# Class 11 RD Sharma Solutions – Chapter 15 Linear Inequations – Exercise 15.2 | Set 2

### Question 11. Solve each of the following system of equations in R: 4x – 1 ≤ 0, 3 – 4x < 0

**Solution:**

Let the first equation be 4x – 1 ≤ 0

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⇒ x ≤ 1/4

and the second equation be 3 – 4x < 0

⇒ 4x > 3

⇒ x > 3/4

According to first equation, x lies in range ( -∞, 1/4 ] and according to second equation, x lies in range ( 3/4, -∞ )

Calculating the intersection of these two intervals we get no value for x. Therefore, given set of inequations has no solution.

### Question 12. Solve each of the following system of equations in R: x + 5 > 2(x+1), 2 – x < 3(x+2)

**Solution:**

Let the first equation be x + 5 > 2(x+1)

⇒ x + 5 > 2x + 2

⇒ x < 3

and the second equation be 2 – x < 3(x+2)

⇒ 2 – x < 3x + 6

⇒ -4 < 4x

⇒ x > -1

Hence using above equations, we know x lies in range (-1,3)

### Question 13. Solve each of the following system of equations in R: 2(x – 6) < 3x – 7, 11 – 2x < 6 – x

**Solution:**

Let the first equation be 2(x – 6) < 3x – 7

⇒ 2x – 12 < 3x – 7

⇒ -x < 5

⇒ x > -5

and the second equation be 11 – 2x < 6 – x

⇒ -x < -5

⇒ x > 5

Hence using above equations, we know x lies in range (5,∞)

### Question 14. Solve each of the following system of equations in R: 5x – 7 < 3 (x + 3), 1 – ≥ x – 4

**Solution:**

Let the first equation be 5x – 7 < 3 (x + 3)

⇒ 5x-7 < 3x + 9

⇒ 2x < 16

⇒ x < 8

and the second equation be1 – ≥ x – 4

⇒ x + ≤ 5

⇒ < 5

⇒ \frac{x}{2} ≤ 1

⇒ x ≤ 2

Hence using above equations, we know x lies in range [-∞, 2 ]

### Question 15. Solve each of the following system of equations in R: -2 ≥ – 6, 2(2x + 3) < 6(x – 2) + 10

**Solution:**

Let the first equation be -2 ≥ – 6

⇒ – ≥ – 6 + 2

⇒ ≥ -4

⇒ 6x – 9 -16x ≥ -48

⇒ 10x ≤ 39

⇒ x ≤

and the second equation be 2(2x + 3) < 6(x – 2) + 10

⇒ 4x + 6 < 6x -12 + 10

⇒ -2x < -8

⇒ x > 4

According to first equation, x lies in range ( -∞, 39/10 ] and according to second equation, x lies in range ( 4, ∞ )

Calculating the intersection of these two intervals we get no value for x. Therefore, given set of inequations has no solution.

### Question 16. Solve each of the following system of equations in R: < -3, + 11 < 0

**Solution:**

Let the first equation be < -3

⇒ 7x – 1 < -6

⇒ 7x < -5

⇒ x < -5/7

and the second equation be + 11 < 0

⇒ 3x + 8 < -55

⇒ 3x < -63

⇒ x < -21

Hence using above equations, we know x lies in range ( -∞, -21)

### Question 17. Solve each of the following system of equations in R: > 5, > 2

**Solution:**

Let the first equation be > 5

⇒ 2x + 1 > 5 (7x -1)

⇒ 2x – 35x > -6

⇒ – 33x > -6

⇒ x < 2/11

Also, 7x -1 > 0 ⇒ x > 1/7

using first equation we get, x lies in range

and let the second equation be > 2

⇒ x + 7 > 2x – 16

⇒ 23 > x

⇒ x < 23

Also, x – 8 > 0 ⇒ x > 8

Hence using above equations, we know x lies in range ( 8, -23 )

Calculating the intersection of the two intervals we get after solving equation 1 and equation 2 we get no value for x. Therefore, given set of inequations has no solution.

### Question 18. Solve each of the following system of equations in R: 0 < < 3

**Solution:**

Using the equation, 0 < < 3

⇒ 0 < -x < 6

⇒ 0 > x > -6

⇒ x > -5

Hence using above equation, we know x lies in range ( -6, 0 )

### Question 19. Solve each of the following system of equations in R: 10 ≤ -5 (x – 2) < 20

**Solution:**

Let the first equation be 10 ≤ -5 (x – 2) < 20

⇒ 10 ≤ -5x + 10 < 20

⇒ 0 ≤ -5x < 10

⇒ 0 ≤ x < -2

Hence using above equations, we know x lies in range ( -2, 0 ]

### Question 20. Solve each of the following system of equations in R: -5 < 2x -3 < 5

**Solution:**

Using the equation -5 < 2x -3 < 5

⇒ -2 < 2x < 8

⇒ -1 < x < 4

Hence using above equations, we know x lies in range ( -1, 4 )

### Question 21. Solve each of the following system of equations in R: ≤ 3 ≤ , x>0

**Solution:**

Using the equation ≤ 3 ≤ , x>0

⇒ 4 ≤ 3 (x+1) ≤ 6

⇒ 4 ≤ 3x + 3 ≤ 6

⇒ 4 ≤ 3x + 3 ≤ 6

⇒ 1 ≤ 3x ≤ 3

⇒ 1/3 ≤ x ≤ 1

Hence using above equations, we know x lies in range