Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 1

Question 1. In each of the following systems of equations determine whether the system has a unique solution, no solution, or infinite solutions. In case there is a unique solution, find it from 1 to 4:

x − 3y − 3 = 0, 3x − 9y − 2 = 0.

Solution: 

Given that,

x − 3y − 3 = 0     ...(1)  

3x − 9y − 2 = 0     ...(2)  

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get 

a₁ = 1, b₁ = −3, c₁ = −3

a₂ = 3, b₂ = −9, c₂ = −2

Let's check the equation's,

a₁/a₂ = 1/3

b₁/b₂ = -3/-9 = 1/3

c₁/c₂ = -3/-2 = 3/2

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Hence, the given set of equations has no solution.

Question 2. In each of the following systems of equations 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: 

Given that,

2x + y − 5 = 0     ...(1)  

4x + 2y − 10 = 0    ...(2)  

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0    ...(3)  

a₂x + b₂y + c₂ = 0    ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get 

a₁ = 2, b₁ = 1, c₁ = −5 and

a₂ = 4, b₂ = 2, c₂ = −10

Lets check the equation's,

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 1/2

and c₁/c₂ = -5/-10 = 1/2

Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, the given set of equations has infinitely many solutions.

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: 

Given that,

3x − 5y = 20    ...(1) 

6x − 10y = 40    ...(2)  

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get 

a₁ = 3, b₁ = −5, c₁ = − 20

a₂ = 6, b₂ = −10, c₂ = − 40

Lets check the equation's,

a₁/a₂ = 3/6 = 1/2

b₁/b₂ = -5/-10 = 1/2 and

c₁/c₂ = -20/-40 = 1/2 

Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, the given set of equations has infinitely many solutions.

Question 4. 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 − 2y − 8 = 0, 5x − 10y − 10 = 0

Solution: 

Given that,

x − 2y − 8 = 0    ...(1)  

5x − 10y − 10 = 0    ...(2)  

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get 

a₁ = 1, b₁ = −2, c₁ = −8

a₂ = 5, b₂ = −10, c₂ = −10

Lets check the equation's,

a₁/a₂ = 1/5

b₁/b₂ = -2/-10 and

c₁/c₂ = -8/-10

Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Hence, the given set of equations 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: 

Given that,

kx + 2y − 5 = 0    ...(1)  

3x + y − 1 = 0    ...(2)  

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = k, b₁ = 2, c₁ = −5

a₂ = 3, b₂ = 1, c₂ = −1

For unique solution,

a₁/a₂ ≠ b₁/b₂

k/3 ≠ 2/1

k ≠ 6 

So, the given set of equations 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: 

Given that,

4x + ky + 8 = 0    ...(1)  

2x + 2y + 2 = 0    ...(2)  

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = 4, b₁ = k, c₁ = 8

a₂ = 2, b₂ = 2, c₂ = 2

For unique solution, 

a₁/a₂ ≠ b₁/b₂

4/2 ≠ k/2

k ≠ 4

So, the given set of equations 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: 

Given that,

4x − 5y − k = 0    ...(1)    

2x − 3y − 12 = 0    ...(2)  

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = 4, b₁ = −5, c₁ = −k

a₂ = 2, b₂ = -3, c₂ = -12

For unique solution, 

a₁/a₂ ≠ b₁/b₂

4/2 ≠ -5/-3

Here, k can have any real values.

Hence, the given set of equations 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: 

Given that,

x + 2y = 3    ...(1)  

5x + ky + 7 = 0    ...(2) 

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = 1, b₁ = 2, c₁ = −3

a₂ = 5, b₂ = k, c₂ = 7

For unique solution,

a₁/a₂ ≠ b₁/b₂

1/5 ≠ 2/k

k ≠ 10

So, the given set of equations 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 solutions:  

2x + 3y − 5 = 0, 6x − ky − 15 = 0

Solution: 

Given that,

2x + 3y − 5 = 0   ...(1) 

6x − ky − 15 = 0  ...(2) 

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = 2, b₁ = 3, c₁ = −5

a₂ = 6, b₂ = - k, c₂ = −15

For infinitely many solutions, 

We have

a₁/a₂ = b₁/b₂ = c₁/c₂

2/6 = 3/k

k = 9

Hence, when k = 9 the given set of equations will have infinitely many solutions.

Question 10. Find the value of k for which each of the following system of equations having infinitely many solutions:

4x + 5y = 3, kx + 15y = 9

Solution: 

Given that,

4x + 5y = 3   ...(1) 

kx +15y = 9  ...(2)

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = 4, b₁ = 5, c₁ = 3

a₂ = k, b₂ = 15, c₂ = 9

For infinitely many solutions,

We have

a₁/a₂ = b₁/b₂ = c₁/c₂

4/k = 5/15 = -3/-9

4/k = 1/3

k = 12

Hence, when k = 12 the given set of equations will have infinitely many solutions.

Question 11. Find the value of k for which each of the following system of equations having infinitely many solutions:  

kx − 2y + 6 = 0, 4x - 3y + 9 = 0

Solution: 

Given that,

kx − 2y + 6 = 0 ...(1)

4x - 3y + 9 = 0 ...(2)

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = k, b₁ = −2, c₁ = 6

a₂ = 4, b₂ = −3, c₂ = 9

For infinitely many solutions,

We have

a₁/a₂ = b₁/b₂ = c₁/c₂

k/4 = -2/-3 = 2/3

k = 8/3

Hence, when k = 8/3 the given set of equations will have infinitely many solutions.

Question 12. Find the value of k for which each of the following system of equations having infinitely many solutions:

8x + 5y = 9, kx + 10y = 18

Solution: 

Given that,

8x + 5y - 9 ...(1)

kx + 10y - 19 ...(2)

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = 8, b₁ = 5, c₁ = −9

a₂ = k, b₂ = 10, c₂ = −18

For infinitely many solutions

We have

a₁/a₂ = b₁/b₂ = c₁/c₂

8/k = 5/10 = -9 / -18 , k = 16

Hence, when k = 16 the given set of equations will have infinitely many solutions.

Question 13. Find the value of k for which each of the following system of equations having infinitely many solutions:

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

Solution: 

Given that,

2x − 3y - 7 ...(1)

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

So, the given equations are in the form of:

a₁x + b₁y + c₁ = 0     ...(3)  

a₂x + b₂y + c₂ = 0     ...(4)  

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a₁ = 2, b₁ = −3, c₁ = −7

a₂ = k + 2, b₂ = − (2k + 1), c₂ = −3(2k − 1)

For infinitely many solutions

We have

a₁/a₂ = b₁/b₂ = c₁/c₂

= 2/(k + 2) = -3/-(2k + 1) = -7/-3(2k - 1)

= 2/(k + 2) = -3/-(2k + 1) and -3/-(2k + 1) = -7/-3(2k - 1) 

= 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

Hence, when k = 4 the given set of equations will have infinitely many solutions.