A given system of equations may or may not have a solution. Sometimes it can also have infinitely many solutions. All these conditions for solvability are studied in this exercise. The RD Sharma Solutions Class 10 prepared by experts at BYJUâ€™S can help students get a strong conceptual knowledge on the subject. Also, the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.5 PDF given below is available for students for further clarifications.

## RD Sharma Solutions for Class 10 Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.5 Download PDF

### Access RD Sharma Solutions for Class 10 Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.5

**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, find it from 1 to 4:**

**1.Â x âˆ’ 3y = 3**

**Â Â Â 3x âˆ’ 9y = 2**

**Solution:**

The given system of equations is:

x âˆ’ 3y â€“ 3 = 0

3x âˆ’ 9y âˆ’ 2 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 1, b_{1 }= âˆ’3, c_{1 }= âˆ’3

a_{2} = 3, b_{2} = âˆ’9, c_{2} = âˆ’2

So according to the question, we get

a_{1} / a_{2 }= 1/3

b_{1} / b_{2} = âˆ’3/ âˆ’9 = 1/3 and,

c_{1 }/ c_{2} = âˆ’3/ âˆ’2 = 3/2

â‡’ a_{1 }/_{ }a_{2} = b_{1}/ b2 â‰ c_{1 }/ c_{2}

Hence, we can conclude that the given system of equation has no solution.

**2. 2x + y = 5**

** 4x + 2y = 10 **

**Solution: **

** **

** **The given system of equations is:

2x + y â€“ 5 = 0

4x + 2y â€“ 10 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 2, b_{1 }= 1, c_{1 }= âˆ’5

a_{2} = 4, b_{2} = 2, c_{2} = âˆ’10

So according to the question, we get

a_{1} / a_{2 }= 2/4 = 1/2

b_{1} / b_{2} = 1/ 2 and,

c_{1 }/ c_{2} = âˆ’5/ âˆ’10 = 1/2

â‡’ a_{1 }/_{ }a_{2} = b_{1}/ b2 = c_{1 }/ c_{2}

Hence, we can conclude that the given system of equation has infinity many solutions.

**3. 3x – 5y = 20 **

** 6x â€“ 10y = 40 **

**Solution: **

The given system of equations is:

3x – 5y â€“ 20 = 0

6x – 10y â€“ 40 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 3, b_{1 }= -5, c_{1 }= âˆ’20

a_{2} = 6, b_{2} = -10, c_{2} = âˆ’40

So according to the question, we get

a_{1} / a_{2 }= 3/6 = 1/2

b_{1} / b_{2} = -5/ -10 = 1/2 and,

c_{1 }/ c_{2} = -20/ âˆ’40 = 1/2

â‡’ a_{1 }/_{ }a_{2} = b_{1}/ b2 = c_{1 }/ c_{2}

Hence, we can conclude that the given system of equation has infinity many solutions.

**4. x â€“ 2y = 8**

** 5x â€“ 10y = 10 **

**Solution: **

The given system of equations is:

x â€“ 2y – 8 = 0

5x â€“ 10y – 10 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 1, b_{1 }= -2, c_{1 }= âˆ’8

a_{2} = 5, b_{2} = -10, c_{2} = -10

So according to the question, we get

a_{1} / a_{2 }= 1/5

b_{1} / b_{2} = -2/ -10 = 1/5 and,

c_{1 }/ c_{2} = -8/ âˆ’10 = 4/5

â‡’ a_{1 }/_{ }a_{2} = b_{1}/ b2 â‰ c_{1 }/ c_{2}

Hence, we can conclude that the given system of equation has no solution.

