Ex-3.5 Pair Of Linear Equations In Two Variables (Part - 2), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

(21) x + 2y = 0

    2x + ky = 5

Sol.

The given system may be written as

x2y = 0

2x + ky = 5

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 1,b₁ = 2,c₁ = 0

a₂ = 2,b₂ = k,c₂ = − 5

For no solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

1/2 = 2/k ≠ 2/7

⇒ k = 4

Therefore, the given system of equations will have no solutions, if k = 4.

(22) 3x − 4y + 7 = 0

    kx + 3y − 5 = 0

Sol.

The given system may be written as

3x − 4y + 7 = 0

kx + 3y − 5 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 3,b₁ = − 4,c₁ = 7

a₂ = k,b₂ = 3,c₂ = − 5

For no solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

3/k = − 4/3

⇒ k = − 9/4

Therefore, the given system of equations will have no solutions, if k = − 9/4.
 

(23) 2x − ky + 3 = 0

    3x + 2y − 1 = 0

Sol.

The given system may be written as

2x − ky + 3 = 0

3x + 2y − 1 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = − k,c₁ = 3

a₂ = 3,b₂ = 2,c₂ = − 1

For no solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

2/3 = − k/2

⇒ k = − 4/3

Therefore, the given system of equations will have no solutions, if k = − 4/3.

(24) 2x + ky − 11 = 0

    5x − 7y − 5 = 0

Sol.

The given system may be written as

2x + ky − 11 = 0

5x − 7y − 5 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = k,c₁ = − 11

a₂ = 5,b₂ = − 7,c₂ = − 5

For no solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

2/5 = − k /− 7

⇒ k = − 14/5

Therefore, the given system of equations will have no solutions, if k = − 14/5.

(25) kx + 3y = 3

    12x + ky = 6

Sol.

The given system may be written as

kx + 3y = 3

12x + ky = 6

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = k,b₁ = 3,c₁ = − 3

a₂ = 12,b₂ = k,c₂ = − 6

For no solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

k/12 = 3/k ≠ 3/6    ……(i)

⇒ k² = 36

⇒ k = + 6or − 6

From (i)

k/12 ≠ 3/6

⇒ k ≠ 6

Therefore, the given system of equations will have no solutions, if k = − 6.

(26) For what value of a, the following system of equation will be inconsistent?

4x + 6y − 11 = 0

2x + ay − 7 = 0

Sol.

The given system may be written as

4x + 6y − 11 = 0

2x + ay − 7 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 4,b₁ = 6,c₁ = − 11

a₂ = 2,b₂ = a,c₂ = − 7

For unique solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

a₁/a₂ = b₁/b₂

4/2 = 6/a

⇒ a = 3

Therefore, the given system of equations will be inconsistent, if a = 3.

(27) For what value of a, the following system of equation have no solution?

ax + 3y = a − 3

12x + ay = a

Sol.

The given system may be written as

ax + 3y = a − 3

12x + ay = a

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = a,b₁ = 3,c₁ = − (a − 3)

a₂ = 12,b₂ = a,c₂ = − a

For unique solution,we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

3/a ≠ frac − (a − 3) − a

⇒ a − 3 ≠ 3

⇒ a ≠ 6

and ,

a/12 = 3/a

⇒ a² = 36

⇒ a = + 6or − 6

∵ a ≠ 6

⇒ a = − 6

Therefore, the given system of equations will have no solution, if a = − 6.

(28) Find the value of a,for which the following system of equation have

(i) Unique solution

(ii) No solution

kx + 2y = 5

3x + y = 1

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

(i) For unique solution, we have

a₁/a₂ ≠ b₁/b₂

k/3 ≠ 2/1

k ≠ 6

Therefore, the given system of equations will have unique solution, if k ≠ 6.

(ii) For no solution, we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

k/3 = 2/1 ≠ − 5/ − 1

k/3 = 2/1

⇒ k = 6

Therefore, the given system of equations will have no solution, if a = 6.

