Ex-3.5 Pair Of Linear Equations In Two Variables (Part - 1), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, find it from 1 to 4:

(1) x − 3y − 3 = 0

   3x − 9y − 2 = 0

Sol.

The given system may be written as

x − 3y − 3 = 0

3x − 9y − 2 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 1,b₁ = − 3,c₁ = − 3

a₂ = 3,b₂ = − 9,c₂ = − 2

We have,

a₁/a₂ = 1/3

b₁/b₂ = − 3/ − 9 = 13

and , c₁/c₂ = − 3/ − 2 = 3/2

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, the given equation has no solution.

(2) 2x + y − 5 = 0

   4x + 2y − 10 = 0

Sol.

The given system may be written as

2x + y − 5 = 0

4x + 2y − 10 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = 1,c₁ = − 5

a₂ = 4,b₂ = 2,c₂ = − 10

We have,

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 1/2

and , c₁/c₂ = − 5/ − 10 = 1/2

So, a₁/a₂ = b₁/b₂ = c₁/c₂

Therefore, the given equation has infinitely many solution.

(3) 3x − 5y = 20

   6x − 10y = 40

Sol.

The given system may be written as

3x − 5y = 20

6x − 10y = 40

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 3,b₁ = − 5,c₁ = − 20

a₂ = 6,b₂ = − 10,c₂ = − 40

We have,

a₁/a₂ = 3/6 = 1/2

b₁/b₂ = − 5/ − 10 = 1/2

and , c₁/c₂ = − 20 /− 40 = 1/2

So, a₁/a₂ = b₁/b₂ = c1/c₂

Therefore, the given equation has infinitely many solution.

(4) x − 2y − 8 = 0

   5x − 10y − 10 = 0

Sol.

The given system may be written as

x − 2y − 8 = 0

5x − 10y − 10 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 1,b₁ = − 2,c₁ = − 8

a₂ = 5,b₂ = − 10,c₂ = − 10

We have,

a₁/a₂ = 1/5

b₁/b₂ = − 2 /− 10 = 1/5

and , c₁/c₂ = − 8/ − 10 = 4/5

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, the given equation has no solution.

Find the value of k for each of the following system of equations which have a unique solution (5-8)

(5) kx + 2y − 5 = 0

     3x + y − 1 = 0

Sol.

The given system may be written as

kx + 2y − 5 = 0

3x + y − 1 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = k,b₁ = 2,c₁ = − 5

a₂ = 3,b₂ = 1,c₂ = − 1

For unique solution,we have

a₁/a₂ ≠ b₁/b₂

k/3 ≠ 2/1

⇒ k ≠ 6

Therefore, the given system will have unique solution for all real values of k other than 6.

(6) 4x + ky + 8 = 0

    2x + 2y + 2 = 0

Sol.

The given system may be written as

4x + ky + 8 = 0

2x + 2y + 2 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 4,b₁ = k,c₁ = 8

a₂ = 2,b₂ = 2,c₂ = 2

For unique solution,we have

a₁/a₂ ≠ b₁/b₂

4/2 ≠ k/2

⇒ k ≠ 4

Therefore, the given system will have unique solution for all real values of k other than 4.

(7) 4x − 5y = k

    2x − 3y = 12

Sol.

The given system may be written as

4x − 5y − k = 0

2x − 3y − 12 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 4,b₁ = − 5,c₁ = − k

a₂ = 2,b₂ = − 3,c₂ = − 12

For unique solution,we have

a₁a₂ ≠ b₁b₂

4/2 ≠ − 5/ − 3

⇒ k can have any real values.

Therefore, the given system will have unique solution for all real values of k.

(8) x + 2y = 3

    5x + ky + 7 = 0

Sol.

The given system may be written as

x + 2y = 3

5x + ky + 7 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 1,b₁ = 2,c₁ = − 3

a₂ = 5,b₂ = k,c₂ = 7

For unique solution,we have

a₁a₂ ≠ b₁b₂

1/5 ≠ 2/k

⇒ k ≠ 10

Therefore, the given system will have unique solution for all real values of k other than 10.

Find the value of k for which each of the following system of equations having infinitely many solution: (9-19)

(9) 2x + 3y − 5 = 0

    6x − ky − 15 = 0

Sol.

The given system may be written as

2x + 3y − 5 = 0

6x − ky − 15 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, 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

Therefore, the given system of equation will have infinitely many solutions, if k = 9.

(10) 4x + 5y = 3

    kx + 15y = 9

Sol.

The given system may be written as

4x + 5y = 3

kx + 15y = 9

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, 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

Therefore, the given system will have infinitely many solutions if k = 12.

(11) kx − 2y + 6 = 0

    4x + 3y + 9 = 0

Sol.

The given system may be written as

kx − 2y + 6 = 0

4x + 3y + 9 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, 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

Therefore, the given system of equations will have infinitely many solutions, if k = 8/3.

(12) 8x + 5y = 9

    kx + 10y = 19

Sol.