**Find the value of k for which the following system of equations has a unique solution: (5-8)**

**5. kx + 2y = 5**

** 3x + y = 1 **

**Solution: **

The given system of equations is:

kx + 2y – 5 = 0

3x + y – 1 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= k, b_{1 }= 2, c_{1 }= âˆ’5

a_{2} = 3, b_{2} = 1, c_{2} = -1

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} â‰ b_{1} / b_{2}

k/3 â‰ 2/1

â‡’ k â‰ 6

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

**6. 4x + ky + 8 = 0**

** 2x + 2y + 2 = 0 **

**Solution: **

The given system of equations is:

4x + ky + 8 = 0

2x + 2y + 2 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 4, b_{1 }= k, c_{1 }= 8

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} â‰ b_{1} / b_{2}

4/2 â‰ k/2

â‡’ k â‰ 4

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

**7. 4x â€“ 5y = k **

** 2x â€“ 3y = 12**

**Solution **

The given system of equations is:

4x â€“ 5y – k = 0

2x â€“ 3y â€“ 12 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 4, b_{1 }= 5, c_{1 }= -k

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} â‰ b_{1} / b_{2}

4/2 â‰ 5/3

â‡’kÂ can have any real values.

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

**8. x + 2y = 3 **

** 5x + ky + 7 = 0 **

**Solution: **

The given system of equations is:

x + 2y – 3 = 0

5x + ky + 7 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 1, b_{1 }= 2, c_{1 }= -3

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} â‰ b_{1} / b_{2}

1/5 â‰ 2/k

â‡’ k â‰ 10

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

**Find the value of k for which each of the following system of equations having infinitely many solution: (9-19)**

**9. 2x + 3y â€“ 5 = 0 **

** 6x + ky â€“ 15 = 0 **

**Solution: **

The given system of equations is:

2x + 3y – 5 = 0

6x + ky – 15 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 2, b_{1 }= 3, c_{1 }= -5

a_{2} = 6, b_{2} = k, c_{2} = -15

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

2/6 = 3/k

â‡’ k = 9

Hence, the given system of equations will have infinitely many solutions, if k = 9.

**10. 4x + 5y = 3**

** kx + 15y = 9**

**Solution: **

The given system of equations is:

4x + 5y – 3= 0

kx + 15y – 9 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 4, b_{1 }= 5, c_{1 }= -3

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

4/ k = 5/ 15 = âˆ’3/ âˆ’9

4/ k = 1/ 3

â‡’k = 12

Hence, the given system of equations will have infinitely many solutions, if k = 12.

**11. kx â€“ 2y + 6 = 0**

** 4x â€“ 3y + 9 = 0 **

**Solution: **

The given system of equations is:

kx â€“ 2y + 6 = 0

4x â€“ 3y + 9 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= k, b_{1 }= -2, c_{1 }= 6

a_{2} = 4, b_{2} = -3, c_{2} = 9

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

k/ 4 = âˆ’2/ âˆ’3 = 2/ 3

â‡’k = 8/ 3

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

**12. 8x + 5y = 9 **

** kx + 10y = 18 **

**Solution: **

The given system of equations is:

8x + 5y – 9 = 0

kx + 10y – 18 = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 8, b_{1 }= 5, c_{1 }= -9

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

8/k = 5/10 = âˆ’9/ âˆ’18 = 1/2

â‡’k=16

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

**13. 2x â€“ 3y = 7 **

** (k+2)x â€“ (2k+1)y = 3(2k-1)**

**Solution: **

The given system of equations is:

2x â€“ 3y – 7 = 0

(k+2)x â€“ (2k+1)y – 3(2k-1) = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 2, b_{1 }= -3, c_{1 }= -7

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

2/ (k+2) = âˆ’3/ âˆ’(2k+1) = âˆ’7/ âˆ’3(2kâˆ’1)

2/(k+2) = âˆ’3/ âˆ’(2k+1)and âˆ’3/ âˆ’(2k+1)= âˆ’7/ âˆ’3(2kâˆ’1

â‡’2(2k+1) = 3(k+2) and 3Ã—3(2kâˆ’1) = 7(2k+1)

â‡’4k+2 = 3k+6 and 18k â€“ 9 = 14k + 7

â‡’k=4 and 4k = 16 â‡’k=4

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

**14. 2x + 3y = 2**

** (k+2)x + (2k+1)y = 2(k-1) **

**Solution: **

The given system of equations is:

