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

Last Updated : 07 Apr, 2021

### (k + 2)x + (2k + 1)y = 2(k âˆ’ 1)

Solution:

Given that,

2x + 3y = 2      …(1)

(k + 2)x + (2k + 1)y = 2(k âˆ’ 1)      …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0       …(3)

a2x + b2y âˆ’ c2 = 0      …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = âˆ’2

a2 = (k + 2), b2 = (2k + 1), c2 = âˆ’2(k âˆ’ 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 âˆ’ 1) = (2k + 1)

= 4k + 2 = 3k + 6 and 3k âˆ’ 3 = 2k + 1

= k = 4 and k = 4

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

### (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 âˆ’ c1 = 0      …(3)

a2x + b2y âˆ’ c2 = 0     …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 1, b1 = (k + 1), c1 = âˆ’4

a2 = (k + 1), b2 = 9, c2 = âˆ’ (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 âˆ’ 8 = 0 and 5k2 + 7k âˆ’ 34 = 0

k2 + 4k âˆ’ 2k âˆ’ 8 = 0 and 5k2 + 17k âˆ’ 10k âˆ’ 34 = 0

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

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

### 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 âˆ’ c1 = 0      …(3)

a2x + b2y âˆ’ c2 = 0     …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = k, b1 = 3, c1 = âˆ’(2k + 1)

a2 = 2(k + 1), b2 = 9, c2 = âˆ’(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 Ã— 2(k + 1) and 3(7k + 1) = 9(2k + 1)

9k âˆ’ 6k = 6 and 21k âˆ’ 18k = 9 âˆ’ 3

3k = 6 â‡’ k = 2 and k = 2

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

### 6x + (2k âˆ’ 1)y = (2k + 5)

Solution:

Given that,

2x +( k âˆ’ 2)y = k    …(1)

6x + (2k âˆ’ 1)y = (2k + 5)     …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0     …(3)

a2x + b2y âˆ’ c2 = 0     …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = (k âˆ’ 2), c1 = âˆ’k

a2 = 6, b2 = (2k âˆ’ 1), c2 = âˆ’(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 âˆ’ 3k = âˆ’6 + 1 and k + k = 10

âˆ’k = âˆ’5 and 2k = 10 = k = 5 and k = 5

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

### (k + 1)x + (2k âˆ’ 1)y = (4k + 1)

Solution:

Given that,

2x + 3y = 7     …(1)

(k + 1)x + (2k âˆ’ 1)y = (4k + 1)    …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0     …(3)

a2x + b2y âˆ’ c2 = 0    …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = âˆ’7

a2 = k + 1, b2 = 2k âˆ’ 1, c2 = âˆ’(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 âˆ’ 2 = 3k + 3 and 12k + 3 = 14k âˆ’ 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.

### (k âˆ’ 1)x + (k + 2)y = 3k

Solution:

Given that,

2x + 3y = k     …(1)

(k âˆ’ 1)x + (k + 2)y = 3k    …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0     …(3)

a2x + b2y âˆ’ c2 = 0    …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = âˆ’k

a2 = k âˆ’ 1, b2 = k + 2, c2 = âˆ’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 âˆ’ 3 and 9 = k + 2

2k + 4 = 3k âˆ’ 3 and 9 = k + 2 â‡’ k = 7 and k = 7

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

### 6x + 2y = 7

Solution:

Given that,

kx âˆ’ 5y = 2    …(1)

6x + 2y = 7   …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0    …(3)

a2x + b2y âˆ’ c2 = 0   …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = k, b1 = âˆ’5, c1 = âˆ’2

a2 = 6 b2 = 2, c2 = âˆ’7

For no solution, we have

a1/a2 = b1/b2 â‰  c1/c2

1/2 = 2/k â‰  2/7

k = 4

2k = -30

k = -15

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

### 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 âˆ’ c1 = 0    …(3)

a2x + b2y âˆ’ 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 = âˆ’5

For no solution, we have

a1/a2 = b1/b2 â‰  c1/c2

k/6 = -5/2 â‰  2/7

k = 4

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

### kx + 3y âˆ’ 5 = 0

Solution:

Given that,

3x âˆ’ 4y + 7 = 0   …(1)

kx + 3y âˆ’ 5 = 0  …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0    …(3)

a2x + b2y âˆ’ c2 = 0  …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 3, b1 = âˆ’4, c1 = 7

a2 = k, b2 = 3, c2 = âˆ’5

For no solution, we have

a1/a2 = b1/b2 â‰  c1/c2

3/k = -4/3

k = -9/4

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

### 3x + 2y âˆ’ 1 = 0

Solution:

Given that,

2x âˆ’ ky + 3 = 0  …(1)

3x + 2y âˆ’ 1 = 0   …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0   …(3)

a2x + b2y âˆ’ c2 = 0   …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = âˆ’k, c1 = 3

a2 = 3, b2 = 2, c2 = âˆ’1

For no solution, we have

a1/a2 = b1/b2 â‰  c1/c2

2/3 = -k/2

k = -4/3

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

### 5x âˆ’ 7y âˆ’ 5 = 0

Solution:

Given that,

2x + ky âˆ’ 11 = 0   …(1)

5x âˆ’ 7y âˆ’ 5 = 0   …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0   …(3)

a2x + b2y âˆ’ c2 = 0   …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = k, c1 = âˆ’11

a2 = 5, b2 = âˆ’7, c2 = âˆ’5

For no solution, we have

a1/a2 = b1/b2 â‰  c1/c2

2/5 = -k/-7

k = -14/5

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

### 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 âˆ’ c1 = 0  …(3)

a2x + b2y âˆ’ c2 = 0  …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = k, b1 = 3, c1 = âˆ’3

a2 = 12, b2 = k, c2 = âˆ’ 6

For no solution, we have

a1/a2 = b1/b2 â‰  c1/c2

k/12 = 3/k â‰  3/6    …(5)

k2 = 36 â‡’ k = + 6 or âˆ’6

From eq (5), we get

k/12 â‰  3/6

k â‰  6

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

### 2x + ay âˆ’ 7 = 0

Solution:

Given that,

4x + 6y âˆ’ 11 = 0 …(1)

2x + ay âˆ’ 7 = 0 …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ c1 = 0 …(3)

a2x + b2y âˆ’ c2 = 0 …(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 4, b1 = 6, c1 = âˆ’11

a2 = 2, b2 = a, c2 = âˆ’7

For inconsistent solution, we have

a1/a2 = b1/b2 â‰  c1/c2

a1/a2 = b1/b2

4/2 = 6/a

a = 3

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

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