Skip to content
Related Articles

Related Articles

Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5 | Set 1
  • Last Updated : 07 Apr, 2021

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, 

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/-9 = 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 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: 

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 unique solution, 

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,

x + 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 unique solution,

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 unique solution

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 = 19

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 = −19

For unique solution

We have

a1/a2 = b1/b2 = c1/c2

8/k = 5/10 = 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, b2 = − (2k + 1), c2 = −3(2k − 1)

Now, for unique solution

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.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :