RD Sharma Solutions for Class 11 Maths Chapter 15 Linear Inequations

RD Sharma Solutions Class 11 Maths Chapter 15 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 15 – Linear Inequations are given here for students to study and secure good marks in the final examination. In this chapter, students will study linear inequations in one and two variables. The knowledge of linear inequations is very helpful in solving problems in science, mathematics, engineering, linear programming, etc. Students who aim to score high marks can refer to RD Sharma Class 11 Maths Solutions PDF from the links given below.

Chapter 15 – Linear Inequations contains six exercises, and RD Sharma Solutions provide accurate answers to the questions present in each exercise. Subject experts at BYJU’S have designed these RD Sharma Solutions, updated for the 2023-24 exam, in a very lucid manner that helps students solve problems in the most efficient possible ways. Now, let us have a look at the concepts discussed in this chapter.

• Solutions of an inequation.
• Solving linear inequations in one variable.
• The solution of a system of linear inequations in one variable.
• Some applications of linear inequations in one variable.
• Graphical solution of linear inequations in two variables.
• The solution of simultaneous linear inequations in two variables.

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EXERCISE 15.1 PAGE NO: 15.10

Solve the following linear Inequations in R.

1. Solve: 12x < 50, when

(i) x ∈ R

(ii) x ∈ Z

(iii) x ∈ N

Solution:

Given:

12x < 50

So when we divide by 12, we get

12x/ 12 < 50/12

x < 25/6

(i) x ∈ R

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

(ii) x ∈ Z

When, 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 {…, –2, –1, 0, 1, 2, 3, 4}.

(iii) x ∈ N

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

2. Solve: -4x > 30, when

(i) x ∈ R

(ii) x ∈ Z

(iii) x ∈ N

Solution:

Given:

-4x > 30

So when we divide by 4, we get

-4x/4 > 30/4

-x > 15/2

x < – 15/2

(i) x ∈ R

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

(ii) x ∈ Z

When, -8 < -15/2 < -7

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

The solution of the given inequation is {…, –11, –10, -9, -8}.

(iii) x ∈ N

As natural numbers start from 1 and can never be negative, when x is a natural number, the solution of the given inequation is ∅.

3. Solve: 4x-2 < 8, when

(i) x ∈ R

(ii) x ∈ Z

(iii) x ∈ N

Solution:

Given:

4x – 2 < 8

4x – 2 + 2 < 8 + 2

4x < 10

So divide by 4 on both sides, and we get,

4x/4 < 10/4

x < 5/2

(i) x ∈ R

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

(ii) x ∈ Z

When, 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 {…, –2, –1, 0, 1, 2}.

(iii) x ∈ N

When, 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, and the solution of the given inequation is {1, 2}.

4. 3x – 7 > x + 1

Solution:

Given:

3x – 7 > x + 1

3x – 7 + 7 > x + 1 + 7

3x > x + 8

3x – x > x + 8 – x

2x > 8

Divide both sides by 2, and we get

2x/2 > 8/2

x > 4

∴ The solution to the given inequation is (4, ∞).

5. x + 5 > 4x – 10

Solution:

Given: x + 5 > 4x – 10

x + 5 – 5 > 4x – 10 – 5

x > 4x – 15

4x – 15 < x

4x – 15 – x < x – x

3x – 15 < 0

3x – 15 + 15 < 0 + 15

3x < 15

Divide both sides by 3, and we get

3x/3 < 15/3

x < 5

∴ The solution to the given inequation is (-∞, 5).

6. 3x + 9 ≥ –x + 19

Solution:

Given: 3x + 9 ≥ –x + 19

3x + 9 – 9 ≥ –x + 19 – 9

3x ≥ –x + 10

3x + x ≥ –x + 10 + x

4x ≥ 10

Divide both sides by 4, and we get

4x/4 ≥ 10/4

x ≥ 5/2

∴ The solution to the given inequation is (5/2, ∞).

