RBSE Solutions for Class 8 Maths Chapter 11 Linear Equations with One Variable Ex 11.1 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Exercise 11.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Linear Equations with One Variable |

Exercise |
Exercise 11.1 |

Number of Questions |
10 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Ex 11.1

**Solve the following equations**

Question 1.

6x + 3 = 4x + 11

Solution

6x + 3 = 4x + 11

⇒ 6x – 4x = 11 – 3

| On transposing 4x and 3

⇒ 2x = 8

\(\frac { 2x }{ 2 } =\frac { 8 }{ 2 } \)

| On dividing(RBSESolutions.com)by 2 on both sides

⇒ x = 4

Question 2.

3(x + 5) = 4x + 9

Solution

3(x + 5) = 4x + 9

⇒ 3x + 15 = 4x + 9

⇒ 3x – 4x = 9 – 15

| On transposing 4x and 15

⇒ – x = – 6

\(\frac { -x }{ -1 } =-\frac { 6 }{ 1 } \)

| On dividing(RBSESolutions.com)by – 1 on both sides

⇒ x = 6

Question 3.

3x + 2(x + 3) = 21

Solution

3x + 2(x + 3) = 21

⇒ 3x + 2x + 6 = 21

⇒ 5x + 6 = 21

⇒ 5x = 21 – 6

| On transposing 6

⇒ 5x = 15

\(\frac { 5x }{ 5 } =\frac { 15 }{ 5 } \)

| On dividing by 5 on both sides

⇒ x = 3

Question 4.

\(\frac { x+1 }{ 2 } +\frac { x+2 }{ 3 } =\frac { 2x-5 }{ 7 } +9\)

Solution.

\(\frac { x+1 }{ 2 } +\frac { x+2 }{ 3 } =\frac { 2x-5 }{ 7 } +9\)

21(x + 1) + 14 (x + 2) = 6(2x – 5) + 9 x 42

Multiplying by 42 as LCM of 2, 3 and 7 is 42

⇒ 21x + 21 + 14x + 28 = 12x – 30 + 378

⇒ 21x + 14x + 21 + 28 = 12x – 30 + 378

⇒ 35x + 49 = 12x + 348

⇒ 35x – 12x = 348 – 49

| On transposing 12x and 49

⇒ 23x = 299

⇒ \(\frac { 23x }{ 23 } =\frac { 299 }{ 23 } \)

| On dividing(RBSESolutions.com)by 23 on both sides

⇒ x = 13

Question 5.

\(\frac { 3x-2 }{ 5 } =4-\left( \frac { x+2 }{ 3 } \right) \)

Solution

\(\frac { 3x-2 }{ 5 } =4-\left( \frac { x+2 }{ 3 } \right) \)

⇒ 3(3x – 2) = 15 × 4 – 5 (x + 2)

Multiplying by 15 as LCM of 5 and 3 is 15

⇒ 9x – 6 = 60 – 5x – 10

⇒ 9x – 6 = – 5x + 60 – 10

⇒ 9x – 6 = – 5x + 50

⇒ 9x + 5x = 50 + 6

| On transposing – 6 and – 5x

⇒ 14x = 56

⇒ \(\frac { 14x }{ 14 } =\frac { 56 }{ 14 } \)

| On dividing by 14 on both sides

⇒ x = 4

Question 6.

\(\frac { x+2 }{ 2 } +\frac { x+4 }{ 3 } =\frac { x+6 }{ 4 } +\frac { x+8 }{ 5 } \)

Solution.

\(\frac { x+2 }{ 2 } +\frac { x+4 }{ 3 } =\frac { x+6 }{ 4 } +\frac { x+8 }{ 5 } \)

⇒ 30(x + 2) + 20(x + 4) = 15(x + 6) + 12(x + 8)

On multiplying by LCM of 2, 3, 4 and 5 i.e. 60 on both sides

⇒ 30x + 60 + 20x + 80 = 15x + 90 + 12x + 96

⇒ 30x + 20x + 60 + 80 = 15x + 12x + 90 + 96

⇒ 50x + 140 = 27x + 186

⇒ 50x – 27x = 186 – 140

| On transposing 140 and 27x

⇒ 23x = 46

⇒ \(\frac { 23x }{ 23 } =\frac { 46 }{ 23 } \)

On dividing(RBSESolutions.com)by 23 on both sides

⇒ x = 2

Question 7.

0.6x + 0.25x = 0.45x + 1.2

Solution

0.6x + 0.25x = 0.45x + 1.2

⇒ \(\frac { 6 }{ 10 } x+\frac { 25 }{ 100 } x=\frac { 45 }{ 100 } x+\frac { 12 }{ 10 } \)

On changing decimal fractions into simple fraction

⇒ 60x + 25x = 45x + 120

On multiplying by 100 since LCM of 10 and 100 is 100

⇒ 85x = 45x + 120

⇒ 85x – 45x = 120

| On transposing 45*

40x = 120

⇒ \(\frac { 40x }{ 40 } =\frac { 120 }{ 40 } \)

| On dividing by 40 on both sides

⇒ x = 3

Question 8.

2.5x – 7 = 0.5x + 13

Solution

2.5x – 7 = 0.5x + 13

⇒ 2.5x – 0.5x = 13 + 7

| On transposing -7 and 0.5x

⇒ 2x = 20

⇒ \(\frac { 2x }{ 2 } =\frac { 20 }{ 2 } \)

| On dividing(RBSESolutions.com)by 2 on both sides

⇒ x = 10

Question 9.

\(\frac { 7x+4 }{ x+2 } =-\frac { 4 }{ 3 } \)

Solution

\(\frac { 7x+4 }{ x+2 } =-\frac { 4 }{ 3 } \)

=> 3(7x + 4) = – 4(x + 2)

| By cross multiplication

=> 21x + 12 = – 4x – 8

=> 21x + 4x = – 8 – 12

| On transposing 12 and – 4x

⇒ 25x = – 20

⇒ \(\frac { 25x }{ 25 } =-\frac { 20 }{ 25 } \)

| On dividing by 25 on both sides

⇒ \(x=-\frac { 4 }{ 5 }\)

Question 10.

\(\frac { 4x+8 }{ 5x+8 } =\frac { 5 }{ 6 } \)

Solution

\(\frac { 4x+8 }{ 5x+8 } =\frac { 5 }{ 6 } \)

⇒ 6(4x + 8) = 5(5x + 8)

| By cross multiplication

⇒ 24x + 48 = 25x + 40

⇒ 24x – 25x = 40 – 48

| On(RBSESolutions.com)transposing 48 and 25x

⇒ – x = – 8

⇒ \(\frac { -x }{ -1 } =\frac { -8 }{ -1 } \)

| On dividing by – 1 on both sides

⇒ x = 8

We hope the given RBSE Solutions for Class 8 Maths Chapter 11 Linear Equations with One Variable Ex 11.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable Exercise 11.1, drop a comment below and we will get back to you at the earliest.

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