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

Question 27. For what value of a, the following system of the equation has no solution:

ax + 3y = a − 3

12x + ay = a

Solution: 

Given that,

ax + 3y = a − 3  …(1)

12x + ay = a …(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 = a, b1 = 3, c1 = – (a − 3) 

a2 = 12, b2 = a, c2 = − a

For unique solution, we have

a1/a2 = b1/b2 ≠ c1/c2

a/12 = 3/a ≠ -(a – 3)/-a

 a-3 ≠ 3

 a ≠ 6

And,

a/12 = 3/a

a2 = 36

a = + 6 or – 6?

a  ≠ 6, a = – 6

Hence, when a = -6 the given set of equations will have no solution.

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: 

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

(i) For unique solution, we have

a1/a2 ≠ b1/b2

k/3 ≠ 2

k ≠ 6

Hence, when k ≠ 6 the given set of equations will have unique solution.

(ii) For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

k/3 = 2/1 ≠ -5/-1

k/3 = 2/1 

k = 6

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

Question 29. For what value of c, the following system of equation have infinitely many solutions (where c ≠ 0 c ≠ 0):

6x + 3y = c − 3

12x + cy = c

Solution: 

Given that,

6x + 3y − (c − 3) = 0  …(1)

12x + cy − c = 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 = 6, b1 = 3, c1 = −(c − 3) 

a2 = 12, b2 = c, c2 = – c

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

6/12 = 3/c = -(c + 3)/-c

6/12 = 3/c and 3/c = -(c + 3)/-c 

c = 6 and c – 3 = 3

c = 6 and c = 6

Hence, when c = 6 the given set of equations will have infinitely many solution.

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: 

Given that,

2x + ky = 1 …(1)

3x − 5y = 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 = 2, b1 = k, c1 = −1 

a2 = 3, b2 = −5, c2 = −7

(i) For unique solution, we have

a1/a2 ≠ b1/b2

2/3 ≠ -k/-5 ≠ -10/3

Hence, when k ≠ -10/3 the given set of equations will have unique solution.

(ii) For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

2/3 = k/-5 ≠ -1/-7

2/3 = k/-5 and k/-5 ≠ -1/-7

k = -10/3 and k ≠ -5/7

k = -10/3

Hence, when k = -10/3 the given set of equations will have no solution.

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

a1/a2 = b1/b2 = c1/c2

2/3 = k/-5 = -1/-7

Clearly a1/a2 ≠ c1/c2

Hence. there is no value of k for which the given set 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: 

Given that,

x + 2y + 7 = 0  …(1)

2x + ky + 14 = 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 = 7 

a2 = 2, b2 = k, c2 = 14

The given system of equation will represent the coincident

lines if they have infinitely many solutions

So, 

a1/a2 = b1/b2 = c1/c2

1/2 = 2/k = 7/14

1/2 = 2/k = 7/14

k = 4

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

Question 32. Find the value of k, for which the following system of equation have unique solution:

ax + by = c,

lx + my = n

Solution: 

Given that,

ax + by − c = 0 …(1)

lx + my − n = 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 = a, b1 = b, c1 = − c

a2 = l, b2 = m, c2 = − n

For unique solution, we have

a1/a2 ≠ b1/b2

a/l ≠ b/m

am ≠ bl

Hence, when am ≠ bl the given set of equations will have unique solution.

Question 33. Find the value of a and b such that the following system of linear equation have infinitely many solutions:

(2a − 1)x + 3y − 5 = 0,

3x + (b − 1)y − 2 = 0

Solution: 

Given that, 

(2a − 1)x + 3y − 5 = 0 

3x + (b − 1)y − 2 = 0

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

a2 = 3, b2 = b − 1, c2 = −2

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

(2a – 1)/3 = 3/(b – 1) = -5/-2

(2a – 1)/3 = 3/(b – 1) and 3/(b – 1) = -5/-2 

2(2a − 1) = 15 and 6 = 5(b − 1)

4a − 2 = 15 and 6 = 5b − 5 

4a = 17 and 5b = 11

So, a = 17/4 and b = 11/5

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: 

Given that, 

2x − 3y − 7 = 0 …(1)

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

a2 = (a + b), b2 = −(a + b − 3), c2 = −(4a + b)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a + b) = -3/-(a + b – 3) = -7/-(4a + b)

2/(a + b) = -3/-(a + b – 3) and -3/-(a + b – 3) = -7/-(4a + b)

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, and we will get,

5( -b – 6) – 4b = -21

− 5b − 30 − 4b = − 21

9b = − 9 ⇒ b = −1

As a = – 6 – b

a = − 6 + 1 = − 5

So, 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: 

Given that, 

2x − 3y − 9 = 0 …(1)

(p + q)x + (2p − q)y − 3(p + q + 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 = 3, c1 = −9 

a2 = (p + q), b2 = (2p − q), c2 = -3(p + q + 1)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(p + q) = 3/(2p – q) = -9/-3(p + q + 1)

2/(p + q) = 3/(2p – q) and 3/(2p – q) = -9/-3(p + q + 1)

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

So, 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

(vii) 2x + 3y = 7,

(a – b)x + (a + b)y = 3a + b + 2

(viii) x + 2y = 1,

(a − b)x + (a + b)y = a + b – 2

(ix) 2x + 3y = 7,

2ax + a y = 28 – by

Solution: 

(i) Given that, 

(2a − 1)x + 3y − 5 = 0 …(1)

3x + (b − 2)y − 3 = 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 = 2a − 1, b1 = 3, c1 = −5 

a2 = 3, b2 = b − 2, c2 = -3(p + q + 1)

For infinitely many solution, we have

a1/a2 = b1/b2 = c1/c2

(2a – 1)/3 = -3/(b – 2) = -5/-3

(2a – 1)/3 = -3/(b – 2) and -3/(b – 2) = -5/-3 

2a – 1 = 5 and – 9 = 5(b – 2)

a = 3 and -9 = 5b – 10  

a = 3 and b = 1/5

So, a = 3 and b = 1/5.