7. 2 (3 – x) ≥ x/5 + 4

Solution:

Given: 2 (3 – x) ≥ x/5 + 4

6 – 2x ≥ x/5 + 4

6 – 2x ≥ (x+20)/5

5(6 – 2x) ≥ (x + 20)

30 – 10x ≥ x + 20

30 – 20 ≥ x + 10x

10 ≥11x

11x ≤ 10

Divide both sides by 11, and we get

11x/11 ≤ 10/11

x ≤ 10/11

∴ The solution to the given inequation is (-∞, 10/11).

8. (3x – 2)/5 ≤ (4x – 3)/2

Solution:

Given:

(3x – 2)/5 ≤ (4x – 3)/2

Multiply both sides by 5, and we get,

(3x – 2)/5 × 5 ≤ (4x – 3)/2 × 5

(3x – 2) ≤ 5(4x – 3)/2

3x – 2 ≤ (20x – 15)/2

Multiply both sides by 2, and we get,

(3x – 2) × 2 ≤ (20x – 15)/2 × 2

6x – 4 ≤ 20x – 15

20x – 15 ≥ 6x – 4

20x – 15 + 15 ≥ 6x – 4 + 15

20x ≥ 6x + 11

20x – 6x ≥ 6x + 11 – 6x

14x ≥ 11

Divide both sides by 14, and we get

14x/14 ≥ 11/14

x ≥ 11/14

∴ The solution to the given inequation is (11/14, ∞).

9. –(x – 3) + 4 < 5 – 2x

Solution:

Given: –(x – 3) + 4 < 5 – 2x

–x + 3 + 4 < 5 – 2x

–x + 7 < 5 – 2x

–x + 7 – 7 < 5 – 2x – 7

–x < –2x – 2

–x + 2x < –2x – 2 + 2x

x < –2

∴ The solution to the given inequation is (–∞, –2).

10. x/5 < (3x-2)/4 – (5x-3)/5

Solution:

Given: x/5 < (3x-2)/4 – (5x-3)/5

x/5 < [5(3x-2) – 4(5x-3)]/4(5)

x/5 < [15x – 10 – 20x + 12]/20

x/5 < [2 – 5x]/20

Multiply both sides by 20, and we get,

x/5 × 20 < [2 – 5x]/20 × 20

4x < 2 – 5x

4x + 5x < 2 – 5x + 5x

9x < 2

Divide both sides by 9, and we get

9x/9 < 2/9

x < 2/9

∴ The solution to the given inequation is (-∞, 2/9).

11. [2(x-1)]/5 ≤ [3(2+x)]/7

Solution:

Given:

(2x – 2)/5 ≤ (6 + 3x)/7

Multiply both sides by 5, and we get,

(2x – 2)/5 × 5 ≤ (6 + 3x)/7 × 5

2x – 2 ≤ 5(6 + 3x)/7

7 (2x – 2) ≤ 5 (6 + 3x)

14x – 14 ≤ 30 + 15x

14x – 14 + 14 ≤ 30 + 15x + 14

14x ≤ 44 + 15x

14x – 44 ≤ 44 + 15x – 44

14x – 44 ≤ 15x

15x ≥ 14x – 44

15x – 14x ≥ 14x – 44 – 14x

x ≥ –44

∴ The solution to the given inequation is (–44, ∞).

12. 5x/2 + 3x/4 ≥ 39/4

Solution:

Given:

5x/2 + 3x/4 ≥ 39/4

By taking LCM,

13x/4 ≥ 39/4

Multiply both sides by 4, and we get,

13x/4 × 4 ≥ 39/4 × 4

13x ≥ 39

Divide both sides by 13, and we get

13x/13 ≥ 39/13

x ≥ 39/13

x ≥ 3

∴ The solution to the given inequation is (3, ∞).