The given system may be written as

8x + 5y = 9

kx + 10y = 19

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 8,b₁ = 5,c₁ = − 9

a₂ = k,b₂ = 10,c₂ = − 18

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

8/k = 5/10 = − 9 /− 18 = 1/2

⇒ k = 16

Therefore, the given system of equations will have infinitely many solutions, if k = 16.

(13) 2x − 3y = 7

    (k + 2)x − (2k + 1)y = 3(2k − 1)

Sol.

The given system may be written as

2x − 3y = 7

(k + 2)x − (2k + 1)y = 3(2k − 1)

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = − 3,c₁ = − 7

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

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

 and 

⇒ 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

Therefore, the given system of equations will have infinitely many solutions, if k = 4.

(14) 2x + 3y = 2

    (k + 2)x + (2k + 1)y = 2(k − 1)

Sol.

The given system may be written as

2x + 3y = 2

(k + 2)x + (2k + 1)y = 2(k − 1)

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = 3,c₁ = − 2

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

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

 and 

⇒ 2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1)

⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1

⇒ k = 4 and k = 4

Therefore, the given system of equations will have infinitely many solutions, if k = 4.

(15) x + (k + 1)y = 4

    (k + 1)x + 9y = (5k + 2)

Sol.

The given system may be written as

x + (k + 1)y = 4

(k + 1)x + 9y = (5k + 2)

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 1,b₁ = (k + 1),c₁ = − 4

a₂ = (k + 1),b₂ = 9,c₂ = − (5k + 2)

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 9 = (k + 1)² and (k + 1)(5k + 2) = 36

⇒ 9 = k² + 2k + 1 and 5k² + 2k + 5k + 2 = 36

⇒ k² + 2k − 8 = 0 and 5k² + 7k − 34 = 0

⇒ k² + 4k − 2k − 8 = 0 and 5k² + 17k − 10k − 34 = 0

⇒ k(k + 4) − 2(k + 4) = 0 and (5k + 17) − 2(5k + 17) = 0

⇒ (k + 4)(k − 2) = 0 and (5k + 17)(k − 2) = 0

⇒ k = − 4 ork = 2 and k = − 17/5 ork = 2

thus, k = 2 satisfies both the condition.

Therefore, the given system of equations will have infinitely many solutions, if k = 2.

(16) kx + 3y = 2k + 1

    2(k + 1)x + 9y = (7k + 1)

Sol.

The given system may be written as

kx + 3y = 2k + 1

2(k + 1)x + 9y = (7k + 1)

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = k,b₁ = 3,c₁ = − (2k + 1)

a₂ = 2(k + 1),b₂ = 9,c₂ = − (7k + 1)

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 9k = 3×2(k + 1) and 3(7k + 1) = 9(2k + 1)

⇒ 9k − 6k = 6 and 21k − 18k = 9 − 3

⇒ 3k = 6 and 3k = 6

⇒ k = 2 and k = 2

Therefore, the given system of equations will have infinitely many solutions, if k = 2.

(17) 2x + (k − 2)y = k

    6x + (2k − 1)y = (2k + 5)

Sol.

The given system may be written as

2x + (k − 2)y = k

6x + (2k − 1)y = (2k + 5)

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = (k − 2),c₁ = − k

a₂ = 6,b₂ = (2k − 1),c₂ = − (2k + 5)

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

 and 2k² + 5k − 4k − 10 = 2k² − k

⇒ 2k − 3k = − 6 + 1 and k + k = 10

⇒ − k = − 5 and 2k = 10

⇒ k = 5 and k = 5

Therefore, the given system of equations will have infinitely many solutions, if k = 5.

(18) 2x + 3y = 7

    (k + 1)x + (2k − 1)y = (4k + 1)

Sol.

The given system may be written as

2x + 3y = 7

(k + 1)x + (2k − 1)y = (4k + 1)

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = 3,c₁ = − 7

a₂ = k + 1,b₂ = 2k − 1,c₂ = − (4k + 1)

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

Extra close brace or missing open brace

⇒ 4k − 2 = 3k + 3 and 12k + 3 = 14k − 7

⇒ k = 5 and 2k = 10

⇒ k = 5 and k = 5

Therefore, the given system of equations will have infinitely many solutions, if k = 5.

(19) 2x + 3y = k

    (k − 1)x + (k + 2)y = 3k

Sol.

The given system may be written as

2x + 3y = k

(k − 1)x + (k + 2)y = 3k

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = 3,c₁ = − k

a₂ = k − 1,b₂ = k + 2,c₂ = − 3k

For unique solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

Extra close brace or missing open brace

⇒ 2k + 4 = 3k − 3 and 9 = k + 2

⇒ k = 7 and k = 7

Therefore, the given system of equations will have infinitely many solutions, if k = 7.

Find the value of k for which the following system of equation has no solution : (20-25)

(20) kx − 5y = 2

    6x + 2y = 7

Sol.

The given system may be written as

kx − 5y = 2

6x + 2y = 7

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = k,b₁ = − 5,c₁ = − 2

a₂ = 6,b₂ = 2,c₂ = − 7

For no solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

k/6 = − 5/2 ≠ 2/7

⇒ 2k = − 30

⇒ k = − 15

Therefore, the given system of equations will have no solutions, if k = − 15.