2x + 3y – 2= 0

(k+2)x + (2k+1)y – 2(k-1) = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 2, b_{1 }= 3, c_{1 }= -5

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

2/ (k+2) = 3/ (2k+1) = âˆ’2/ âˆ’2(kâˆ’1)

2/ (k+2) = 3/ (2k+1) and 3(2k+1) = 22(kâˆ’1

â‡’2(2k+1) = 3(k+2) and 3(kâˆ’1) = (2k+1)

â‡’4k+2 = 3k+6 and 3kâˆ’3 = 2k+1

â‡’k = 4 and k = 4

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

**15. x + (k+1)y = 4**

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

**Solution: **

The given system of equations is:

x + (k+1)y – 4= 0

(k+1)x + 9y â€“ (5k + 2) = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 1, b_{1 }= (k+1), c_{1 }= -4

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

1/ k+1 = (k+1)/ 9 = âˆ’4/ âˆ’(5k+2)

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

â‡’9 = (k+1)^{2 }and (k+1)(5k+2) = 36

â‡’9 = k^{2 }+ 2k + 1 and 5k^{2}+2k+5k+2 = 36

â‡’k^{2}+2kâˆ’8 = 0 and 5k^{2}+7kâˆ’34 = 0

â‡’k^{2}+4kâˆ’2kâˆ’8 = 0 and 5k^{2}+17kâˆ’10kâˆ’34 = 0

â‡’k(k+4)âˆ’2(k+4) = 0 and (5k+17)âˆ’2(5k+17) = 0

â‡’(k+4)(kâˆ’2) = 0 and (5k+17) (kâˆ’2) = 0

â‡’k = âˆ’4 or k = 2 and k = âˆ’17/5 or k = 2

Its seen that k=2 satisfies both the condition.

Hence, the given system of equations will have infinitely many solutions, if k = 9.

**16. kx + 3y = 2k + 1**

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

**Solution: **

The given system of equations is:

kx + 3y â€“ (2k + 1) = 0

2(k+1)x + 9y â€“ (7k + 1) = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= k, b_{1 }= 3, c_{1 }= – (2k+1)

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

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

3k = 2k +2 and 7k+1 = 3(2k+1) = 6k + 3

k = 2 and k = 2

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

**17. 2x + (k-2)y = k**

** 6x + (2k-1)y = 2k + 5 **

**Solution: **

The given system of equations is:

2x + (k-2)y – k = 0

6x + (2k-1)y â€“ (2k+5) = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 2, b_{1 }= k-2, c_{1 }= – k

a_{2} = 6, b_{2} = 2k-1, c_{2} = -2k-5

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

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

4k -2 = 6k -12 and (k-2)(2k+5) = k(2k-1)

2k = 10 and 2k^{2 }– 4k + 5k â€“ 10 = 2k^{2} â€“ k

â‡’ k = 5 and 2k = 10 â‡’ k = 5

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

**18. 2x + 3y = 7**

** (k+1)x + (2k-1)y = 4k+1**

**Solution: **

The given system of equations is:

2x + 3y – 7= 0

(k+1)x + (2k-1)y â€“ (4k+1) = 0

The above equations are of the form

a_{1 }x + b_{1 }y âˆ’ c_{1 }= 0

a_{2 }x + b_{2 }y âˆ’ c_{2} = 0

Here,Â a_{1 }= 2, b_{1 }= 3, c_{1 }= – 7

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

So according to the question,

For unique solution, the condition is

a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}

2/ (k+1) = 3/ (2kâˆ’1) = âˆ’7/ âˆ’(4k+1)

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

2(2kâˆ’1) = 3(k+1) and 3(4k+1) = 7(2kâˆ’1)

â‡’4kâˆ’2 = 3k+3 and 12k + 3 = 14k âˆ’ 7

â‡’k = 5 and 2k = 10

â‡’k = 5 and k = 5

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