13. (x – 1)/3 + 4 < (x – 5)/5 – 2

Solution:

Given:

(x – 1)/3 + 4 < (x – 5)/5 – 2

Subtract both sides by 4, and we get,

(x – 1)/3 + 4 – 4 < (x – 5)/5 – 2 – 4

(x – 1)/3 < (x – 5)/5 – 6

(x – 1)/3 < (x – 5 – 30)/5

(x – 1)/3 < (x – 35)/5

Cross multiply, and we get,

5 (x – 1) < 3 (x – 35)

5x – 5 < 3x – 105

5x – 5 + 5 < 3x – 105 + 5

5x < 3x – 100

5x – 3x < 3x – 100 – 3x

2x < –100

Divide both sides by 2, and we get

2x/2 < -100/2

x < -50

∴ The solution to the given inequation is (-∞, -50).

14. (2x + 3)/4 – 3 < (x – 4)/3 – 2

Solution:

Given:

(2x + 3)/4 – 3 < (x – 4)/3 – 2

Add 3 on both sides, and we get,

(2x + 3)/4 – 3 + 3 < (x – 4)/3 – 2 + 3

(2x + 3)/4 < (x – 4)/3 + 1

(2x + 3)/4 < (x – 4 + 3)/3

(2x + 3)/4 < (x – 1)/3

Cross multiply, and we get,

3 (2x + 3) < 4 (x – 1)

6x + 9 < 4x – 4

6x + 9 – 9 < 4x – 4 – 9

6x < 4x – 13

6x – 4x < 4x – 13 – 4x

2x < –13

Divide both sides by 2, and we get

2x/2 < -13/2

x < -13/2

∴ The solution of the given inequation is (-∞, -13/2).

EXERCISE 15.2 PAGE NO: 15.15

Solve each of the following systems of equations in R.

1. x + 3 > 0, 2x < 14

Solution:

Given: x + 3 > 0 and 2x < 14

Let us consider the first inequality.

x + 3 > 0

x + 3 – 3 > 0 – 3

x > –3

Now, let us consider the second inequality.

2x < 14

Divide both sides by 2, and we get,

2x/2 < 14/2

x < 7

∴ The solution of the given system of inequation is (–3, 7).

2. 2x – 7 > 5 – x, 11 – 5x ≤ 1

Solution:

Given:

2x – 7 > 5 – x and 11 – 5x ≤ 1

Let us consider the first inequality.

2x – 7 > 5 – x

2x – 7 + 7 > 5 – x + 7

2x > 12 – x

2x + x > 12 – x + x

3x > 12

Divide both sides by 3, and we get,

3x/3 > 12/3

x > 4

∴ x ∈ ( 4, ∞) … (1)

Now, let us consider the second inequality.

11 – 5x ≤ 1

11 – 5x – 11 ≤ 1 – 11

–5x ≤ –10

Divide both sides by 5, and we get,

-5x/5 ≤ -10/5

–x ≤ –2

x ≥ 2

∴ x ∈ (2, ∞) … (2)

From (1) and (2), we get

x ∈ (4, ∞) ∩ (2, ∞)

x ∈ (4, ∞)

∴ The solution to the given system of inequations is (4, ∞).

3. x – 2 > 0, 3x < 18

Solution:

Given:

x – 2 > 0 and 3x < 18

Let us consider the first inequality.

x – 2 > 0

x – 2 + 2 > 0 + 2

x > 2

∴ x ∈ ( 2, ∞) …(1)

Now, let us consider the second inequality.

3x < 18

Divide both sides by 3, and we get,

3x/3 < 18/3

x < 6

∴ x ∈ (–∞, 6) …(2)

From (1) and (2), we get

x ∈ (2, ∞) ∩ (–∞, 6)

x ∈ ( 2, 6)

∴ The solution to the given system of inequations is (2, 6).

4. 2x + 6 ≥ 0, 4x – 7 < 0

Solution:

Given:

2x + 6 ≥ 0 and 4x – 7 < 0

Let us consider the first inequality.

