In this chapter, we provide RD Sharma Class 10 Ex 3.5 Solutions Chapter 3 Pair of Linear Equations in Two Variables for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 3.5 Solutions Chapter 3 Pair of Linear Equations in Two Variables pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 3 |

Chapter Name | Pair of Linear Equations in Two Variables |

Exercise | 3.5 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **3 Pair of Linear Equations in Two Variables Ex 3.5 Download PDF**

**Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.5**

**Question: 1**

In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,

x − 3y − 3 = 0, x − 3y − 3 = 0

3x − 9y − 2 = 0, 3x − 9y − 2 = 0

**Solution:**

The given system may be written as

x − 3y − 3 = 0

3x − 9y − 2 = 0

The given system of equation is of the form

a_{2}x + b_{2}y − c_{2} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 1, b_{1} = −3, c_{1} = −3

a_{2} = 3, b_{2} = −9, c_{2} = −2

We have,

Therefore, the given equation has no solution.

**Question: 2**

In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,

2x + y − 5 = 0

4x + 2y − 10 = 0

**Solution:**

The given system may be written as

2x + y − 5 = 0 4x + 2y − 10 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = 1, c_{1} = −5

a_{2} = 4, b_{2} = 2, c_{2} = −10

We have,

Therefore, the given equation has infinitely many solution.

**Question: 3**

In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,

3x − 5y = 20

6x − 10y = 40

**Solution:**

The given system may be written as

3x − 5y = 20 6x − 10y = 40

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 3, b_{1} = −5, c_{1} = − 20

a_{2} = 6, b_{2} = −10, c_{2 }= − 40

We have,

Therefore, the given equation has infinitely many solution.

**Question: 4**

x − 2y − 8 = 0

5x − 10y − 10 = 0

**Solution:**

The given system may be written as

x − 2y − 8 = 0 5x − 10y − 10 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 1, b_{1} = −2, c_{1} = −8

a_{2} = 5, b_{2} = −10, c_{2} = −10

We have,

Therefore, the given equation has no solution.

**Question: 5**

Find the value of k for each of the following system of equations which have a unique solution

kx + 2y − 5 = 0

3x + y − 1 = 0

**Solution:**

The given system may be written as

kx + 2y − 5 = 0

3x + y − 1 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0

a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1 }= k, b_{1 }= 2, c_{1} = −5

a_{2} = 3, b_{2} = 1, c_{2} = −1

For unique solution, we have

Therefore, the given system will have unique solution for all real values of k other than 6.

**Question: 6**

Find the value of k for each of the following system of equations which have a unique solution

4x + ky + 8 = 0

2x + 2y + 2 = 0

**Solution:**

The given system may be written as

4x + ky + 8 = 0 2x + 2y + 2 = 0

The given system of equation is of the form

a_{1}x +b_{1}y − c_{1 }= 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 4, b_{1} = k, c_{1} = 8

a_{2 }= 2, b_{2} = 2, c_{2} = 2

For unique solution, we have

Therefore, the given system will have unique solution for all real values of k other than 4.

**Question: 7**

Find the value of k for each of the following system of equations which have a unique solution

4x − 5y = k

2x − 3y = 12

**Solution:**

The given system may be written as

4x − 5y − k = 0

2x − 3y − 12 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1 }= 4, b_{1} = −5, c_{1} = −k

a_{2} = 2, b_{2 }= -3, c_{2 }= -12

For unique solution, we have

⇒ k can have any real values.

Therefore, the given system will have unique solution for all real values of k.

**Question: 8**

Find the value of k for each of the following system of equations which have a unique solution

x + 2y = 3

5x + ky + 7 = 0

**Solution:**

The given system may be written as

x + 2y = 3

5x + ky + 7 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where a_{1 }= 1, b_{1} = 2, c_{1} = −3

a_{2 }= 5, b_{2} = k, c_{2} = 7

For unique solution, we have

Therefore, the given system will have unique solution for all real values of k other than 10.

