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

Question 15. Solve:\frac{5−2x}{3} <\frac{x}{6} − 5 in R.

Solution:

Given:\frac{5−2x}{3} <\frac{x}{6} − 5

⇒ \frac{5−2x}{3} <\frac{x−30}{6}

⇒ 6(5−2x) < 3(x−30)

⇒ 30 − 12x < 3x − 90

⇒ 15x > 120

⇒ x > 8

Thus, the solution set is (8, ∞).

Question 16. Solve:\frac{4+2x}{3} ≥\frac{x}{2} − 3.

Solution:

Given:\frac{4+2x}{3} ≥\frac{x}{2} − 3.

⇒\frac{4+2x}{3} ≥\frac{x−6}{2}

⇒ 2(4+2x) ≥ 3(x−60)

⇒ 8 + 4x ≥ 3x − 180

⇒ x ≥ −26

Thus, the solution set is [−26, ∞).

Question 17. Solve:\frac{2x+3}{5} − 2 <\frac{3(x−2)}{5} .

Solution:

Given:\frac{2x+3}{5} − 2 <\frac{3(x-2)}{5}

⇒\frac{2x+3−10}{5} <\frac{3x−6}{5}

⇒ 2x + 3 − 10 < 3x − 6

⇒ x > −1

Thus, the solution set is (−1, ∞).

Question 18. Solve: x−2 ≤\frac{5x+8}{3}

Solution:

Given: x−2 ≤\frac{5x+8}{3}

⇒ 3(x−2) ≤ 5x+8

⇒ 3x − 6 ≤ 5x + 8

⇒ 2x ≥ −14

⇒ x ≥ −7

Thus, the solution set is [−7, ∞).

Question 19. Solve:\frac{6x−5}{4x+1} < 0.

Solution:

Given:\frac{6x−5}{4x+1} < 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:\frac{2x−3}{3x−7} > 0.

Solution:

Given:\frac{2x−3}{3x−7} > 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:\frac{3}{x−2} < 1.

Solution:

Given:\frac{3}{x−2} < 1

⇒\frac{3}{x−2} −1 < 0

⇒\frac{3−x+2}{x−2} < 0

⇒\frac{x−5}{x−2} > 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:\frac{1}{x−1} ≤ 2.

Solution:

Given:\frac{1}{x−1} ≤ 2

⇒\frac{1}{x−1} − 2 ≤ 0

⇒\frac{1−2x+2}{x−1} ≤ 0

⇒\frac{3−2x}{x−1} ≤ 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:\frac{4x+3}{2x−5} < 6

Solution:

Given:\frac{4x+3}{2x−5} < 6

⇒\frac{4x+3}{2x−5} −6 < 0

⇒\frac{4x+3−12x+30}{2x−5} < 0

⇒\frac{8x−33}{2x−5} < 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:\frac{5x−6}{x+6} < 1.

Solution:

Given:\frac{5x−6}{x+6} < 1

⇒\frac{5x−6}{x+6} − 1 < 0

⇒\frac{5x−6−x−6}{x+6} < 0

⇒\frac{4x−12}{x+6} < 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:\frac{5x+8}{4−x} < 2.

Solution:

Given:\frac{5x+8}{4−x} < 2

⇒\frac{5x+8}{4−x} − 2 < 0

⇒\frac{5x+8−8+2x}{4−x} < 0

⇒\frac{7x}{4−x} < 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:\frac{x−1}{x+3} > 2.

Solution:

Given:\frac{x−1}{x+3} > 2.

⇒\frac{x−1}{x+3} − 2 > 0

⇒\frac{x−1−2x−6}{x+3} > 0

⇒\frac{x+7}{x+3} < 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:\frac{7x−5}{8x+3} > 4.

Solution:

Given:\frac{7x−5}{8x+3} > 4

⇒\frac{7x−5}{8x+3} − 4 > 0

⇒\frac{7x−5−32x−12}{8x+3} > 0

⇒\frac{−25x−17}{8x+3} > 0

⇒\frac{25x+17}{8x+3} < 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:\frac{x}{x−5} > 1/2.

Solution:

Given:\frac{x}{x−5} > 1/2.

⇒\frac{x}{x−5} − 1/2 > 0

⇒\frac{x+5}{2x−10} > 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, ∞).