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USE CODE: SELF10

Chapter 13: Linear Equation in Two Variables Exercise – 13.2

Quesiton: 1

Write two solutions for each of the following equations:

(i) 5x - 2y = 7 

(ii) x = 6y 

(iii)  x + πy = 4 

(iv) (2/3)x - y = 4. 

Solution:

(i) We are given,

3x + 4y = 7

Substituting x = 1

In the given equation, We get

3 ×1 + 4y = 7

4y = 7- 3 - 4 = 4Y

Y = 1

Thus x = 1 and y = 1 is the solution of 3x + 4y = 7

Substituting x = 2 in the given equation, we get

3× 2 + 4y = 7

4y = 7 - 6

y = 1/4

Thus x = 2 and y = 1/4 is the solution of 3x + 4y = 7 
(ii) We are given, x = 6y

Substituting x = 0 in the given equation, we get 0 = 6y

6 y = 0

y = 0

Thus x = 0, ⟹ Solution (0, 0)

Substituting x = 6

6 = 6y

y = 6/6

y = 1

⟹ Solution (6, 1) 

(iii) We are given x + πy = 4 

Substituting x = 0 in the given equation,

We get 0 + πy = 4 

πy = 4

y = 4/π 

⟹ Solution = (0, 4/π)

Substituting y = 0 in the given equation, we get x + 0 = 4

x = 4

⟹ Solution = (4, 0)
(iv) We are given (2/3) x – y = 4 

Substituting x = 0 in the given equation, we get 0 - y = 4

y = - 4

Thus x = 0 and y = - 4 is a solution

Substituting x = 3 in the given equation, we get (2/3) × 3 − y = 4 

2 - y = 4

y = 2 - 4

y = -2

Thus x = 3 and y = – 2 is a solution 

Quesiton: 2

Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations:

(i) 5x - 2y =10

(ii) - 4x + 3y =12

(iii) 2x + 3y = 24 

Solution:

(i) We are given, 5x - 2y = 10

Substituting x = 0 in the given equation, We get;

5× 0 - 2y = 10 - 2y = 10 - y = 10/2

y = - 5

Thus x = 0 and y = - 5 is the solution of 5x - 2y = 10

Substituting y = 0 in the given equation, we get 5x – 2 × 0 = 10

5x = 10

x = 10/5

x = 2

Thus x = 2 and y = 0 is a solution of 5x - 2y = 10

(ii) We are given, - 4x + 3y = 12

Substituting x = 0 in the given equation, we get;

- 4 × 0 + 3y = 12

3y = 12

y = 4

Thus x = 0 and y = 4 is a solution of the - 4x + 3y = 12

Substituting y = 0 in the given equation, we get;

- 4x + 3 × 0 = 12 – 4x = 12

x = -12/4

x = - 3

Thus x = - 3 and y = 0 is a solution of - 4x + 3y = 12

(iii) We are given, 2x + 3y = 24

Substituting x = 0 in the given equation, we get; 2 × 0 + 3y = 24

3y = 24

y = 24/3

y = 8

Thus x = 0 and y = 8 is a solution of 2x+ 3y = 24

Substituting y = 0 in the given equation, we get;

2x + 3 × 0 = 24

2x = 24

x = 24/2

x =12

Thus x = 12 and y = 0 is a solution of 2x + 3y = 24 

Quesiton: 3

Check which of the following are solutions of the equation 2x - y = 6 and Which are not: 

(i) (3, 0)

(ii) (0, 6)

(iii) (2, - 2)

(iv) (√3, 0)

(v) (1/2 , - 5)

Solution:

We are given, 2x - y = 6

(i) In the equation 2x - y = 6,

We have L.H.S = 2x - y and R.H.S = 6

Substituting x = 3 and y = 0 in 2x - y,

We get L.H.S = 2 × 3 - 0 = 6 ⟹ L.H.S = R.H.S

⟹ (3, 0) is a solution of 2x - y = 6.

(ii) In the equation 2x - y = 6, We have L.H.S= 2x - y and R.H.S = 6

Substituting x = 0 and y = 6 in 2x - y

We get L.H.S = 2 × 0 - 6 = - 6

⟹ L.H.S ≠ R.H.S

⟹ (0, 6) is not a solution of 2x - y = 6.

(iii) In the equation 2x - y = 6,

We have L.H.S = 2x - y and R.H.S = 6

Substituting x = 2 and y = - 2 in 2x - y,

We get L.H.S = 2 × 2 - (-2) = 6

⟹ L.H.S = R.H.S

⟹ (2, - 2) is a solution of 2x - y = 6.

(iv) In the equation 2x - y = 6,

We have L.H.S = 2x- y and R.H.S = 6

Substituting x = √3 and y = 0 in 2x - y,

We get L. H. S = 2 × √3 - 0

⟹ L.H.S ≠ R.H.S

⟹ (√3, 0) is not a solution of 2x - y = 6.

(v) In the equation 2x - y = 6,

We have L.H.S = 2x - y and R.H.S = 6

Substituting x = 1/2 and y = in 2x - y, we get L.H.S = 2 × (1/2) - (-5)

⟹ 1 + 5 = 6

⟹ L.H.S = R.H.S

⟹ (12, - 5) is a solution of 2x - y = 6. 

Quesiton: 4

If x = – 1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k. 

Solution:

We are given, 3x + 4y = k

Given that, (-1, 2) is the solution of equation 3x + 4y = k.

Substituting x = -1 and y = 2 in 3x + 4y = k,

We get; 3x - 1 + 4 × 2 = k

K = - 3 + 8 k = 5 

Quesiton: 5

Find the value of λ, if x = -λ and y = 5/2 is a solution of the equation x + 4y – 7 = 0 

Solution:

We are given, x + 4y - 7 = 0 (- λ, - 5) is a solution of equation 3x + 4y = k

Substituting x = - λ and y = 5/2 in x + 4y - 7 = 0,

We get; - λ + 4 × (5/2) - 7 = 0 -λ + 4 × 5/2 - 7 = 0 

λ = 10 - 7 

λ = 3 

Quesiton: 6

If x = 2a + 1 and y = a -1 is a solution of the equation 2x - 3y + 5 = 0, find the value of a. 

Solution:

We are given, 2x - 3y + 5 = 0 (2a + 1, a - 1) is the solution of equation 2x - 3y + 5 = 0.

Substituting x = 2a + 1 and y = a - 1 in 2x - 3y + 5 = 0,

We get 2 × 2a + (1- 3) x a - 1 + 5 = 0

⟹ 4a + 2 - 3a + 3 + 5 = 0

⟹ a + 10 = 0

⟹ a = - 10 

Quesiton: 7

If x = 1 and y = 6 is a solution of the equation 8x - ay + a2 = 0, find the values of a. 

Solution:

We are given, 8x - ay + a2 = 0 (1, 6) is a solution of equation 8x - ay + a2 = 0

Substituting x = 1 and y = 6 in 8x - ay + a2 = 0, we get 8 × 1 - a × 6 + a2 = 0

⟹ a2 - 6a + 8 = 0

Using quadratic factorization a2 - 4a - 2a + 8 = 0 a(a - 4 ) - 2(a - 4) = 0 (a - 2)(a - 4) = 0

a = 2, 4 

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