2x + 6 ≥ 0

2x + 6 – 6 ≥ 0 – 6

2x ≥ –6

Divide both sides by 2, and we get,

2x/2 ≥ -6/2

x ≥ -3

∴ x ∈ (–3, ∞) …(1)

Now, let us consider the second inequality.

4x – 7 < 0

4x – 7 + 7 < 0 + 7

4x < 7

Divide both sides by 4, and we get,

4x/4 < 7/4

x < 7/4

∴ x ∈ (–∞, 7/4) …(2)

From (1) and (2), we get

x ∈ (-3, ∞) ∩ (–∞, 7/4)

x ∈ (-3, 7/4)

∴ The solution to the given system of inequations is (-3, 7/4).

5. 3x – 6 > 0, 2x – 5 > 0

Solution:

Given:

3x – 6 > 0 and 2x – 5 > 0

Let us consider the first inequality.

3x – 6 > 0

3x – 6 + 6 > 0 + 6

3x > 6

Divide both sides by 3, and we get,

3x/3 > 6/3

x > 2

∴ x ∈ ( 2, ∞)… (1)

Now, let us consider the second inequality.

2x – 5 > 0

2x – 5 + 5 > 0 + 5

2x > 5

Divide both sides by 2, and we get,

2x/2 > 5/2

x > 5/2

∴ x ∈ (5/2, ∞)… (2)

From (1) and (2), we get

x ∈ (2, ∞) ∩ (5/2, ∞)

x ∈ (5/2, ∞)

∴ The solution to the given system of inequations is (5/2, ∞).

6. 2x – 3 < 7, 2x > –4

Solution:

Given:

2x – 3 < 7 and 2x > –4

Let us consider the first inequality.

2x – 3 < 7

2x – 3 + 3 < 7 + 3

2x < 10

Divide both sides by 2, and we get,

2x/2 < 10/2

x < 5

∴ x ∈ ( –∞, 5)… (1)

Now, let us consider the second inequality.

2x > –4

Divide both sides by 2, and we get,

2x/2 > -4/2

x > –2

∴ x ∈ (–2, ∞)… (2)

From (1) and (2), we get

x ∈ (–∞, 5) ∩ (–2, ∞)

x ∈ (–2, 5)

∴ The solution to the given system of inequations is (–2, 5).

7. 2x + 5 ≤ 0, x – 3 ≤ 0

Solution:

Given:

2x + 5 ≤ 0 and x – 3 ≤ 0

Let us consider the first inequality.

2x + 5 ≤ 0

2x + 5 – 5 ≤ 0 – 5

2x ≤ –5

Divide both sides by 2, and we get,

2x/2 ≤ –5/2

x ≤ – 5/2

∴ x ∈ (–∞, -5/2)… (1)

Now, let us consider the second inequality.

x – 3 ≤ 0

x – 3 + 3 ≤ 0 + 3

x ≤ 3

∴ x ∈ (–∞, 3)… (2)

From (1) and (2), we get

x ∈ (–∞, -5/2) ∩ (–∞, 3)

x ∈ (–∞, -5/2)

∴ The solution to the given system of inequations is (–∞, -5/2).

8. 5x – 1 < 24, 5x + 1 > –24

Solution:

Given:

5x – 1 < 24 and 5x + 1 > –24

Let us consider the first inequality.

5x – 1 < 24

5x – 1 + 1 < 24 + 1

5x < 25

Divide both sides by 5, and we get,

5x/5 < 25/5

x < 5

∴ x ∈ (–∞, 5)… (1)

Now, let us consider the second inequality.

5x + 1 > –24

5x + 1 – 1 > –24 – 1

5x > –25

Divide both sides by 5, and we get,

5x/5 > -25/5

x > -5

∴ x ∈ (–5, ∞)… (2)

From (1) and (2), we get

x ∈ (–∞, 5) ∩ (–5, ∞)

x ∈ (–5, 5)

∴ The solution of the given system of inequations is (–5, 5).

