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

Last Updated : 28 Apr, 2021

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

Solution:

Given:\frac{5āˆ’2x}{3} <\frac{x}{6} āˆ’ 5

ā‡’ \frac{5āˆ’2x}{3} <\frac{xāˆ’30}{6}

ā‡’ 6(5āˆ’2x) < 3(xāˆ’30)

ā‡’ 30 āˆ’ 12x < 3x āˆ’ 90

ā‡’ 15x > 120

ā‡’ x > 8

Thus, the solution set is (8, āˆž).

Question 16. Solve:\frac{4+2x}{3} ā‰„\frac{x}{2} āˆ’ 3.

Solution:

Given:\frac{4+2x}{3} ā‰„\frac{x}{2} āˆ’ 3.

ā‡’\frac{4+2x}{3} ā‰„\frac{xāˆ’6}{2}

ā‡’ 2(4+2x) ā‰„ 3(xāˆ’60)

ā‡’ 8 + 4x ā‰„ 3x āˆ’ 180

ā‡’ x ā‰„ āˆ’26

Thus, the solution set is [āˆ’26, āˆž).

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

Solution:

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

ā‡’\frac{2x+3āˆ’10}{5} <\frac{3xāˆ’6}{5}

ā‡’ 2x + 3 āˆ’ 10 < 3x āˆ’ 6

ā‡’ x > āˆ’1

Thus, the solution set is (āˆ’1, āˆž).

Question 18. Solve: xāˆ’2 ā‰¤\frac{5x+8}{3}

Solution:

Given: xāˆ’2 ā‰¤\frac{5x+8}{3}

ā‡’ 3(xāˆ’2) ā‰¤ 5x+8

ā‡’ 3x āˆ’ 6 ā‰¤ 5x + 8

ā‡’ 2x ā‰„ āˆ’14

ā‡’ x ā‰„ āˆ’7

Thus, the solution set is [āˆ’7, āˆž).

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

Solution:

Given:\frac{6xāˆ’5}{4x+1} < 0.

Case I: When 6x āˆ’ 5 > 0 and 4x +1 < 0

ā‡’ x > 5/6 and x < āˆ’1/4, which is clearly impossible.

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

ā‡’ x < 5/6 and x > āˆ’1/4

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

Question 20. Solve:\frac{2xāˆ’3}{3xāˆ’7} > 0.

Solution:

Given:\frac{2xāˆ’3}{3xāˆ’7} > 0.

Case I: When 2xāˆ’3 > 0 and 3xāˆ’7 > 0

ā‡’ x > 3/2 and x > 7/3

ā‡’ x > 7/3 ....(a)

Case II: When 2xāˆ’3 < 0 and 3xāˆ’7 < 0

ā‡’ x < 3/2 and x < 7/3

ā‡’ x < 3/2 ....(b)

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

The solution set is (āˆ’ āˆž, 3/2)āˆŖ (7/3, āˆž).

Question 21. Solve:\frac{3}{xāˆ’2} < 1.

Solution:

Given:\frac{3}{xāˆ’2} < 1

ā‡’\frac{3}{xāˆ’2} āˆ’1 < 0

ā‡’\frac{3āˆ’x+2}{xāˆ’2} < 0

ā‡’\frac{xāˆ’5}{xāˆ’2} > 0

Case I: When xāˆ’5 > 0 and xāˆ’2 > 0

ā‡’ x > 5 and x > 2

ā‡’ x > 5 ....(a)

Case II: When xāˆ’5 < 0 and xāˆ’2 < 0

ā‡’ x < 5 and x < 2

ā‡’ x < 2 ....(b)

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

The solution set is (āˆ’ āˆž, 2)āˆŖ (5, āˆž).

Question 22. Solve:\frac{1}{xāˆ’1} ā‰¤ 2.

Solution:

Given:\frac{1}{xāˆ’1} ā‰¤ 2

ā‡’\frac{1}{xāˆ’1} āˆ’ 2 ā‰¤ 0

ā‡’\frac{1āˆ’2x+2}{xāˆ’1} ā‰¤ 0

ā‡’\frac{3āˆ’2x}{xāˆ’1} ā‰¤ 0

Case I: When 3āˆ’2x ā‰„ 0 and xāˆ’1 < 0

ā‡’ x ā‰„ 3/2 and x < 1

ā‡’ x < 1 .....(a)

Case II: 3āˆ’2x ā‰¤ 0 and xāˆ’1 > 0

ā‡’ x ā‰„ 3/2 and x > 1

ā‡’ x ā‰„ 3/2 ....(b)

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

The solution set is (āˆ’ āˆž, 1)āˆŖ (3/2, āˆž).

