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# Class 11 RD Sharma Solutions – Chapter 15 Linear Inequations – Exercise 15.1 | Set 1

• Last Updated : 28 Apr, 2021

### (i) x âˆˆ R

Solution:

Given: 12x < 50

Dividing both sides by 12, we get

12x/ 12 < 50/12

â‡’ x < 25/6

When x is a real number, the solution of the given inequation is (-âˆž, 25/6).

### (ii) x âˆˆ Z

Solution:

Since, 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 {â€¦, â€“2, â€“1, 0, 1, 2, 3, 4}.

### (iii) x âˆˆ N

Solution:

Since 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.

Hence the solution of the given inequation is {1, 2, 3, 4}.

### (i) x âˆˆ R

Solution:

Given: âˆ’ 4x > 30

So when we divide by 4, we get

â‡’ âˆ’ 4x/4 > 30/4

â‡’ âˆ’ x > 15/2

â‡’ x < â€“ 15/2

When x is a real number, the solution of the given inequation is (-âˆž, âˆ’15/2).

### (ii) x âˆˆ Z

Solution:

Since, âˆ’ 8 < âˆ’ 15/2 < âˆ’ 7

So, when x is an integer, the maximum possible value of x is âˆ’ 8.

The solution of the given inequation is {â€¦, â€“11, â€“10, âˆ’ 9, âˆ’8}.

### (iii) x âˆˆ N

Solution:

As natural numbers start from 1 and can never be negative.

Hence when x is a natural number, the solution of the given inequation is âˆ….

### (i) x âˆˆ R

Solution:

Given: 4x â€“ 2 < 8

4x â€“ 2 + 2 < 8 + 2

â‡’ 4x < 10

So dividing by 4 on both sides we get,

4x/4 < 10/4

â‡’ x < 5/2

When x is a real number, the solution of the given inequation is (-âˆž, 5/2).

### (ii) x âˆˆ Z

Solution:

Since, 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 {â€¦, â€“2, â€“1, 0, 1, 2}.

### (iii) x âˆˆ N

Solution:

Since, 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}.

### Question 4. Solve: 3x â€“ 7 > x + 1

Solution:

Given:

3x â€“ 7 > x + 1

â‡’ 3x â€“ 7 + 7 > x + 1 + 7

â‡’ 3x > x + 8

â‡’ 3x â€“ x > x + 8 â€“ x

â‡’ 2x > 8

Dividing both sides by 2, we get

2x/2 > 8/2

â‡’ x > 4

âˆ´ The solution of the given inequation is (4, âˆž).

### Question 5. Solve: x + 5 > 4x â€“ 10

Solution:

Given: x + 5 > 4x â€“ 10

â‡’ x + 5 â€“ 5 > 4x â€“ 10 â€“ 5

â‡’ x > 4x â€“ 15

â‡’ 4x â€“ 15 < x

â‡’ 4x â€“ 15 â€“ x < x â€“ x

â‡’ 3x â€“ 15 < 0

â‡’ 3x â€“ 15 + 15 < 0 + 15

â‡’ 3x < 15

Dividing both sides by 3, we get

3x/3 < 15/3

â‡’ x < 5

âˆ´ The solution of the given inequation is (-âˆž, 5).

### Question 6. Solve: 3x + 9 â‰¥ â€“x + 19

Solution:

Given: 3x + 9 â‰¥ â€“x + 19

â‡’ 3x + 9 â€“ 9 â‰¥ â€“x + 19 â€“ 9

â‡’ 3x â‰¥ â€“x + 10

â‡’ 3x + x â‰¥ â€“x + 10 + x

â‡’ 4x â‰¥ 10

Dividing both sides by 4, we get

4x/4 â‰¥ 10/4

â‡’ x â‰¥ 5/2

âˆ´ The solution of the given inequation is [5/2, âˆž).

### Question 7. Solve: 2 (3 â€“ x) â‰¥ x/5 + 4

Solution:

Given: 2 (3 â€“ x) â‰¥ x/5 + 4

â‡’ 6 â€“ 2x â‰¥ x/5 + 4

â‡’ 6 â€“ 2x â‰¥ (x+20)/5

â‡’ 5(6 â€“ 2x) â‰¥ (x + 20)

â‡’ 30 â€“ 10x â‰¥ x + 20

â‡’ 30 â€“ 20 â‰¥ x + 10x

â‡’ 10 â‰¥11x

â‡’ 11x â‰¤ 10

Dividing both sides by 11, we get

11x/11 â‰¤ 10/11

â‡’ x â‰¤ 10/11

âˆ´ The solution of the given inequation is (-âˆž, 10/11].