9. 3x – 1 ≥ 5, x + 2 > -1

Solution:

Given:

3x – 1 ≥ 5 and x + 2 > –1

Let us consider the first inequality.

3x – 1 ≥ 5

3x – 1 + 1 ≥ 5 + 1

3x ≥ 6

Divide both sides by 3, and we get,

3x/3 ≥ 6/3

x ≥ 2

∴ x ∈ (2, ∞)… (1)

Now, let us consider the second inequality.

x + 2 > –1

x + 2 – 2 > –1 – 2

x > –3

∴ x ∈ (–3, ∞)… (2)

From (1) and (2), we get

x ∈ (2, ∞) ∩ (–3, ∞)

x ∈ (2, ∞)

∴ The solution to the given system of inequations is (2, ∞).

10. 11 – 5x > –4, 4x + 13 ≤ –11

Solution:

Given:

11 – 5x > –4 and 4x + 13 ≤ –11

Let us consider the first inequality.

11 – 5x > –4

11 – 5x – 11 > –4 – 11

–5x > –15

Divide both sides by 5, and we get,

-5x/5 > -15/5

–x > –3

x < 3

∴ x ∈ (–∞, 3) (1)

Now, let us consider the second inequality.

4x + 13 ≤ –11

4x + 13 – 13 ≤ –11 – 13

4x ≤ –24

Divide both sides by 4, and we get,

4x/4 ≤ –24/4

x ≤ –6

∴ x ∈ (–∞, –6) (2)

From (1) and (2), we get

x ∈ (–∞, 3) ∩ (–∞, –6)

x ∈ (–∞, –6)

∴ The solution to the given system of inequations is (–∞, –6).

EXERCISE 15.3 PAGE NO: 15.22

Solve each of the following systems of equations in R.

1. |x + 1/3| > 8/3

Solution:

Let ‘r’ be a positive real number and ‘a’ be a fixed real number. Then,

|x + a| > r ⟺ x > r – a or x < – (a + r)

Here, a = 1/3 and r = 8/3

x > 8/3 – 1/3 or x < – (8/3 + 1/3)

x > (8-1)/3 or x < – (8+1)/3

x > 7/3 or x < – 9/3

x > 7/3 or x < – 3

x ∈ (7/3, ∞) or x ∈ (–∞, -3)

∴ x ∈ (–∞, -3) ∪ (7/3, ∞)

2. |4 – x| + 1 < 3

Solution:

|4 – x | + 1 < 3

Let us subtract 1 from both sides, and we get

|4 – x| + 1 – 1 < 3 – 1

|4 – x| < 2

Let ‘r’ be a positive real number and ‘a’ be a fixed real number. Then,

|a – x| < r ⟺ a – r < x < a + r

Here, a=4 and r=2

4 – 2 < x < 4 + 2

2 < x < 6

∴ x ∈ (2, 6)

3. | (3x – 4)/2 | ≤ 5/12

Solution:

Given:

| (3x – 4)/2 | ≤ 5/12

We can rewrite it as

| 3x/2 – 4/2 | ≤ 5/12

| 3x/2 – 2 | ≤ 5/12

Let ‘r’ be a positive real number and ‘a’ be a fixed real number. Then,

|x – a| ≤ r ⟺ a – r ≤ x ≤ a + r

Here, a = 2 and r = 5/12

2 – 5/12 ≤ 3x/2 ≤ 2 + 5/12

(24-5)/12 ≤ 3x/2 ≤ (24+5)/12

19/12 ≤ 3x/2 ≤ 29/12

Now, multiplying the whole inequality by 2 and dividing by 3, we get

19/18 ≤ x ≤ 29/18

∴ x ∈ [19/18, 29/18]

4. |x – 2| / (x – 2) > 0

Solution:

Given:

|x – 2| / (x – 2) > 0

Clearly, it states x≠2, so two cases arise:

Case1: x–2>0

x>2

In this case, |x–2| = x – 2

x ϵ (2, ∞)….(1)

Case 2: x–2<0

x<2

In this case, |x–2|= – (x–2)

– (x – 2) / (x – 2) > 0

– 1 > 0

Inequality doesn’t get satisfied.