**Question: 9**

Find the value of k for which each of the following system of equations having infinitely many solution:

2x + 3y − 5 = 0

6x − ky − 15 = 0

**Solution:**

The given system may be written as

2x + 3y − 5 = 0 6x − ky − 15 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = 3, c_{1} = −5

a_{2} = 6, b_{2} = k, c_{2} = −15

For unique solution, we have

Therefore, the given system of equation will have infinitely many solutions, if k = 9.

**Question: 10**

Find the value of k for which each of the following system of equations having infinitely many solution:

4x + 5y = 3

x + 15y = 9

**Solution:**

The given system may be written as

4x + 5y = 3

kx +15y = 9

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 4, b_{1} = 5, c_{1} = 3

a_{2} = k, b_{2 }= 15, c_{2} = 9

For unique solution, we have

Therefore, the given system will have infinitely many solutions if k = 12.

**Question: 11**

Find the value of k for which each of the following system of equations having infinitely many solution:

kx − 2y + 6 = 0

4x + 3y + 9 = 0

**Solution:**

The given system may be written as

kx − 2y + 6 = 0 4x + 3y + 9 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = k, b_{1} = −2, c_{1} = 6

a_{2} = 4, b_{2} = −3, c_{2} = 9

For unique solution, we have

Therefore, the given system of equations will have infinitely many solutions, if k = 8/3.

**Question: 12**

8x + 5y = 9

kx + 10y = 19

**Solution:**

The given system may be written as

8x + 5y = 9 kx + 10y = 19

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }=0

Where, a_{1} = 8, b_{1} = 5, c_{1 }= −9

a_{2} = k, b_{2} = 10, c_{2} = −18 a_{2} = k, b_{2} = 10,c_{2 }= −18

For unique solution, we have

Therefore, the given system of equations will have infinitely many solutions, if k = 16.

**Question: 13**

2x − 3y = 7

(k + 2)x − (2k + 1)y = 3(2k − 1)

**Solution:**

The given system may be written as

2x − 3y = 7 (k + 2)x − (2k + 1)y = 3(2k − 1)

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = −3, c_{1} = −7

a_{2 }= k, b_{2} = − (2k + 1), c_{2} = −3(2k − 1)

For unique solution, we have

⇒ 2(2k + 1) = 3(k + 2) and 3 × 3(2k − 1) = 7(2k + 1)

⇒ 4k + 2 = 3k + 6 and 18k − 9 = 14k + 7

⇒ k = 4 and 4k = 16

⇒ k = 4

Therefore, the given system of equations will have infinitely many solutions, if k = 4.

**Question: 14**

2x + 3y = 2

(k + 2)x + (2k + 1)y = 2(k − 1)

**Solution:**

The given system may be written as

2x + 3y = 2 (k + 2)x + (2k + 1)y = 2(k − 1)

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = 3,c_{1} = −2

a_{2} = (k + 2), b_{2} = (2k + 1),c_{2} = −2(k − 1)

For unique solution, we have

⇒ 2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1)

⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1

⇒ k = 4 and k = 4

Therefore, the given system of equations will have infinitely many solutions, if k = 4.

**Question: 15**

x + (k + 1)y = 4

(k + 1)x + 9y = (5k + 2)

**Solution:**

The given system may be written as

x + (k + 1)y = 4 (k + 1)x + 9y = (5k + 2)

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 1, b_{1} = (k + 1), c_{1} = −4

a_{2} = (k + 1), b_{2} = 9, c_{2} = − (5k + 2)

For unique solution, we have

⇒ 9 = (k + 1)^{2} and (k + 1)(5k + 2) = 36

⇒ 9 = k^{2} + 2k + 1 and 5k^{2} + 2k + 5k + 2 = 36

⇒ k^{2 }+ 2k − 8 = 0 and 5k^{2} + 7k − 34 = 0

⇒ k^{2} + 4k − 2k − 8 = 0 and 5k^{2} + 17k − 10k − 34 = 0

⇒ k(k + 4) −2 (k + 4) = 0 and (5k + 17) − 2 (5k + 17) = 0

⇒ (k + 4)(k − 2) = 0 and (5k + 17)(k − 2) = 0

⇒ k = – 4 or k = 2 and k = -17/5 or k = 2

Thus, k = 2 satisfies both the condition.

