solution:
2x – 3 < x + 2 ≤ 3x + 5, x ∈ R 2x – 3 < x + 2
⇒ 2x – x < 2 + 3 ⇒ x < 5 … (i)
and x + 2 ≤ 3x + 5 ⇒ 2 – 5 ≤ 3x – x
⇒ – 2 ≤ 2x ⇒ – \(\frac { 3 }{ 2 }\) ≤ x … (ii)
From (i) and (ii)
Solution: \(\left\{\frac{-3}{2} \leq x<5, x \in \mathrm{R}\right\}\)
On number line
Question 2.
Solve the inequation : 12 + 1 \(\frac { 5 }{ 6 }\) x ≤ 5 + 3x, x ∈ R. Represent the solution on a number line.
Solution:
Question 3.
Solve the inequation : – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line.
Solution:
– 3 ≤ 3 – 2x < 9, x ∈ R
– 3 ≤ 3 – 2x
– 3 – 3 ≤ – 2x
⇒ – 6 ≤ – 2x
⇒ 2x ≤ 6 ⇒ x ≤ \(\frac { 6 }{ 2 }\)
⇒ x ≤ 3 … (i)
and 3 – 2x < 9
⇒ – 2x < 9 – 3 ⇒ – 2x < 6 ⇒ x > \(\frac { 6 }{ – 2 }\) ⇒ x > – 3
⇒ – 3 < x … (ii)
From (i) and (ii)
– 3 < x ≤ 3
∴ Solution = {x : – 3 < x ≤ 3}
On the number line
Question 4.
Find the value of x, which satisfies the inequation – 2 ≤ \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ 1\(\frac { 5 }{ 6 }\), x ∈ N.
Graph the solution on the number line.
Solution:
Question 5.
Solve the following inequation, and graph the solution on the number line
(2x – 5 ≤ 5x + 4 < 11, x ∈ R)
Solution:
2x – 5 ≤ 5x + 4 < 11, x ∈ R
Now 2x – 5 ≤ 5x + 4
⇒ – 5 – 4 ≤ 5x – 2x ⇒ – 9 ≤ 3x
⇒ \(\frac { – 9 }{ 3 }\) ≤ x ⇒ – 3 ≤ x … (i)
and 5x + 4 < 11 ⇒ 5x < 11 – 4
⇒ 5x < 7 ⇒ x \(\frac { 7 }{ 5 }\) ⇒ x < 1.4 … (ii)
From (i) and (ii)
Solution = {- 3 < x < 1.4, x ∈ R}
On number line
Question 6.
Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on a number line.
Solution:
2 ≤ 2x – 3 ≤ 5, x ∈ R
Now 2 ≤ 2x – 3 ⇒ 2 + 3 ≤ 2x
⇒ 5 ≤ 2x ⇒ \(\frac { 5 }{ 2 }\) ≤ x ⇒ 2.5 ≤ x … (i)
and 2x – 3 ≤ 5 ⇒ 2x ≤ 5 + 3
⇒ 2x ≤ 8 ⇒ x ≤ \(\frac { 8 }{ 2 }\) ⇒ x ≤ 4 … (ii)
From (i) and (ii)
Solution = {2.5 ≤ x ≤ 4, x ∈ R}
On number line
Question 7.
Given that x ∈ I, solve the inequation and graph the solution on the number line. 3 ≥ \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ≥ 2.
