Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 2

Chapter 3 of RD Sharma's Class 10 Mathematics textbook deals with the "Pair of Linear Equations in the Two Variables." This chapter explores methods for solving systems of linear equations involving the two variables. Understanding these methods is crucial for solving the problems in algebra and geometry. Exercise 3.5 focuses on applying these techniques to solve specific problems.

Question 14. Find the value of k for which each of the following system of equations having infinitely many solutions:  

2x + 3y = 2,

(k + 2)x + (2k + 1)y = 2(k − 1)

Solution: 

Given that,

2x + 3y = 2      ...(1) 

(k + 2)x + (2k + 1)y = 2(k − 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 = −2

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

For unique solution, we have

a1/a2 = b1/b2 = c1/c2

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

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

= 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

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

Question 15. Find the value of k for which each of the following system of equations having infinitely many solutions:  

x + (k + 1)y = 4,

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

Solution: 

Given that,

x + (k + 1)y = 4      ...(1) 

(k + 1)x + 9y = (5k + 2)     ...(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 = (k + 1), c1 = −4 

a2 = (k + 1), b2 = 9, c2 = − (5k + 2)

For unique solution, we have

a1/a2 = b1/b2 = c1/c2

1/(k + 1) = (k + 1)/9 = -4/-(5k + 2)

1/(k + 1) = (k + 1)/9 and (k + 1)/9 = 4/(5k + 2) 

9 = (k + 1)² and (k + 1)(5k + 2) = 36

9 = k2 + 2k + 1 and 5k2 + 2k + 5k + 2 = 36

k² + 2k − 8 = 0 and 5k² + 7k − 34 = 0

k² + 4k − 2k − 8 = 0 and 5k² + 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

Hence, k = 2 satisfies both the condition.

So, when k = 2 the given set of equations will have infinitely many solutions.

Question 16. Find the value of k for which each of the following system of equations having infinitely many solutions:

kx + 3y = 2k + 1,

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

Solution: 

Given that,

kx + 3y = 2k + 1     ...(1) 

2(k + 1)x + 9y = (7k + 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 = k, b1 = 3, c1 = −(2k + 1) 

a2 = 2(k + 1), b2 = 9, c2 = −(7k + 1)

For unique solution, we have

a1/a2 = b1/b2 = c1/c2

1/2(k + 2) = 3/9 = -2k + 1/-(7k + 1)

1/2(k + 2) = 3/9 and 3/9 = 2k + 1/(7k + 1) 

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

Hence, when k = 2 the given set of equations will have infinitely many solutions.

Question 17. Find the value of k for which each of the following system of equations having infinitely many solutions:

2x + (k − 2)y = k,

6x + (2k − 1)y = (2k + 5)

Solution: 

Given that,

2x +( k − 2)y = k    ...(1)  

6x + (2k − 1)y = (2k + 5)     ...(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 = (k − 2), c1 = −k 

a2 = 6, b2 = (2k − 1), c2 = −(2k + 5)

For unique solution, we have

a1/a2 = b1/b2 = c1/c2

2/6 = k - 1/(2k - 1) = -k/-2(2k + 5)

2/6 = k - 1/(2k - 1) and (k - 1) / (2k - 1) = k / 2(2k + 5)

2k − 3k = −6 + 1 and k + k = 10

−k = −5 and 2k = 10 = k = 5 and k = 5

Hence, when k = 5 the given set of equations will have infinitely many solutions.

Question 18. Find the value of k for which each of the following system of equations having infinitely many solutions:

2x + 3y = 7,

(k + 1)x + (2k − 1)y = (4k + 1)

Solution: 

Given that,

2x + 3y = 7     ...(1) 

(k + 1)x + (2k − 1)y = (4k + 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 + 1, b2 = 2k − 1, c2 = −(4k + 1)

For unique solution, we have

a1/a2 = b1/b2 = c1/c2

2/(k + 1) = 3/(2k - 1) = -7/-(4k + 1)

2/(k + 1) = 3/(2k - 1) and 3/(2k - 1) = -7/-(4k + 1)

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

Hence, when k = 5 the given set of equations will have infinitely many solutions.

Question 19. Find the value of k for which each of the following system of equations having infinitely many solutions:

2x + 3y = k,

(k − 1)x + (k + 2)y = 3k

Solution: 

Given that,

2x + 3y = k     ...(1) 

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

a2 = k − 1, b2 = k + 2, c2 = −3k

For unique solution, we have

a1/a2 = b1/b2 = c1/c2

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

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

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

Hence, when k = 7 the given set of equations will have infinitely many solutions.

Question 20. Find the value of k for which the following system of equation has no solution:

kx − 5y = 2,

6x + 2y = 7

Solution: 

Given that,

kx − 5y = 2    ...(1)   

6x + 2y = 7   ...(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 = −5, c1 = −2 

a2 = 6 b2 = 2, c2 = −7

For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

1/2 = 2/k ≠ 2/7

k = 4

2k = -30

k = -15

Hence, when k = -15 the given set of equations will have no solutions.

Question 21. Find the value of k for which the following system of equation has no solution:

x + 2y = 0,

2x + ky = 5

Solution: 

Given that,

x + 2y = 0   ...(1)   

2x + ky = 5   ...(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 = 0 

a2 = 2, b2 = k, c2 = −5

For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

k/6 = -5/2 ≠ 2/7

k = 4

Hence, when k = 4 the given set of equations will have no solutions.

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: 

Given that,

3x − 4y + 7 = 0   ...(1)  

kx + 3y − 5 = 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 = 3, b1 = −4, c1 = 7 

a2 = k, b2 = 3, c2 = −5

For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

3/k = -4/3

k = -9/4

Hence, when k = -9/4 the given set of equations will have no solutions.

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: 

Given that,

2x − ky + 3 = 0  ...(1)  

3x + 2y − 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 = 2, b1 = −k, c1 = 3 

a2 = 3, b2 = 2, c2 = −1

For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

2/3 = -k/2

k = -4/3

Hence, when k = -4/3 the given set of equations will have no solutions.

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: 

Given that,

2x + ky − 11 = 0   ...(1) 

5x − 7y − 5 = 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 = k, c1 = −11 

a2 = 5, b2 = −7, c2 = −5

For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

2/5 = -k/-7

k = -14/5

Hence, when k = -14/5 the given set of equations will have no solutions.

Question 25. Find the value of k for which the following system of equation has no solution:

kx + 3y = 3,

12x + ky = 6

Solution: 

Given that, 

kx + 3y = 3  ...(1)

12x + ky = 6  ...(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 = 3, c1 = −3 

a2 = 12, b2 = k, c2 = − 6

For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

k/12 = 3/k ≠ 3/6    ...(5)

k² = 36 ⇒ k = + 6 or −6

From eq (5), we get

k/12 ≠ 3/6 

k ≠ 6

Hence, when k = -6 the given set of equations will have no solutions.

Question 26. For what value of a, the following system of equation will be inconsistent?

4x + 6y − 11 = 0,

2x + ay − 7 = 0

Solution: 

Given that,

4x + 6y − 11 = 0 ...(1)

2x + ay − 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 = 4, b1 = 6, c1 = −11 

a2 = 2, b2 = a, c2 = −7

For inconsistent solution, we have

a1/a2 = b1/b2 ≠ c1/c2

a1/a2 = b1/b2

4/2 = 6/a

a = 3

Hence, when a = 3 the given set of equations will be inconsistent.

Read More:

• Pair of Linear Equations in Two Variables
• Linear Equation in Two Variables