Therefore, the given system of equations will have infinitely many solutions, if k = 2.

**Question: 16**

kx + 3y = 2k + 1

2(k + 1)x + 9y = (7k + 1)

**Solution:**

The given system may be written as

kx + 3y = 2k + 1 2(k + 1)x + 9y = (7k + 1)

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1 }= k, b_{1} = 3, c_{1} = −(2k + 1)

a_{2} = 2(k + 1), b_{2} = 9, c_{2} = −(7k + 1)

For unique solution, we have

⇒ 9k = 3 × 2(k + 1) and 3(7k + 1) = 9(2k + 1)

⇒ 9k − 6k = 6 and 21k − 18k = 9 − 3

⇒ 3k = 6 ⇒ k = 2 and k = 2

Therefore, the given system of equations will have infinitely many solutions, if k = 2.

**Question: 17**

2x + (k − 2)y = k

6x + (2k − 1)y = (2k + 5)

**Solution:**

The given system may be written as

2x +( k − 2)y = k 6x + (2k − 1)y = (2k + 5)

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2,b_{1} = (k − 2),c_{1} = −k

a_{2} = 6,b_{2} = (2k − 1),c_{2} = −(2k + 5)

For unique solution, we have

⇒ 2k − 3k = −6 + 1 and k + k = 10

⇒ −k = −5 and 2k = 10 ⇒ k = 5 and k = 5

Therefore, the given system of equations will have infinitely many solutions, if k = 5.

**Question: 18**

2x + 3y = 72x + 3y = 7

(k + 1)x + (2k − 1)y = (4k + 1)

**Solution:**

The given system may be written as

2x + 3y = 7 (k + 1)x + (2k − 1)y = (4k + 1)

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }= 0

Where, a_{1} = 2, b_{1} = 3, c_{1} = −7

a_{2 }= k + 1, b_{2} = 2k − 1, c_{2} = −(4k + 1)

For unique solution, we have

Extra close brace or missing open brace

⇒ 4k − 2 = 3k + 3 and 12k + 3 = 14k − 7

⇒ k = 5 and 2k = 10 ⇒ k = 5 and k = 5

Therefore, the given system of equations will have infinitely many solutions, if k = 5.

**Question: 19**

2x + 3y = k

(k − 1)x + (k + 2)y = 3k

**Solution:**

The given system may be written as

2x + 3y = k (k − 1)x + (k + 2)y = 3k

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1 }= 2,b_{1 }= 3, c_{1} = −k

a_{2} = k − 1, b_{2 }= k + 2, c_{2} = −3k

For unique solution, we have

Extra close brace or missing open brace

⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ k = 7 and k = 7

Therefore, the given system of equations will have infinitely many solutions, if k = 7.

**Question: 20**

Find the value of k for which the following system of equation has no solution:

kx − 5y = 2

6x + 2y = 7

**Solution:**

The given system may be written as

kx − 5y = 2 6x + 2y = 7

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1 }= k, b_{1} = −5, c_{1 }= −2

a_{2} = 6 b_{2 }= 2, c_{2} = −7

For no solution, we have

⇒ 2k = -30 ⇒ k = -15

Therefore, the given system of equations will have no solutions, if k = −15.

**Question: 21**

Find the value of k for which the following system of equation has no solution:

x + 2y = 0

2x + ky = 5

**Solution:**

The given system may be written as

x_{2}y = 0 2x + ky = 5

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 1, b_{1} = 2, c_{1} = 0

a_{2} = 2, b_{2} = k, c_{2} = −5

For no solution, we have

⇒ k = 4

Therefore, the given system of equations will have no solutions, if k = 4.

**Question: 22**

Find the value of k for which the following system of equation has no solution:

3x − 4y + 7 = 0

kx + 3y − 5 = 0

**Solution:**

The given system may be written as

3x − 4y + 7 = 0 kx + 3y − 5 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 3, b_{1} = −4, c_{1} = 7

a_{2} = k, b_{2} = 3, c_{2} = −5

For no solution, we have

Therefore, the given system of equations will have no solutions, if k = – 9 /4.

**Question: 23**

Find the value of k for which the following system of equation has no solution:

2x − ky + 3 = 0

3x + 2y − 1 = 0

**Solution:**

The given system may be written as

2x − ky + 3 = 0 3x + 2y − 1 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = −k, c_{1} = 3

a_{2} = 3, b_{2} = 2, c_{2} = −1

For no solution, we have

Therefore, the given system of equations will have no solutions, if k = – 4/3.

**Question: 24**

Find the value of k for which the following system of equation has no solution:

2x + ky − 11 = 0

5x − 7y − 5 = 0

**Solution:**

The given system may be written as

2x + ky − 11 = 0 5x − 7y − 5 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = k, c_{1} = −11

a_{2} = 5, b_{2} = −7, c_{2} = −5a_{2} = 5, b_{2 }= −7, c_{2} = − 5

For no solution, we have

Therefore, the given system of equations will have no solutions, if k = -14/5.

**Question: 25**

Find the value of k for which the following system of equation has no solution:

kx + 3y = 3

12x + ky = 6

**Solution:**

The given system may be written as

kx + 3y = 3 12x + ky = 6

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = k, b_{1} = 3, c_{1} = −3

a_{2 }= 12, b_{2} = k, c_{2} = − 6

For no solution, we have

⇒ k^{2} = 36 ⇒ k = + 6 or −6

From (i)

Therefore, the given system of equations will have no solutions, if k = − 6.

**Question: 26**

For what value of a, the following system of equation will be inconsistent?

4x + 6y − 11 = 0

2x + ay − 7 = 0

**Solution:**

The given system may be written as

4x + 6y − 11 = 0 2x + ay − 7 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 4, b_{1} = 6, c_{1} = −11

a_{2} = 2, b_{2} = a, c_{2} = −7

For unique solution, we have

Therefore, the given system of equations will be inconsistent, if a = 3.

**Question: 27**

For what value of a, the following system of equation have no solution?

ax + 3y = a − 3

12x + ay = a

**Solution:**

The given system may be written as

ax + 3y = a − 3 12x + ay = a

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = a, b_{1 }= 3, c_{1} = – (a − 3)

a_{2} = 12, b_{2} = a, c_{2 }= − a

For unique solution, we have

And,

⇒ a^{2 }= 36

⇒ a = + 6 or – 6?

a ≠ 6 ⇒ a = – 6

Therefore, the given system of equations will have no solution, if a = − 6.

**Question: 28**

Find the value of a, for which the following system of equation have

(i) Unique solution

(ii) No solution

kx + 2y = 5

3x + y = 1

**Solution:**

The given system may be written as

kx + 2y − 5 = 0 3x + y − 1 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = k, b_{1} = 2, c_{1} = −5

a_{2} = 3, b_{2} = 1, c_{2} = −1

(i) For unique solution, we have

Therefore, the given system of equations will have unique solution, if k ≠ 6 k ≠ 6.

(ii) For no solution, we have

Therefore, the given system of equations will have no solution, if a = 6.

**Question: 29**

For what value of c, the following system of equation have infinitely many solution (where c ≠ 0 c ≠ 0)?

6x + 3y = c − 3

12x + cy = c

**Solution:**

The given system may be written as

6x + 3y − (c − 3) = 0 12x + cy − c = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 6, b_{1} = 3, c_{1} = −(c − 3)

a_{2 }= 12, b_{2} = c, c_{2} = – c

For infinitely many solution, we have

⇒ c = 6 and c – 3 = 3

⇒ c = 6 and c = 6

Therefore, the given system of equations will have infinitely many solution, if c = 6.

**Question: 30**

Find the value of k, for which the following system of equation have

(i) Unique solution

(ii) No solution

(iii) Infinitely many solution

2x + ky = 1

3x − 5y = 7

**Solution:**

The given system may be written as

2x + ky = 1 3x − 5y = 7

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = k, c_{1} = −1a_{1 }= 2, b_{1} = k, c_{1} = −1

a_{2 }= 3, b_{2} = −5, c_{2} = −7

(i) For unique solution, we have

Therefore, the given system of equations will have unique solution, if k ≠ -10/3.

(ii) For no solution, we have

Therefore, the given system of equations will have no solution, if k = -10)/3.