Solution:
3 ≥ \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ≥ 2, x ∈ 1
Now 3 ≥ \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ⇒ 3 ≥ \(\frac{3 x-12+2 x}{6}\)
⇒ 18 ≥ 5x – 12 ⇒ 18 + 12 ≥ 5x
⇒ 30 ≥ 5x ⇒ \(\frac { 30 }{ 5 }\) ≥ x
6 ≥ x ⇒ x ≤ 6 … (i)
and \(\frac { x-4 }{ 2 }\) + \(\frac { x }{ 3 }\) ≥ 2
\(\frac{3 x-12+2 x}{6}\) ≥ 2 ⇒ 5x – 12 ≥ 12
⇒ 5x ≥ 12 + 12 ⇒ 5x ≥ 24
⇒ x ≥ \(\frac { 24 }{ 5 }\) ⇒ x ≥ 4.8
⇒ 4.8 ≤ x … (ii)
From (i) and (ii)
4.8 ≤ x ≤ 6
∵ x ∈ I
∴ Solution set = {5, 6}
On the number line
Question 8.
Give that x ∈ R, solve the following inequality and graph the solution on the number line inequality : – 1 ≤ 3 + 4x < 23.
Solution:
– 1 ≤ 3 + 4x < 23, x ∈ R
Now – 1 ≤ 3 + 4x
⇒ – 1 – 3 ≤ 4x
⇒ – 4 ≤ 4x
⇒ \(\frac { -4 }{ 4 }\) ≤ x ⇒ – 1 ≤ x … (i)
and 3 + 4x < 23 ⇒ 4x < 23 – 3
⇒ 4x < 20 ⇒ x < \(\frac { 20 }{ 4 }\) … (ii)
From (i) and (ii)
Solution : {-1 < x < 5, x ∈ R}
On the number line
Question 9.
Solve the following inequation and graph the solution on the number line :
-2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 2 }{ 3 }\) < 3\(\frac { 1 }{ 3 }\); x ∈ R.
Solution:
-2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 2 }{ 3 }\) < 3\(\frac { 1 }{ 3 }\); x ∈ R.
\(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < \(\frac { 10 }{ 3 }\)
Now \(\frac { -8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) ⇒ \(\frac { -8 }{ 3 }\) – \(\frac { 1 }{ 3 }\) ≤ x
Question 10.
Solve the given inequation and graph the solution on the number line :
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R.
Solution:
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
Now 2y – 3 < y + 1
⇒ 2y – y < 1 + 3 ⇒ y < 4 … (i)
and y + 1 ≤ 4y + 7 ⇒ 1 – 7 ≤ 4y – y
⇒ – 6 ≤ 3 y ⇒ \(\frac { -6 }{ 3 }\) ≤ y
⇒ – 2 ≤ y … (ii)
From (i) and (ii)
Solution = {- 2 ≤ y < 4, y ∈ R}
On the number line
Question 11.
Solve the inequation and represent the solution set on the number line
– 3 + x ≤ \(\frac { 8x }{ 3 }\) + 2 ≤ \(\frac { 14 }{ 3 }\) + 2x, where x ∈ I.
Solution:
Given :
– 3 + x ≤ \(\frac { 8x }{ 3 }\) + 2 ≤ \(\frac { 14 }{ 3 }\) + 2x, where x ∈ I
⇒ – 3 + x ≤ \(\frac { 8x }{ 3 }\) + 2 … (i)
and \(\frac { 8x }{ 3 }\) + 2 ≤ \(\frac { 14 }{ 3 }\) + 2x … (ii)
⇒ – 5 ≤ \(\frac { 5x }{ 3 }\) and \(\frac { 2x }{ 3 }\) ≤ \(\frac { 8x }{ 3 }\)
⇒ x ≥ – 3 and x ≤ 4
∴ – 3 ≤ x ≤ 4
Solution set = {-3, -2, -1, 0, 1,2, 3, 4}
Number line
Question 12.
Solve the following inequation and represent the solution set on the number line.
– 3 < – \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ \(\frac { 5 }{ 6 }\), x ∈ R.
Solution:
– 3 < – \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ \(\frac { 5 }{ 6 }\), x ∈ R.
