Class 11 RD Sharma Solutions – Chapter 15 Linear Inequations – Exercise 15.1 | Set 2

Question 15. Solve:<− 5 in R.

Solution:

Given:<− 5

⇒ <

⇒ 6(5−2x) < 3(x−30)

⇒ 30 − 12x < 3x − 90

⇒ 15x > 120

⇒ x > 8

Thus, the solution set is (8, ∞).

Question 16. Solve:≥− 3.

Solution:

Given:≥− 3.

⇒≥

⇒ 2(4+2x) ≥ 3(x−60)

⇒ 8 + 4x ≥ 3x − 180

⇒ x ≥ −26

Thus, the solution set is [−26, ∞).

Question 17. Solve:− 2 <.

Solution:

Given:− 2 <

⇒<

⇒ 2x + 3 − 10 < 3x − 6

⇒ x > −1

Thus, the solution set is (−1, ∞).

Question 18. Solve: x−2 ≤

Solution:

Given: x−2 ≤

⇒ 3(x−2) ≤ 5x+8

⇒ 3x − 6 ≤ 5x + 8

⇒ 2x ≥ −14

⇒ x ≥ −7

Thus, the solution set is [−7, ∞).

Question 19. Solve:< 0.

Solution:

Given:< 0.

Case I: When 6x − 5 > 0 and 4x +1 < 0

⇒ x > 5/6 and x < −1/4, which is clearly impossible.

Case II: When 6x − 5 < 0 and 4x +1 > 0

⇒ x < 5/6 and x > −1/4

Thus, the solution set is (−1/4, 5/6).

Question 20. Solve:> 0.

Solution:

Given:> 0.

Case I: When 2x−3 > 0 and 3x−7 > 0

⇒ x > 3/2 and x > 7/3

⇒ x > 7/3 ….(a)

Case II: When 2x−3 < 0 and 3x−7 < 0

⇒ x < 3/2 and x < 7/3

⇒ x < 3/2 ….(b)

From (a) and (b), we get:

The solution set is (− ∞, 3/2)∪ (7/3, ∞).

Question 21. Solve:< 1.

Solution:

Given:< 1

⇒−1 < 0

⇒< 0

⇒> 0

Case I: When x−5 > 0 and x−2 > 0

⇒ x > 5 and x > 2

⇒ x > 5 ….(a)

Case II: When x−5 < 0 and x−2 < 0

⇒ x < 5 and x < 2

⇒ x < 2 ….(b)

From (a) and (b), we get:

The solution set is (− ∞, 2)∪ (5, ∞).

Question 22. Solve:≤ 2.

Solution:

Given:≤ 2

⇒− 2 ≤ 0

⇒≤ 0

⇒≤ 0

Case I: When 3−2x ≥ 0 and x−1 < 0

⇒ x ≥ 3/2 and x < 1

⇒ x < 1 …..(a)

Case II: 3−2x ≤ 0 and x−1 > 0

⇒ x ≥ 3/2 and x > 1

⇒ x ≥ 3/2 ….(b)

From (a) and (b), we get:

The solution set is (− ∞, 1)∪ (3/2, ∞).

Question 23. Solve:< 6

Solution:

Given:< 6

⇒−6 < 0

⇒< 0

⇒< 0

Case I: When 8x−33 > 0 and 2x−5 > 0

⇒ x > 33/8 and x > 5/2

⇒ x > 33/8 ….(a)

Case II: When 8x−33 < 0 and 2x−5 < 0

⇒ x < 33/8 and x <5/2

⇒ x < 5/2 ….(b)

From (a) and (b), we get:

The solution set is (− ∞, 5/2)∪ (33/8, ∞).

Question 24. Solve:< 1.

Solution:

Given:< 1

⇒− 1 < 0

⇒< 0

⇒< 0

Case I: When 4x−12 > 0 and x+6 < 0

⇒ x > −3 and x < −6, which is clearly not possible.

Case II: When 4x−12 < 0 and x+6 > 0

⇒ x < −3 and x > −6

The solution set is (− 3, 6).

Question 25. Solve:< 2.

Solution:

Given:< 2

⇒− 2 < 0

⇒< 0

⇒< 0

Case I: When 7x > 0 and 4−x < 0

⇒ x > 0 and x > 4

⇒ x > 4 ….(a)

Case II: When 7x < 0 and 4−x > 0

⇒ x < 0 and x > 4

⇒ x < 0 ….(b)

From (a) and (b), we get:

The solution set is (− ∞, 0)∪ (4, ∞).

Question 26. Solve:> 2.

Solution:

Given:> 2.

⇒− 2 > 0

⇒> 0

⇒< 0

Case I: When x+7 > 0 and x+3 < 0

⇒ x > −7 and x < −3

Case II: When x+7 < 0 and x+3 > 0

⇒ x < −7 and x > −3, which is clearly not possible.

The solution set is (−7, −3).

Question 27. Solve:> 4.

Solution:

Given:> 4

⇒− 4 > 0

⇒> 0

⇒> 0

⇒< 0

Case I: When 25x+17 > 0 and 8x+3 < 0

⇒ x > −17/25 and x < −3/8

Case II: When 25x+17 < 0 and 8x+3 > 0

⇒ x < −17/25 and x > −3/8, which is not clearly possible.

Hence the solution set is (−17/25, −3/8).

Question 28. Solve:> 1/2.

Solution:

Given:> 1/2.

⇒− 1/2 > 0

⇒> 0

Case I: When x+5 > 0 and 2x−10 > 0

⇒ x > −5 and x > 5

⇒ x > 5 ….(a)

Case II: When x+5 < 0 and 2x−10 < 0

⇒ x < −5 and x < 5

⇒ x < −5 ….(b)

From (a) and (b), we get:

The solution set is (− ∞, −5)∪ (5, ∞).