(iii) For the given system to have infinitely many solution, we have

So there is no value of k for which the given system of equation has infinitely many solution.

**Question: 31**

For what value of k, the following system of equation will represent the coincident lines?

x + 2y + 7 = 0

2x + ky + 14 = 0

**Solution:**

The given system may be written as

x + 2y + 7 = 0 2x + ky + 14 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 1, b_{1} = 2, c_{1} = 7

a_{2} = 2, b_{2} = k, c_{2} = 14

The given system of equation will represent the coincident lines if they have infinitely many solution.

Therefore, the given system of equations will have infinitely many solution, if k = 4.

**Question: 32**

Find the value of k, for which the following system of equation have unique solution.

ax + by = c

lx + my = n

**Solution:**

The given system may be written as

ax + by − c = 0 lx + my − n = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2 }= 0

Where, a_{1} = a, b_{1} = b, c_{1 }= − c

a_{2 }= l, b_{2} = m, c_{2} = − n

For unique solution, we have

Therefore, the given system of equations will have unique solution, if am ≠ bl.

**Question: 33**

Find the value of a and b such that the following system of linear equation have infinitely many solution:

(2a − 1)x + 3y − 5 = 0

3x + (b − 1)y − 2 = 0

**Solution:**

The given system of equation may be written as,

(2a − 1)x + 3y − 5 = 0 3x + (b − 1)y − 2 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = (2a − 1), b_{1} = 3, c_{1} = −5

a_{2} = 3, b_{2} = b − 1, c_{2} = −2

The given system of equation will have infinitely many solution, if

⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)

⇒ 4a − 2 = 15 and 6 = 5b − 5 ⇒ 4a = 17 and 5b = 11

**Question: 34**

Find the value of a and b such that the following system of linear equation have infinitely many solution:

2x − 3y = 7

(a + b)x − (a + b − 3)y = 4a + b

**Solution:**

The given system of equation may be written as,

2x − 3y − 7 = 0 (a + b)x − (a + b − 3)y − (4a + b) = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = −3, c_{1} = −7

a_{2} = (a + b), b_{2} = −(a + b − 3), c_{2 }= −(4a + b)

The given system of equation will have infinitely many solution, if

⇒ 2(a + b − 3) = 3(a + b) and 3(4a + b) = 7(a + b − 3)

⇒ 2a + 2b − 6 = 3a + 3b and 12a + 3b = 7a + 7b − 21

⇒ a + b = −6 and 5a − 4b = −21

⇒ a = − 6 − b

Substituting the value of a in 5a − 4b = −21 we have

5( – b – 6) – 4b = – 21

⇒ − 5b − 30 − 4b = − 21

⇒ 9b = − 9 ⇒ b = −1

As a = – 6 – b

⇒ a = − 6 + 1 = − 5

Hence the given system of equation will have infinitely many solution if

a = – 5 and b = –1.

**Question: 35**

Find the value of p and q such that the following system of linear equation have infinitely many solution:

2x − 3y = 9

(p + q)x + (2p − q)y = 3(p + q + 1)

**Solution:**

The given system of equation may be written as,

2x − 3y − 9 = 0 (p + q)x + (2p − q)y − 3(p + q + 1) = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} =2, b_{1} = 3, c_{1 }= −9

a_{2} = (p + q), b_{2} = (2p − q), c_{2} = -3(p + q + 1)

The given system of equation will have infinitely many solution, if

2(2p – q) = 3(p + q) and (p + q + 1) = 2p – q

⇒ 4p – 2q = 3p + 3q and -p + 2q = -1

⇒ p = 5q and p – 2q = 1

Substituting the value of p in p – 2q = 1, we have

3q = 1

⇒ q = 1/3

Substituting the value of p in p = 5q we have

p = 5/3

Hence the given system of equation will have infinitely many solution if

p = 5/3 and q = 1/3.