This case gets nullified.

∴ x ∈ (2, ∞) from (1)

5. 1 / (|x| – 3) < ½

Solution:

We know that if we take the reciprocal of any inequality, we need to change the inequality as well.

Also, |x| – 3 ≠ 0

|x| > 3 or |x| < 3

For |x| < 3

–3 < x < 3

x ∈ (–3, 3) …. (1)

The equation can be rewritten as

|x| – 3 > 2

Let us add 3 on both sides, and we get

|x| – 3 + 3 > 2 + 3

|x| > 5

Let ‘a’ be a fixed real number. Then,

|x | > a ⟺ x < –a or x > a

Here, a = 5

x < –5 or x > 5 …. (2)

From (1) and (2)

x ∈ (–∞,–5 ) or x ∈ (5, ∞)

∴ x ∈ (–∞,–5 ) ⋃ (–3, 3) ⋃ (5, ∞)

6. (|x + 2| – x) / x < 2

Solution:

Given:

(|x + 2| – x) / x < 2

Let us rewrite the equation as

|x + 2|/x – x/x < 2

|x + 2|/x – 1 < 2

By adding 1 on both sides, we get

|x + 2|/x – 1 + 1 < 2 + 1

|x + 2|/x < 3

By subtracting 3 on both sides, we get

|x + 2|/x – 3 < 3 – 3

|x + 2|/x – 3 < 0

Clearly, it states x ≠ -2, so two cases arise:

Case1: x + 2 > 0

x > –2

In this case |x+2| = x + 2

x + 2/x – 3 < 0

(x + 2 – 3x)/x < 0

– (2x – 2)/x < 0

(2x – 2)/x < 0

Let us consider only the numerators, we get

2x – 2 > 0

x>1

x ϵ (1, ∞) ….(1)

Case 2: x + 2 < 0

x < –2

In this case, |x+2| = – (x + 2)

-(x+2)/x – 3 < 0

(-x – 2 – 3x)/x < 0

– (4x + 2)/x < 0

(4x + 2)/x < 0

Let us consider only the numerators, and we get

4x + 2 > 0

x > – ½

But x < -2

From the denominator, we have,

x ∈ (–∞ , 0) …(2)

From (1) and (2)

∴ x ∈ (–∞ , 0) ⋃ (1, ∞)

EXERCISE 15.4 PAGE NO: 15.24

1. Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.

Solution:

Let ‘x’ be the smaller of the two consecutive odd positive integers. Then the other odd integer is x + 2.

Given:

Both the integers are smaller than 10, and their sum is more than 11.

So,

x + 2 < 10 and x + (x + 2) > 11

x < 10 – 2 and 2x + 2 > 11

x < 8 and 2x > 11 – 2

x < 8 and 2x > 9

x < 8 and x > 9/2

9/2 < x < 8

Note the odd positive integers lying between 4.5 and 8.

x = 5, 7 [Since, x is an odd integer]

x < 10 [From the given statement]

9/2 < x < 10

Note that the upper limit here has shifted from 8 to 10. Now, x is the odd integer from 4.5 to 10.

So, the odd integers from 4.5 to 10 are 5, 7 and 9.

Now, let us find pairs of consecutive odd integers.

Let x = 5, then (x + 2) = (5 + 2) = 7.

Let x = 7, then (x + 2) = (7 + 2) = 9.

Let x = 9, then (x + 2) = (9 + 2) = 11. But 11 is greater than 10.

