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.