(ii) Given that, 

2x − (2a + 5)y = 5 …(1)

(2b + 1)x − 9y = 15 …(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 = – (2a + 5), c1 = −5 

a2 = (2b + 1), b2 = −9, c2 = −15

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(2b + 1) = (-2a + 5)/-9 = -5/-15

2/(2b + 1) = (-2a + 5)/-9 and (-2a + 5)/-9 = -5/-15

6 = 2b + 1 and 2a + 5

3 = b = 5/2 and a = -1 

So, a = – 1 and b = 5/2.

(iii) Given that, 

(a − 1)x + 3y = 2 …(1)

6x + (1 − 2b)y = 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 = a – 1, b1 = 3, c1 = −2 

a2 = 6, b2 = 1 − 2b, c2 = −6

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

(a – 1)/6 = 3/(1 – 2b) = 2/6

(a – 1)/6 = 3/(1 – 2b) and 3/(1 – 2b) = 2/6

a – 1 = 2 and 1 – 2b = 9

a – 1 = 2 and 1 – 2b = 9  

a = 3 and b = -4

a = 3 and b = -4

So, a = 3 and b = −4.

(iv) Given that, 

3x + 4y − 12 = 0 …(1)

(a + b)x + 2(a − b)y − (5a − 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 = 3, b1 = 4, c1 = −12 

a2 = (a + b), b2 = 2(a − b), c2 = – (5a − 1)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

3/(a + b) = 4/2(a – b) = 12/(5a – 1)

3/(a + b) = 4/2(a – b) and 4/2(a – b) = 12/(5a – 1)

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

a = 5b and -2a = -12b + 2

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

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

−10b = −12b + 2 ⇒ b = 1

Thus a = 5

So, a = 5 and b = 1.

(v) Given that, 

2x + 3y − 7 = 0 …(1)

(a − 1)x + (a + 1)y − (3a − 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 = 3, c1 = −7 

a2 = (a − 1), b2 = (a + 1), c2 = – (3a − 1)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a – b) = 3/(a + 1) = -7/(3a – 1)

2/(a – b) = 3/(a + 1) and 3/(a + 1) = -7/(3a – 1)  

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

So, a = 5 and b = 1.

(vi) Given that, 

2x + 3y − 7 = 0 …(1)

(a − 1)x + (a + 2)y − 3a = 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 = −7 

a2 = (a − 1), b2 = (a + 2), c2 = −3a

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a – b) = 3/(a + 2) = -7/-3a

2/(a – b) = 3/(a + 2) and 3/(a + 2) = -7/-3a

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

So, a = 7 and b = 1.

(vii) 2x + 3y – 7 = 0,  …(1)

(a – b)x + (a + b)y – 3a – b + 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 = 2, b1 = 3, c1 = −7 

a2 = (a − b), b2 = (a + b), c2 = −(3a + b – 2)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a – b) = 3/(a + b) = -7/−(3a + b – 2)

2/(a – b) = 3/(a + b) and 3/(a + b) = -7/−(3a + b – 2)

2(a + b) = 3(a – b) and 3(3a + b – 2) = 7(a + b)

2a + 2b = 3a – 3b and 9a + 3b – 6 = 7a + 7b

So, a = 5 and b = 1.

(viii) x + 2y – 1 = 0  …(1)

(a − b)x + (a + b)y – a – b + 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 = 2, c1 = −1 

a2 = (a − b), b2 = (a + b), c2 = −(a + b – 2)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

1/(a – b) = 2/(a + b) = -1/−(a + b – 2)

1/(a – b) = 2/(a + b) and 2/(a + b) = -1/−(a + b – 2)

(a + b) = 2(a – b) and 2(a + b – 2) = (a + b)

a + b = 2a – 2b and 2a + 2b – 4 = a + b

So, a = 3 and b = 1.

(ix) 2x + 3y – 7 = 0 …(1)

2ax + ay – 28 + by = 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 = −7 

a2 = 2a, b2 = (a + b), c2 = −28

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/2a = 3/(a + b) = -7/−28

2/2a = 3/(a + b) and 3/(a + b) = 7/28

2(a + b) = 6a and 84 = 7(a + b)

So, a = 4 and b = 8.

Question 37. For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have

(i) no solution?

(ii) infinitely many solutions?

(iii) a unique solution? 

Solution: 

Given that, 

λx + y = λ2 …(1)

x + λy = 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 = λ, b1 = 1, c1 = -λ2 

a2 = 1, b2 = λ, c2 = -λ2 

(i) For no solution,

a1/a2 = b1/b2 ≠ c1/c2

λ = 1/λ ≠ -λ2/-1

λ2 – 1 = 0

λ = 1, -1

Here we will take only λ = -1 because at λ = 1 the linear 

equation will have infinite many solutions.

(ii) For infinite many solutions,

 a1/a2 = b1/b2 = c1/c2

λ = 1/λ = λ2 /1

λ(λ – 1) = 0

When λ ≠ 0 then λ = 1

(iii) For Unique solution,

a1/a2 ≠ b1/b2

λ ≠ 1/λ

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