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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

### 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.

### 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.

### 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.

### 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.

### 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.

### 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.

### 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

### (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.

### (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.

### (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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