(29) For what value of c, the following system of equation have infinitely many solution (where c ≠ 0 )?

6x + 3y = c − 3

12x + cy = c

Sol.

The given system may be written as

6x + 3y − (c − 3) = 0

12x + cy − c = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 6,b₁ = 3,c₁ = − (c − 3)

a₂ = 12,b₂ = c,c₂ = − c

For infinitely many solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ c = 6 and c − 3 = 3

⇒ c = 6 and c = 6

Therefore, the given system of equations will have infinitely many solution, if c = 6.

(30) Find the value of k,for which the following system of equation have

(i) Unique solution

(ii) No solution

(iii) Infinitely many solution

2x + ky = 1

3x − 5y = 7

Sol.

The given system may be written as

2x + ky = 1

3x − 5y = 7

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, 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

k ≠ − 10/3

Therefore, the given system of equations will have unique solution, if k ≠ − 10/3.

(ii) For no solution, we have

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

2/3 = k/ − 5 neq − 1/ − 7

2/3 = k/ − 5 and k /− 5 neq1 7

⇒ k = − 10/3 and k neq − 5/7

⇒ k = − 10/3

Therefore, the given system of equations will have no solution, if k = − 10/3.

(iii) For the given system to have infinitely many solution,we have

a₁/a₂ = b₁/b₂ = c₁/c₂

2/3 = k/ − 5 = − 1/ − 7

Clearly a₁a₂ ≠ c₁c₂ ,

So there is no value of k for which the given system of equation has infinitely many solution.

(31) For what value of k, the following system of equation will represent the coincident lines?

x + 2y + 7 = 0

2x + ky + 14 = 0

Sol.

The given system may be written as

x + 2y + 7 = 0

2x + ky + 14 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, 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 solution.

a₁/a₂ = b₁/b₂ = c₁/c₂

1/2 = 2/k = 7/14

1/2 = 2/k = 1/2

⇒ k = 4

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

(32) (30) Find the value of k,for which the following system of equation have unique solution.

ax + by = c

lx + my = n

Sol.

The given system may be written as

ax + by − c = 0

lx + my − n = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = a,b₁ = b,c₁ = − c

a₂ = l,b₂ = m,c₂ = − n

For unique solution, we have

a₁/a₂ ≠ b_1/b₂

⇒ a/l ≠ b/m

⇒ a/m ≠ b/l

Therefore, the given system of equations will have unique solution, if a/m ≠ b/l.

(33) Find the value of a and b such that the following system of linear equation have infinitely many solution:

(2a − 1)x + 3y − 5 = 0

3x + (b − 1)y − 2 = 0

Sol.

The given system of equation may be written as,

(2a − 1)x + 3y − 5 = 0

3x + (b − 1)y − 2 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = (2a − 1),b₁ = 3,c₁ = − 5

a₂ = 3,b₂ = b − 1,c₂ = − 2

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)

⇒ 4a − 2 = 15 and 6 = 5b − 5

⇒ 4a = 17 and 5b = 11

⇒ a = 17/4 and b = 11/5

(34) Find the value of a and b such that the following system of linear equation have infinitely many solution:

2x − 3y = 7

(a + b)x − (a + b − 3)y = 4a + b

Sol.

The given system of equation may be written as,

2x − 3y − 7 = 0

(a + b)x − (a + b − 3)y − (4a + b) = 0

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 = (a + b),b₂ = − (a + b − 3),c₂ = − (4a + b)

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 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 + b = -6

⇒ a = − 6 − b

Substituting the value of a in 5a − 4b = − 21 we have

5(-b-6)-4b = -21

⇒ − 5b − 30 − 4b = − 21

⇒ 9b = − 9

⇒ b = − 1

As a = -6-b

⇒ a = − 6 + 1 = − 5

Hence the given system of equation will have infinitely many solution if

a = -5 and b = -1.

