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

### 3x − 9y − 2 = 0

Solution:

Given that,

x − 3y − 3 = 0     …(1)

3x − 9y − 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 = −3, c1 = −3

a2 = 3, b2 = −9, c2 = −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 ≠ c1/c2

Hence, the given set of equations has no solution.

### 4x + 2y − 10 = 0

Solution:

Given that,

2x + y − 5 = 0     …(1)

4x + 2y − 10 = 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 = 1, c1 = −5 and

a2 = 4, b2 = 2, c2 = −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 − 10y = 40

Solution:

Given that,

3x − 5y = 20    …(1)

6x − 10y = 40    …(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 = −5, c1 = − 20

a2 = 6, b2 = −10, c2 = − 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 − 10y − 10 = 0

Solution:

Given that,

x − 2y − 8 = 0    …(1)

5x − 10y − 10 = 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 = −8

a2 = 5, b2 = −10, c2 = −10

Lets check the equation’s,

a1/a2 = 1/5

b1/b2 = -2/-10 and

c1/c2 = -8/-10

Therefore, a1/a2 = b1/b2 ≠ c1/c2

Hence, the given set of equations has no solution.

### 3x + y − 1 = 0

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

For unique solution,

a1/a2 ≠ b1/b2

k/3 ≠ 2/1

k ≠ 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 − c1 = 0     …(3)

a2x + b2y − 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 ≠ b1/b2

4/2 ≠ k/2

k ≠ 4

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

### 2x − 3y = 12

Solution:

Given that,

4x − 5y − k = 0    …(1)

2x − 3y − 12 = 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 = 4, b1 = −5, c1 = −k

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

For unique solution,

a1/a2 ≠ b1/b2

4/2 ≠ -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 − 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 = −3

a2 = 5, b2 = k, c2 = 7

For unique solution,

a1/a2 ≠ b1/b2

1/5 ≠ 2/k

k ≠ 10

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

### 6x − ky − 15 = 0

Solution:

Given that,

2x + 3y − 5 = 0   …(1)

6x − ky − 15 = 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 = −5

a2 = 6, b2 = k, c2 = −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 − c1 = 0   …(3)

a2x + b2y − 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 − 2y + 6 = 0 …(1)

4x + 3y + 9 = 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 = 6

a2 = 4, b2 = −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 − c1 = 0 …(3)

a2x + b2y − c2 = 0 …(4)

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

a1 = 8, b1 = 5, c1 = −9

a2 = k, b2 = 10, c2 = −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 − (2k + 1)y = 3(2k − 1)

Solution:

Given that,

2x − 3y = 7 …(1)

(k + 2)x − (2k + 1)y = 3(2k − 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 = 2, b1 = −3, c1 = −7

a2 = k, b2 = − (2k + 1), c2 = −3(2k − 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 × 3(2k − 1) = 7(2k + 1)

= 4k + 2 = 3k + 6 and 18k − 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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