# Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5 | Set 1

Last Updated : 07 Apr, 2021

### 3x âˆ’ 9y âˆ’ 2 = 0

Solution:

Given that,

x âˆ’ 3y âˆ’ 3 = 0     …(1)

3x âˆ’ 9y âˆ’ 2 = 0     …(2)

So, the given equations are in the form of:

a1x + b1y âˆ’ 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 = âˆ’3, c1 = âˆ’3

a2 = 3, b2 = âˆ’9, c2 = âˆ’2

Let’s check the equation’s,

a1/a2 = 1/3

b1/b2 = -3/-9 = 1/3

c1/c2 = -3/-9 = 3/2

a1/a2 = b1/b2 â‰  c1/c2

Hence, the given set of equations has no solution.

### 4x + 2y âˆ’ 10 = 0

Solution:

Given that,

2x + y âˆ’ 5 = 0     …(1)

4x + 2y âˆ’ 10 = 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 = 1, c1 = âˆ’5 and

a2 = 4, b2 = 2, c2 = âˆ’10

Lets check the equation’s,

a1/a2 = 2/4 = 1/2

b1/b2 = 1/2

and c1/c2 = -5/-10 = 1/2

Therefore, a1/a2 = b1/b2 = c1/c2

Hence, the given set of equations has infinitely many solutions.

### 6x âˆ’ 10y = 40

Solution:

Given that,

3x âˆ’ 5y = 20    …(1)

6x âˆ’ 10y = 40    …(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’5, c1 = âˆ’ 20

a2 = 6, b2 = âˆ’10, c2 = âˆ’ 40

Lets check the equation’s,

a1/a2 = 3/6 = 1/2

b1/b2 = -5/-10 – 1/2 and

c1/c2 = -20/-40 = 1/2

Therefore, a1/a2 = b1/b2 = c1/c2

Hence, the given set of equations has infinitely many solutions.

### 5x âˆ’ 10y âˆ’ 10 = 0

Solution:

Given that,

x âˆ’ 2y âˆ’ 8 = 0    …(1)

5x âˆ’ 10y âˆ’ 10 = 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 = 1, b1 = âˆ’2, c1 = âˆ’8

a2 = 5, b2 = âˆ’10, c2 = âˆ’10

Lets check the equation’s,

a1/a2 = 1/5

b1/b2 = -2/-10 and

c1/c2 = -8/-10

Therefore, a1/a2 = b1/b2 â‰  c1/c2

Hence, the given set of equations has no solution.

### 3x + y âˆ’ 1 = 0

Solution:

Given that,

kx + 2y âˆ’ 5 = 0    …(1)

3x + y 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 = k, b1 = 2, c1 = âˆ’5

a2 = 3, b2 = 1, c2 = âˆ’1

For unique solution,

a1/a2 â‰  b1/b2

k/3 â‰  2/1

k â‰  6

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

### 2x + 2y + 2 = 0

Solution:

Given that,

4x + ky + 8 = 0    …(1)

2x + 2y + 2 = 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 = k, c1 = 8

a2 = 2, b2 = 2, c2 = 2

For unique solution,

a1/a2 â‰  b1/b2

4/2 â‰  k/2

k â‰  4

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

### 2x âˆ’ 3y = 12

Solution:

Given that,

4x âˆ’ 5y âˆ’ k = 0    …(1)

2x âˆ’ 3y âˆ’ 12 = 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 = âˆ’5, c1 = âˆ’k

a2 = 2, b2 = -3, c2 = -12

For unique solution,

a1/a2 â‰  b1/b2

4/2 â‰  -5/-3

Here, k can have any real values.

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

### 5x + ky + 7 = 0

Solution:

Given that,

x + 2y = 3    …(1)

5x + ky + 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 = 1, b1 = 2, c1 = âˆ’3

a2 = 5, b2 = k, c2 = 7

For unique solution,

a1/a2 â‰  b1/b2

1/5 â‰  2/k

k â‰  10

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

### 6x âˆ’ ky âˆ’ 15 = 0

Solution:

Given that,

2x + 3y âˆ’ 5 = 0   …(1)

6x âˆ’ ky âˆ’ 15 = 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 = 3, c1 = âˆ’5

a2 = 6, b2 = k, c2 = âˆ’15

For unique solution,

We have

a1/a2 = b1/b2 = c1/c2

2/6 = 3/k

k = 9

Hence, when k = 9 the given set of equations will have infinitely many solutions.

### x + 15y = 9

Solution:

Given that,

4x + 5y = 3   …(1)

kx +15y = 9  …(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 = 5, c1 = 3

a2 = k, b2 = 15, c2 = 9

For unique solution,

We have

a1/a2 = b1/b2 = c1/c2

4/k = 5/15 = -3/-9

4/k = 1/3

k = 12

Hence, when k = 12 the given set of equations will have infinitely many solutions.

### 4x + 3y + 9 = 0

Solution:

Given that,

kx âˆ’ 2y + 6 = 0 …(1)

4x + 3y + 9 = 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 = k, b1 = âˆ’2, c1 = 6

a2 = 4, b2 = âˆ’3, c2 = 9

For unique solution

We have

a1/a2 = b1/b2 = c1/c2

k/4 = -2/-3 = 2/3

k = 8/3

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

### kx + 10y = 19

Solution:

Given that,

8x + 5y = 9 …(1)

kx + 10y = 19 …(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 = 8, b1 = 5, c1 = âˆ’9

a2 = k, b2 = 10, c2 = âˆ’19

For unique solution

We have

a1/a2 = b1/b2 = c1/c2

8/k = 5/10 = k = 16

Hence, when k = 16 the given set of equations will have infinitely many solutions.

### (k + 2)x âˆ’ (2k + 1)y = 3(2k âˆ’ 1)

Solution:

Given that,

2x âˆ’ 3y = 7 …(1)

(k + 2)x âˆ’ (2k + 1)y = 3(2k 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 = âˆ’3, c1 = âˆ’7

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

Now, for unique solution

We have

a1/a2 = b1/b2 = c1/c2

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

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

= 2(2k + 1) = 3(k + 2) and 3 Ã— 3(2k âˆ’ 1) = 7(2k + 1)

= 4k + 2 = 3k + 6 and 18k âˆ’ 9 = 14k + 7

= k = 4 and 4k = 16

= k = 4

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

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