Question 23. Solve:\frac{4x+3}{2xāˆ’5} < 6

Solution:

Given:\frac{4x+3}{2xāˆ’5} < 6

ā‡’\frac{4x+3}{2xāˆ’5} āˆ’6 < 0

ā‡’\frac{4x+3āˆ’12x+30}{2xāˆ’5} < 0

ā‡’\frac{8xāˆ’33}{2xāˆ’5} < 0

Case I: When 8xāˆ’33 > 0 and 2xāˆ’5 > 0

ā‡’ x > 33/8 and x > 5/2

ā‡’ x > 33/8 ....(a)

Case II: When 8xāˆ’33 < 0 and 2xāˆ’5 < 0

ā‡’ x < 33/8 and x <5/2

ā‡’ x < 5/2 ....(b)

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

The solution set is (āˆ’ āˆž, 5/2)āˆŖ (33/8, āˆž).

Question 24. Solve:\frac{5xāˆ’6}{x+6} < 1.

Solution:

Given:\frac{5xāˆ’6}{x+6} < 1

ā‡’\frac{5xāˆ’6}{x+6} āˆ’ 1 < 0

ā‡’\frac{5xāˆ’6āˆ’xāˆ’6}{x+6} < 0

ā‡’\frac{4xāˆ’12}{x+6} < 0

Case I: When 4xāˆ’12 > 0 and x+6 < 0

ā‡’ x > āˆ’3 and x < āˆ’6, which is clearly not possible.

Case II: When 4xāˆ’12 < 0 and x+6 > 0

ā‡’ x < āˆ’3 and x > āˆ’6

The solution set is (āˆ’ 3, 6).

Question 25. Solve:\frac{5x+8}{4āˆ’x} < 2.

Solution:

Given:\frac{5x+8}{4āˆ’x} < 2

ā‡’\frac{5x+8}{4āˆ’x} āˆ’ 2 < 0

ā‡’\frac{5x+8āˆ’8+2x}{4āˆ’x} < 0

ā‡’\frac{7x}{4āˆ’x} < 0

Case I: When 7x > 0 and 4āˆ’x < 0

ā‡’ x > 0 and x > 4

ā‡’ x > 4 ....(a)

Case II: When 7x < 0 and 4āˆ’x > 0

ā‡’ x < 0 and x > 4

ā‡’ x < 0 ....(b)

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

The solution set is (āˆ’ āˆž, 0)āˆŖ (4, āˆž).

Question 26. Solve:\frac{xāˆ’1}{x+3} > 2.

Solution:

Given:\frac{xāˆ’1}{x+3} > 2.

ā‡’\frac{xāˆ’1}{x+3} āˆ’ 2 > 0

ā‡’\frac{xāˆ’1āˆ’2xāˆ’6}{x+3} > 0

ā‡’\frac{x+7}{x+3} < 0

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

ā‡’ x > āˆ’7 and x < āˆ’3

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

ā‡’ x < āˆ’7 and x > āˆ’3, which is clearly not possible.

The solution set is (āˆ’7, āˆ’3).

Question 27. Solve:\frac{7xāˆ’5}{8x+3} > 4.

Solution:

Given:\frac{7xāˆ’5}{8x+3} > 4

ā‡’\frac{7xāˆ’5}{8x+3} āˆ’ 4 > 0

ā‡’\frac{7xāˆ’5āˆ’32xāˆ’12}{8x+3} > 0

ā‡’\frac{āˆ’25xāˆ’17}{8x+3} > 0

ā‡’\frac{25x+17}{8x+3} < 0

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

ā‡’ x > āˆ’17/25 and x < āˆ’3/8

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

ā‡’ x < āˆ’17/25 and x > āˆ’3/8, which is not clearly possible.

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

Question 28. Solve:\frac{x}{xāˆ’5} > 1/2.

Solution:

Given:\frac{x}{xāˆ’5} > 1/2.

ā‡’\frac{x}{xāˆ’5} āˆ’ 1/2 > 0

ā‡’\frac{x+5}{2xāˆ’10} > 0

Case I: When x+5 > 0 and 2xāˆ’10 > 0

ā‡’ x > āˆ’5 and x > 5

ā‡’ x > 5 ....(a)

Case II: When x+5 < 0 and 2xāˆ’10 < 0

ā‡’ x < āˆ’5 and x < 5

ā‡’ x < āˆ’5 ....(b)

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

The solution set is (āˆ’ āˆž, āˆ’5)āˆŖ (5, āˆž).

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