In earlier classes, we studied equations in one and two variables. An equation can be defined as a statement involving variables and a sign of equality. Similarly, inequation can be defined as a statement involving variables and a sign of inequality. In Exercise 15.1, we shall discuss problems on solving linear inequations in one variable. Students can understand the concept of inequations by using the solutions prepared by the tutors at BYJU’S. Solutions are provided with precision in order to help students to sort out their difficulties and reduce the fear of exams. The PDF of RD Sharma Class 11 Solutions is provided here, which students can easily download and start practising offline.
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Solve the following linear Inequations in R.
1. Solve: 12x < 50, when
(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N
Solution:
Given:
12x < 50
So when we divide by 12, we get
12x/ 12 < 50/12
x < 25/6
(i) x ∈ R
When x is a real number, the solution of the given inequation is (-∞, 25/6).
(ii) x ∈ Z
When, 4 < 25/6 < 5
So, when x is an integer, the maximum possible value of x is 4.
The solution of the given inequation is {…, –2, –1, 0, 1, 2, 3, 4}.
(iii) x ∈ N
When, 4 < 25/6 < 5
So, when x is a natural number, the maximum possible value of x is 4. We know that the natural numbers start from 1, the solution of the given inequation is {1, 2, 3, 4}.
2. Solve: -4x > 30, when
(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N
Solution:
Given:
-4x > 30
So when we divide by 4, we get
-4x/4 > 30/4
-x > 15/2
x < – 15/2
(i) x ∈ R
When x is a real number, the solution of the given inequation is (-∞, -15/2).
(ii) x ∈ Z
When, -8 < -15/2 < -7
So, when x is an integer, the maximum possible value of x is -8.
The solution of the given inequation is {…, –11, –10, -9, -8}.
(iii) x ∈ N
As natural numbers start from 1 and can never be negative, when x is a natural number, the solution of the given inequation is ∅.
3. Solve: 4x-2 < 8, when
(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N
Solution:
Given:
4x – 2 < 8
4x – 2 + 2 < 8 + 2
4x < 10
So dividing by 4 on both sides, we get,
4x/4 < 10/4
x < 5/2
(i) x ∈ R
When x is a real number, the solution of the given inequation is (-∞, 5/2).
(ii) x ∈ Z
When, 2 < 5/2 < 3
So, when x is an integer, the maximum possible value of x is 2.
The solution of the given inequation is {…, –2, –1, 0, 1, 2}.
(iii) x ∈ N
When, 2 < 5/2 < 3
So, when x is a natural number, the maximum possible value of x is 2. We know that the natural numbers start from 1, the solution of the given inequation is {1, 2}.
4. 3x – 7 > x + 1
Solution:
Given:
3x – 7 > x + 1
3x – 7 + 7 > x + 1 + 7
3x > x + 8
3x – x > x + 8 – x
2x > 8
Dividing both sides by 2, we get
2x/2 > 8/2
x > 4
∴ The solution of the given inequation is (4, ∞).
5. x + 5 > 4x – 10
Solution:
Given: x + 5 > 4x – 10
x + 5 – 5 > 4x – 10 – 5
x > 4x – 15
4x – 15 < x
4x – 15 – x < x – x
3x – 15 < 0
3x – 15 + 15 < 0 + 15
3x < 15
Dividing both sides by 3, we get
3x/3 < 15/3
x < 5
∴ The solution of the given inequation is (-∞, 5).
6. 3x + 9 ≥ –x + 19
Solution:
Given: 3x + 9 ≥ –x + 19
3x + 9 – 9 ≥ –x + 19 – 9
3x ≥ –x + 10
3x + x ≥ –x + 10 + x
4x ≥ 10
Dividing both sides by 4, we get
4x/4 ≥ 10/4
x ≥ 5/2
∴ The solution of the given inequation is [5/2, ∞).
7. 2 (3 – x) ≥ x/5 + 4
Solution:
Given: 2 (3 – x) ≥ x/5 + 4
6 – 2x ≥ x/5 + 4
6 – 2x ≥ (x+20)/5
5(6 – 2x) ≥ (x + 20)
30 – 10x ≥ x + 20
30 – 20 ≥ x + 10x
10 ≥11x
11x ≤ 10
Dividing both sides by 11, we get
11x/11 ≤ 10/11
x ≤ 10/11
∴ The solution of the given inequation is (-∞, 10/11].