⇒ – 3 < \(\frac { -3-4x }{ 6 }\) ≤ \(\frac { 5 }{ 6 }\)
⇒ – 18 < – 3 – 4x ≤ 5
⇒ – 18 + 3 < – 4x ≤ 5 + 3
⇒ – 15 < – 4x ≤ 8 ⇒ \(\frac { -15 }{ -4 }\) > x ≥ \(\frac { 8 }{ -4 }\)
⇒ \(\frac { 15 }{ 4 }\) > x ≥ – 2
⇒ 3.75 > x ≥ – 2
Question 13.
Solve the following inequation and represent the solution set on the number line
2x – 5 ≤ 5x + 4 < 11, where x ∈ 1.
Solution:
Question 14.
Solve the following inequation and represent the solution set on the number line:
4x – 19 < \(\frac { 3x }{ 5 }\) – 2 ≤ – \(\frac { 2 }{ 5 }\) + x, x ∈ R.
Solution:
we have 4x – 19 < \(\frac { 3x }{ 5 }\) – 2 ≤ – \(\frac { 2 }{ 5 }\) + x, x ∈ R.
⇒ 4x – 19 < \(\frac { 3x }{ 5 }\) – 2 and \(\frac { 3x }{ 5 }\) – 2 ≤ \(\frac { – 2 }{ 5 }\) + x, x ∈ R.
⇒ 4x – \(\frac { 3x }{ 5 }\) < 17 and – 2 + \(\frac { 2 }{ 5 }\) ≤ x – \(\frac { 3x }{ 5 }\), x ∈ R.
⇒ \(\frac { 20x – 3x }{ 5 }\) < 17 and \(\frac{-10+2}{5} \leq \frac{5 x-3 x}{5}\), x ∈ R.
⇒ \(\frac { 17x }{ 5 }\) < 17 and \(\frac { -8 }{ 5 }\) ≤ \(\frac { 2x }{ 5 }\) + x, x ∈ R.
⇒ x < 5 and – 4 ≤ x, x ∈ R.
⇒ – 4 ≤ x < 5, x ∈ R
Hence, solution set is {x: -4 ≤ x ≤ 5, x ∈ R} The solution set is represented on the number line as below:
Question 15.
Solve the following inequation, write the solution set and represent it on the number line:
\(-\frac{x}{3} \leq \frac{x}{2}-1 \frac{1}{3}<\frac{1}{6}, x \in R .\)
Solution:
Question 16.
Find the values of x, which satisfy the inequation \(-2 \frac{5}{6}<\frac{1}{2}-\frac{2 x}{3} \leq 2\), x ∈ W. Graph the solution set on the number line.
Solution:
(Solution set is shown by dots on the number line).
Question 17.
Solve the following inequation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R Represent the solution set on a real number line.
Solution:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R
13x – 5 < 15x + 4 and 15x + 4 < 7x + 12
13x < 15x + 9 and 15x < 7x + 8
0 < 2x + 9 and 8x < 8
– 9 < 2x and x < 1
– \(\frac { 9 }{ 2 }\) < x and x < 1 9
∴ – \(\frac { 9 }{ 2 }\) < x < 1
i.e. – 4.5 < x < 1
Hence, the solution set is {x : – 4.5 < x < 1, x ∈ R}
The required line is
Question 18.
Solve the following inequation, write the solution set and represent it on the number line. – 3(x – 7) ≥ 15 – 7x > \(\frac { x+1 }{ 3 }\), x ∈ R.
Solution:
– 3(x – 7) ≥ 15 – 7x > \(\frac { x+1 }{ 3 }\)
⇒ – 3(x – 7) ≥ 15 – 7x > and 15 – 7x > \(\frac { x+1 }{ 3 }\)
⇒ – 3x + 21 ≥ 15 – 21 and 45 – 21x > x + 1
⇒ – 3x + 7x ≥ 15 – 21 and 45 – 1 > x + 2x
⇒ 4x ≥ – 6 and 44 > 22x
⇒ x ≥ \(\frac { – 3 }{ 2 }\) and 2 > x
Hence, the solution set is