**Question: 36**

Find the values of a and b for which the following system of equation has infinitely many solution:

(i) (2a − 1)x + 3y = 5

3x + (b − 2)y = 3

(ii) 2x − (2a + 5)y = 5

(2b + 1)x − 9y = 15

(iii) (a − 1)x + 3y = 2

6x + (1 − 2b)y = 6

(iv) 3x + 4y = 12

(a + b)x + 2(a − b)y = 5a – 1

(v) 2x + 3y = 7

(a − 1)x + (a + 1)y = 3a − 1

(vi) 2x + 3y = 7

(a − 1)x + (a + 2)y = 3a

**Solution:**

(i) The given system of equation may be written as,

(2a − 1)x + 3y − 5 = 0 3x + (b − 2)y − 3 = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2a − 1, b_{1 }= 3, c_{1} = −5

a_{2} = 3, b_{2} = b − 2, c_{2 }= -3(p + q + 1)

The given system of equation will have infinitely many solution, if

2a – 1 = 5 and – 9 = 5(b – 2)

⇒ a = 3 and -9 = 5b – 10

a = 3 and b = 1/5

Hence the given system of equation will have infinitely many solution if

a = 3 and b = 1/5.

(ii) The given system of equation may be written as,

2x − (2a + 5)y = 5 (2b + 1)x − 9y = 15

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = – (2a + 5), c_{1} = −5

a_{2} = (2b + 1), b_{2} = −9, c_{2} = −15

The given system of equation will have infinitely many solution, if

Hence the given system of equation will have infinitely many solution if

a = – 1 and b = 5/2.

(iii) The given system of equation may be written as,

(a − 1)x + 3y = 2 6x + (1 − 2b)y = 6

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = a-1, b_{1} = 3, c_{1} = −2

a_{2} = 6, b_{2} = 1 − 2b, c_{2} = −6

The given system of equation will have infinitely many solution, if

⇒ a – 1 = 2 and 1 – 2b = 9

⇒ a – 1 = 2 and 1 – 2b = 9

⇒ a = 3 and b = -4

⇒ a = 3 and b = -4

Hence the given system of equation will have infinitely many solution if

a = 3 and b = −4.

(iv) The given system of equation may be written as,

3x + 4y − 12 = 0 (a + b)x + 2(a − b)y − (5a − 1) = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 3, b_{1} = 4, c_{1} = −12

a_{2 }= (a + b), b_{2} = 2(a − b), c_{2} = – (5a − 1)

The given system of equation will have infinitely many solution, if

⇒ 3(a – b) = 2a + 2b and 2(5a – 1) = 12(a – b)

⇒ a = 5b and -2a = -12b + 2

Substituting a = 5b in -2a = -12b + 2, we have

-2(5b) = -12b + 2

⇒ −10b = −12b + 2 ⇒ b = 1

Thus a = 5

Hence the given system of equation will have infinitely many solution if

a = 5 and b = 1.

(v) The given system of equation may be written as,

2x + 3y − 7 = 0 (a − 1)x + (a + 1)y − (3a − 1) = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1} = 2, b_{1} = 3, c_{1} = −7

a_{2} = (a − 1), b_{2} = (a + 1), c_{2} = – (3a − 1)

The given system of equation will have infinitely many solution, if

⇒ 2(a + 1) = 3(a – 1) and 3(3a – 1) = 7(a + 1)

⇒ 2a – 3a = -3 – 2 and 9a – 3 = 7a + 7

⇒ a = 5 and a = 5

Hence the given system of equation will have infinitely many solution if

a = 5 and b = 1.

(vi) The given system of equation may be written as,

2x + 3y − 7 = 0 (a − 1)x + (a + 2)y − 3a = 0

The given system of equation is of the form

a_{1}x + b_{1}y − c_{1} = 0 a_{2}x + b_{2}y − c_{2} = 0

Where, a_{1 }= 2, b_{1} = 3, c_{1} = −7

a_{2} = (a − 1), b_{2 }= (a + 2), c_{2} = −3a

The given system of equation will have infinitely many solution, if

⇒ 2(a + 2) = 3(a – 1) and 3(3a) = 7(a + 2)

⇒ 2a + 4 = 3a – 3 and 9a = 7a + 14

⇒ a = 7 and a = 7

Hence the given system of equation will have infinitely many solution if

a = 7and b = 1.

**All Chapter RD Sharma Solutions For Class10 Maths**

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