(35) Find the value of p and q such that the following system of linear equation have infinitely many solution:

2x − 3y = 9

(p + q)x + (2p − q)y = 3(p + q + 1)

Sol.

The given system of equation may be written as,

2x − 3y − 9 = 0

(p + q)x + (2p − q)y − 3(p + q + 1) = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = 3,c₁ = − 9

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

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

 and 

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

Hence the given system of equation will have infinitely many solution if

p = 5/3  and  q = 1/3.

(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

Sol.

The given system of equation may be written as,

(2a − 1)x + 3y − 5 = 0

3x + (b − 2)y − 3 = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b2y − c₂ = 0

Where, a₁ = 2a − 1,b₁ = 3,c₁ = − 5

a₂ = 3,b₂ = b − 2,c₂ = − 3(p + q + 1)

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

2a − 1 = 5 and − 9 = 5(b − 2)

⇒ a = 3 and − 9 = 5b − 10

⇒ a = 3 and b = 1/5

Hence the given system of equation will have infinitely many solution if

a = 3  and  b = 1/5.

(ii) 2x − (2a + 5)y = 5

(2b + 1)x − 9y = 15

Sol.

The given system of equation may be written as,

2x − (2a + 5)y = 5

(2b + 1)x − 9y = 15

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 2,b₁ = − (2a + 5),c₁ = − 5

a₂ = (2b + 1),b₂ = − 9,c₂ = − 15

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 6 = 2b + 1 and 2a + 5 = 3

⇒ b = 5/2 and a = − 1

Hence the given system of equation will have infinitely many solution if

a = − 1  and  b = 5/2.

(iii) (a − 1)x + 3y = 2

6x + (1 − 2b)y = 6

Sol.

The given system of equation may be written as,

(a − 1)x + 3y = 2

6x + (1 − 2b)y = 6

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = a − 1,b₁ = 3,c₁ = − 2

a₂ = 6,b₂ = 1 − 2b,c₂ = − 6

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ a − 1 = 2 and 1 − 2b = 9

⇒ a = 3 and b = − 4

Hence the given system of equation will have infinitely many solution if

a = 3  and  b = − 4.

(iv) 3x + 4y = 12

(a + b)x + 2(a − b)y = 5a − 1

Sol.

The given system of equation may be written as,

3x + 4y − 12 = 0

(a + b)x + 2(a − b)y − (5a − 1) = 0

The given system of equation is of the form

a₁x + b₁y − c₁ = 0

a₂x + b₂y − c₂ = 0

Where, a₁ = 3,b₁ = 4,c₁ = − 12

a₂ = (a + b),b₂ = 2(a − b),c₂ = − (5a − 1)

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 3(a − b) = 2a + 2b and 2(5a − 1) = 12(a − b)

⇒ a = 5b and − 2a = − 12b + 2

Substituting a = 5b in -2a = -12b + 2 , we have

-2(5b) = -12b + 2

⇒ − 10b = − 12b + 2

⇒ b = 1

Thus a = 5

Hence the given system of equation will have infinitely many solution if

a = 5  and  b = 1.

(v) 2x + 3y = 7

(a − 1)x + (a + 1)y = 3a − 1

Sol.

The given system of equation may be written as,

2x + 3y − 7 = 0

(a − 1)x + (a + 1)y − (3a − 1) = 0

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 = (a − 1),b₂ = (a + 1),c₂ = − (3a − 1)

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 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

Hence the given system of equation will have infinitely many solution if

a = 5  and  b = 1.

(vi) 2x + 3y = 7

(a − 1)x + (a + 2)y = 3a

Sol.

The given system of equation may be written as,

2x + 3y − 7 = 0

(a − 1)x + (a + 2)y − 3a = 0

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 = (a − 1),b₂ = (a + 2),c₂ = − 3a

The given system of equation will have infinitely many solution, if

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ 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

Hence the given system of equation will have infinitely many solution if

a = 7  and  b = 1..