### Question 8. Solve:â‰¤

Solution:

Given:â‰¤

Multiplying both the sides by 5 we get,

Ã— 5 â‰¤Ã— 5

â‡’ (3x â€“ 2) â‰¤ 5(4x â€“ 3)/2

â‡’ 3x â€“ 2 â‰¤ (20x â€“ 15)/2

Multiplying both the sides by 2 we get,

(3x â€“ 2) Ã— 2 â‰¤ (20x â€“ 15)/2 Ã— 2

â‡’ 6x â€“ 4 â‰¤ 20x â€“ 15

â‡’ 20x â€“ 15 â‰¥ 6x â€“ 4

â‡’ 20x â€“ 15 + 15 â‰¥ 6x â€“ 4 + 15

â‡’ 20x â‰¥ 6x + 11

â‡’ 20x â€“ 6x â‰¥ 6x + 11 â€“ 6x

â‡’ 14x â‰¥ 11

Dividing both sides by 14, we get

14x/14 â‰¥ 11/14

â‡’ x â‰¥ 11/14

âˆ´ The solution of the given inequation is [11/14, âˆž).

### Question 9. Solve: â€“(x â€“ 3) + 4 < 5 â€“ 2x

Solution:

Given: â€“(x â€“ 3) + 4 < 5 â€“ 2x

â‡’ â€“x + 3 + 4 < 5 â€“ 2x

â‡’ â€“x + 7 < 5 â€“ 2x

â‡’ â€“x + 7 â€“ 7 < 5 â€“ 2x â€“ 7

â‡’ â€“x < â€“2x â€“ 2

â‡’ â€“x + 2x < â€“2x â€“ 2 + 2x

â‡’ x < â€“2

âˆ´ The solution of the given inequation is (â€“âˆž, â€“2).

### Question 10. Solve:<â€“

Solution:

Given:<â€“

â‡’<

â‡’<

â‡’<

Multiplying both the sides by 20 we get,

Ã— 20 <Ã— 20

â‡’ 4x < 2 â€“ 5x

â‡’ 4x + 5x < 2 â€“ 5x + 5x

â‡’ 9x < 2

Divide both sides by 9, we get

9x/9 < 2/9

â‡’ x < 2/9

âˆ´ The solution of the given inequation is (-âˆž, 2/9).

### Question 11. Solve:â‰¤

Solution:

Given:â‰¤

â‡’â‰¤

Multiply both the sides by 5 we get,

Ã— 5 â‰¤Ã— 5

â‡’ 2x â€“ 2 â‰¤

â‡’ 7 (2x â€“ 2) â‰¤ 5 (6 + 3x)

â‡’ 14x â€“ 14 â‰¤ 30 + 15x

â‡’ 14x â€“ 14 + 14 â‰¤ 30 + 15x + 14

â‡’ 14x â‰¤ 44 + 15x

â‡’ 14x â€“ 44 â‰¤ 44 + 15x â€“ 44

â‡’ 14x â€“ 44 â‰¤ 15x

â‡’ 15x â‰¥ 14x â€“ 44

â‡’ 15x â€“ 14x â‰¥ 14x â€“ 44 â€“ 14x

â‡’ x â‰¥ â€“44

âˆ´ The solution of the given inequation is [â€“44, âˆž).

### Question 12. Solve: 5x/2 + 3x/4 â‰¥ 39/4

Solution:

Given: 5x/2 + 3x/4 â‰¥ 39/4

By taking LCM, we get:

â‰¥ 39/4

â‡’ 13x/4 â‰¥ 39/4

Multiplying both the sides by 4 we get,

13x/4 Ã— 4 â‰¥ 39/4 Ã— 4

â‡’ 13x â‰¥ 39

Divide both sides by 13, we get

13x/13 â‰¥ 39/13

â‡’ x â‰¥ 39/13

â‡’ x â‰¥ 3

âˆ´ The solution of the given inequation is [3, âˆž).

### Question 13. Solve:+ 4 <â€“ 2

Solution:

Given:+ 4 <â€“ 2

Subtracting both sides by 4 we get,

â‡’+ 4 â€“ 4 <â€“ 2 â€“ 4

â‡’<â€“ 6

â‡’<

â‡’<

After Cross multiplying, we get,

5 (x â€“ 1) < 3 (x â€“ 35)

â‡’ 5x â€“ 5 < 3x â€“ 105

â‡’ 5x â€“ 5 + 5 < 3x â€“ 105 + 5

â‡’ 5x < 3x â€“ 100

â‡’ 5x â€“ 3x < 3x â€“ 100 â€“ 3x

â‡’ 2x < â€“100

Divide both sides by 2, we get

2x/2 < -100/2

â‡’ x < -50

âˆ´ The solution of the given inequation is (-âˆž, -50).

### Question 14. Solve:â€“ 3 <â€“ 2

Solution:

Given:â€“ 3 <â€“ 2

Adding 3 on both sides we get,

â€“ 3 + 3 <â€“ 2 + 3

â‡’<+ 1

â‡’<

â‡’<

After Cross multiplying, we get,

3(2x + 3) < 4(x â€“ 1)

â‡’ 6x + 9 < 4x â€“ 4

â‡’ 6x + 9 â€“ 9 < 4x â€“ 4 â€“ 9

â‡’ 6x < 4x â€“ 13

â‡’ 6x â€“ 4x < 4x â€“ 13 â€“ 4x

â‡’ 2x < â€“13

Dividing both sides by 2, we get

2x/2 < -13/2

â‡’ x < -13/2

âˆ´ The solution of the given inequation is (-âˆž, -13/2).

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