∴ The required pairs of odd integers are (5, 7) and (7, 9)

2. Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.

Solution:

Let ‘x’ be the smaller of the two consecutive odd natural numbers. Then the other odd number is x + 2.

Given:

Both the natural numbers are greater than 10, and their sum is less than 40.

So,

x > 10 and x + x + 2 <40

x > 10 and 2x < 38

x > 10 and x < 38/2

x > 10 and x < 19

10 < x < 19

From this inequality, we can say that x lies between 10 and 19.

So, the odd natural numbers lying between 10 and 19 are 11, 13, 15 and 17. (Excluding 19 as x < 19)

Now, let us find pairs of consecutive odd natural numbers.

Let x = 11, then (x + 2) = (11 + 2) = 13

Let x = 13, then (x + 2) = (13 + 2) = 15

Let x = 15, then (x + 2) = (15 + 2) = 17

Let x = 17, then (x + 2) = (17 + 2) = 19.

x = 11, 13, 15, 17 [Since x is an odd number]

∴ The required pairs of odd natural numbers are (11, 13), (13, 15), (15, 17) and (17, 19)

3. Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.

Solution:

Let ‘x’ be the smaller of the two consecutive even positive integers. Then the other even integer is x + 2.

Given:

Both the even integers are greater than 5, and their sum is less than 23.

So,

x > 5 and x + x + 2 < 23

x > 5 and 2x < 21

x > 5 and x < 21/2

5 < x < 21/2

5 < x < 10.5

From this inequality, we can say that x lies between 5 and 10.5.

So, the even positive integers lying between 5 and 10.5 are 6, 8 and 10.

Now, let us find pairs of consecutive even positive integers.

Let x = 6, then (x + 2) = (6 + 2) = 8

Let x = 8, then (x + 2) = (8 + 2) = 10

Let x = 10, then (x + 2) = (10 + 2) = 12.

x = 6, 8, 10 [Since, x is even integer]

∴ The required pairs of even positive integers are (6, 8), (8, 10) and (10, 12)

4. The marks scored by Rohit in the two tests were 65 and 70. Find the minimum marks he should score in the third test to have an average of at least 65 marks.

Solution:

Given:

Marks scored by Rohit in two tests were 65 and 70.

Let marks in the third test be x.

So let us find a minimum x for which the average of all three papers would be at least 65 marks.

That is,

Average marks in three papers ≥ 65 …(i)

Average is given by

Average = (sum of all numbers)/(Total number of items)

= (marks in 1^st two papers + marks in third test)/3

= (65 + 70 + x)/3

= (135 + x)/3

Substituting this value of average in the inequality (i), we get

(135 + x)/3 ≥ 65

(135 + x) ≥ 65 × 3

(135 + x) ≥ 195

x ≥ 195 – 135

x ≥ 60

This inequality means that Rohit should score at least 60 marks on his third test to have an average of at least 65 marks.

So, the minimum mark to get an average of 65 marks is 60.

∴ The minimum mark required in the third test is 60.

5. A solution is to be kept between 86^o and 95^oF. What is the range of temperature in degree Celsius, if the Celsius (C)/Fahrenheit (F) conversion formula is given by F = 9/5C + 32?

Solution:

Let us consider F₁ = 86^o F

And F₂ = 95^o

We know, F = 9/5C + 32

F₁ = 9/5 C₁ + 32

F₁ – 32 = 9/5 C₁

C₁ = 5/9 (F₁ – 32)

= 5/9 (86 – 32)

= 5/9 (54)

= 5 × 6

= 30^o C

Now,

F₂ = 9/5 C₂ + 32

F₂ – 32 = 9/5 C₂

C₂ = 5/9 (F₂ – 32)

= 5/9 (95 – 32)

= 5/9 (63)

= 5 × 7

= 35^o C

∴ The range of temperature of the solution in degree Celsius is 30° C and 35° C.