8. (3x – 2)/5 ≤ (4x – 3)/2
Solution:
Given:
(3x – 2)/5 ≤ (4x – 3)/2
Multiplying both the sides by 5 we get,
(3x – 2)/5 × 5 ≤ (4x – 3)/2 × 5
(3x – 2) ≤ 5(4x – 3)/2
3x – 2 ≤ (20x – 15)/2
Multiplying both the sides by 2 we get,
(3x – 2) × 2 ≤ (20x – 15)/2 × 2
6x – 4 ≤ 20x – 15
20x – 15 ≥ 6x – 4
20x – 15 + 15 ≥ 6x – 4 + 15
20x ≥ 6x + 11
20x – 6x ≥ 6x + 11 – 6x
14x ≥ 11
Dividing both sides by 14, we get
14x/14 ≥ 11/14
x ≥ 11/14
∴ The solution of the given inequation is [11/14, ∞).
9. –(x – 3) + 4 < 5 – 2x
Solution:
Given: –(x – 3) + 4 < 5 – 2x
–x + 3 + 4 < 5 – 2x
–x + 7 < 5 – 2x
–x + 7 – 7 < 5 – 2x – 7
–x < –2x – 2
–x + 2x < –2x – 2 + 2x
x < –2
∴ The solution of the given inequation is (–∞, –2).
10. x/5 < (3x-2)/4 – (5x-3)/5
Solution:
Given: x/5 < (3x-2)/4 – (5x-3)/5
x/5 < [5(3x-2) – 4(5x-3)]/4(5)
x/5 < [15x – 10 – 20x + 12]/20
x/5 < [2 – 5x]/20
Multiplying both the sides by 20 we get,
x/5 × 20 < [2 – 5x]/20 × 20
4x < 2 – 5x
4x + 5x < 2 – 5x + 5x
9x < 2
Dividing both sides by 9, we get
9x/9 < 2/9
x < 2/9
∴ The solution of the given inequation is (-∞, 2/9).
11. [2(x-1)]/5 ≤ [3(2+x)]/7
Solution:
Given:
[2(x-1)]/5 ≤ [3(2+x)]/7(2x – 2)/5 ≤ (6 + 3x)/7
Multiplying both sides by 5 we get,
(2x – 2)/5 × 5 ≤ (6 + 3x)/7 × 5
2x – 2 ≤ 5(6 + 3x)/7
7 (2x – 2) ≤ 5 (6 + 3x)
14x – 14 ≤ 30 + 15x
14x – 14 + 14 ≤ 30 + 15x + 14
14x ≤ 44 + 15x
14x – 44 ≤ 44 + 15x – 44
14x – 44 ≤ 15x
15x ≥ 14x – 44
15x – 14x ≥ 14x – 44 – 14x
x ≥ –44
∴ The solution of the given inequation is [–44, ∞).
12. 5x/2 + 3x/4 ≥ 39/4
Solution:
Given:
5x/2 + 3x/4 ≥ 39/4
By taking LCM
[2(5x)+3x]/4 ≥ 39/413x/4 ≥ 39/4
Multiplying both sides by 4 we get,
13x/4 × 4 ≥ 39/4 × 4
13x ≥ 39
Divide both sides by 13, we get
13x/13 ≥ 39/13
x ≥ 39/13
x ≥ 3
∴ The solution of the given inequation is [3, ∞).
13. (x – 1)/3 + 4 < (x – 5)/5 – 2
Solution:
Given:
(x – 1)/3 + 4 < (x – 5)/5 – 2
Subtract both sides by 4 we get,
(x – 1)/3 + 4 – 4 < (x – 5)/5 – 2 – 4
(x – 1)/3 < (x – 5)/5 – 6
(x – 1)/3 < (x – 5 – 30)/5
(x – 1)/3 < (x – 35)/5
Cross multiply we get,
5 (x – 1) < 3 (x – 35)
5x – 5 < 3x – 105
5x – 5 + 5 < 3x – 105 + 5
5x < 3x – 100
5x – 3x < 3x – 100 – 3x
2x < –100
Divide both sides by 2, we get
2x/2 < -100/2
x < -50
∴ The solution of the given inequation is (-∞, -50).
14. (2x + 3)/4 – 3 < (x – 4)/3 – 2
Solution:
Given:
(2x + 3)/4 – 3 < (x – 4)/3 – 2
Add 3 on both sides we get,
(2x + 3)/4 – 3 + 3 < (x – 4)/3 – 2 + 3
(2x + 3)/4 < (x – 4)/3 + 1
(2x + 3)/4 < (x – 4 + 3)/3
(2x + 3)/4 < (x – 1)/3
Cross multiplying, we get,
3 (2x + 3) < 4 (x – 1)
6x + 9 < 4x – 4
6x + 9 – 9 < 4x – 4 – 9
6x < 4x – 13
6x – 4x < 4x – 13 – 4x
2x < –13
Dividing both sides by 2, we get
2x/2 < -13/2
x < -13/2
∴ The solution of the given inequation is (-∞, -13/2).
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