6. A solution is to be kept between 30^oC and 35^oC. What is the range of temperature in degree Fahrenheit?

Solution:

Let us consider C₁ = 30^o C

And C₂ = 35^o

We know, F = 9/5C + 32

F₁ = 9/5 C₁ + 32

= 9/5 × 30 + 32

= 9 × 6 + 32

= 54 + 32

= 86^o F

Now,

F₂ = 9/5 C₂ + 32

= 9/5 × 35 + 32

= 9 × 7 + 32

= 63 + 32

= 95^o F

∴ The range of temperature of the solution in degree Fahrenheit is 86° F and 95° F.

EXERCISE 15.5 PAGE NO: 15.28

Represent the solution set of the following inequations graphically in a two-dimensional plane.

1. x + 2y – 4 ≤ 0

Solution:

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y – 4 ≤ 0

So when,

2. x + 2y ≥ 6

Solution:

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y ≥ 6

So when,

3. x + 2 ≥ 0

Solution:

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2 ≥ 0

x ≥ -2

As there is only one variable, ‘x,’ and y = 0, which means that x has only one value when considered as an equation.

4. x – 2y < 0

Solution:

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x – 2y < 0

x < 2y

So when,

5. – 3x + 2y ≤ 6

Solution:

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

– 3x + 2y ≤ 6

So when,

EXERCISE 15.6 PAGE NO: 15.30

1. Solve the following systems of linear inequations graphically.
(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0

(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0

(iii) x – y ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0

(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0

(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

Solution:

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 6

So when,

3x + 2y ≤ 6

So when,

x ≥ 0, y ≥ 0

(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 6

So when,

x + 4y ≤ 4

So when,

x ≥ 0, y ≥ 0

(iii) x – y ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x – y ≤ 1

So when,

x + 2y≤ 8

So when,

2x + y ≥ 2

So when,

x ≥ 0, y ≥ 0

(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + y ≥ 1

So when,

7x + 9y ≤ 63

So when,

x ≤ 6, y ≤ 5 and x ≥ 0, y ≥ 0

(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 35

So when,

y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

2. Show that the solution set of the following linear inequations is an empty set:
(i) x – 2y ≥ 0, 2x – y ≤ –2, x ≥ 0, y ≥ 0

(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

Solution:

(i) x – 2y ≥ 0, 2x – y ≤ –2, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x – 2y ≥ 0

So when,

2x – y ≤ –2

So when,

x ≥ 0, y ≥ 0

The lines do not intersect each other for x ≥ 0, y ≥ 0

Hence, there is no solution for the given inequations.

(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality.

You can choose any value but find the two mandatory values, which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y ≤ 3

So when,

3x + 4y ≥ 12

So when,

y ≥ 1, x ≥ 0, y ≥ 0

3. Find the linear inequations for which the shaded area in Fig. 15.41 is the solution set. Draw the diagram of the solution set of the linear inequations.

Solution:

Here, we shall apply the concept of a common solution area to find the signs of inequality by using their given equations and the given common solution area (shaded part).

We know that,

If a line is in the form ax + by = c and c is a positive constant. (In case of negative c, the rule becomes opposite), so there are two cases which are,

If a line is above the origin:

(i) If the shaded area is below the line, then ax + by < c

(ii) If the shaded area is above the line, then ax + by > c

If a line is below the origin, then the rule becomes the opposite.

So, according to the rules,

4. Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42.

Solution:

Here, we shall apply the concept of a common solution area to find the signs of inequality by using their given equations and the given common solution area (shaded part).

We know that,

If a line is in the form ax + by = c and c is a positive constant.

So, according to the rules,

Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 15 – Linear Inequations

Exercise 15.1 Solutions

Exercise 15.2 Solutions

Exercise 15.3 Solutions

Exercise 15.4 Solutions

Exercise 15.5 Solutions

Exercise 15.6 Solutions