RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables Exercise 13.2

Chapter 13 Linear Equations in Two Variables RD Sharma Solutions Exercise 13.2 Class 9 Maths

Chapter Name	RD Sharma Chapter 13 Linear Equations in Two Variables Exercise 13.2
Book Name	RD Sharma Mathematics for Class 10
Other Exercises	Exercise 13.1 Exercise 13.3 Exercise 13.4
Related Study	NCERT Solutions for Class 10 Maths

Exercise 13.2 Solutions

1. Write two solutions for each of the following equations: 

(i) 3x + 4y = 7 
(ii) x = 6y
(iii) x + πy = 4 
(iv) 2x/3 - y = 4

Solution

(i) Given that 3x + 4y = 7 
Substituting x = 0 in this equation, we get 
3 × 0 + 4y = 7 
⇒ y = 7/4 
So, (0, 7/4) is a solution of the given equation substituting x = 1, in given equation, we 
get 
⇒ 3 ×1 + 4y = 7 
⇒ 4y = 7 - 3 = 4 
⇒ y = 1 
So, (1, 1) is a solution of the given equation 

∴ (0, 7/4) and (1, 1) are the solutions for the given equation 

(ii) We have 
x = 6y
Substituting y = 0 in this equation, we get x = 6 × 0 = 0 
So, (0, 0) is a function of the given equation substituting y = 1, in the given equation, we 
set x = 6 × 1 = 6 
So, (6, 1) is a solution of the given equation. 

∴ we obtain (0, 0) and (6, 1) as solutions of the given equation. 

(iii) We have 
x +  πy = 4 
Substituting y = 0 in this equation, we get 
x + π(0) = 4 
⇒ x = 4 
So, (y, 0) is a solution of the give equation. 

∴ We obtain (4, 0) and (4, -x) as solutions of the given equation. 

(iv) Given that 

2x/3 - y = 4
Substituting y = 0 in this equation we get 

2x/3 - 0 = 4
⇒ x = 4×3/2
⇒ x = 6 
So, (6, 0) is a solution of the given equation

Substituting y = 1 in the given equation, we get 

2/3 × -1 = 4 
2x/3 = 5

⇒ x = 15/2 
So, (15/2, 1) is a solution of the given equation. 

∴ We obtain (6, 0) and (15/2, 1) as solutions of the given equation.

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) Given that 

5x - 2y = 10 

Substituting x = 0 in the equation 5x - 2y = 10 

We get 5 ×0 - 2y = 10 
⇒ y = -10/2 = -5
Thus x = 0 and y = -5 is a solution of 5x - 2y = 10 

Substituting y = 0, we get 
⇒ 5x - 2 × 0 = 10
⇒ 5x = 10 
⇒ x = 2 
Thus, x = 2 and y = 0 is a solution of 5x - 2y = 10 
Thus x = 0, y = -5 and x = 2, y = 0 are two solutions of 5x - 2y = 10 

(ii) Given that , 

-4x + 3y = 12 
Substituting x = 0  in the equation 

-4x + 3y = 12, we get 

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

Substituting y = 0 in the equation 

-4x + 3y = 12, we get 

⇒-4x + 3× 0 = 12
⇒ -4x = 12 
⇒ x = 12/-4 = - 3
Thus, x = -3 and y = 0 is a solution of -4x + 3y = 12. 

Thus x = 0, y = 4 and x = -3, y = 0 are two solutions of -4x + 3y = 12 

(iii) Given that 

2x +3y = 24 

Substituting x = 0 in the given equation

2x + 3y = 24, We get 

⇒ 2 × 0 + 3y = 24 
⇒ 3y = 24 
⇒ y = 24/3 = 8 

Thus, x = 0 and y = 8 is a solution of 2x + 3y = 24
Substituting y = 0 in 2x + 3y = 24, we get 2x + 3 × 0 = 24 
⇒ 2x = 24 
⇒ x = 24/2 = 12
Thus x = 12 and y= 0 is a solution of 2x + 3y = 24 
Thus x = 0, y = -8 and x = 12, y = 0 are two solutions of 2x + 3y = 24 

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)

Solution

In the equation 2x - y = 6 we get 

LHS = 2x - y and RHS = 6 

(i) Substituting x = 3 and y = 0 in 2x - y = 6, we get 

LHS = 2 × 3 - 0 = 6 - 0 = 6 = RHS 

So, x = 3, y = 0 or (3, 0) is a solution of 2x - y = 6 

(ii) Substituting x = 0 and  y = 6 in 2x - y = 6, we get 

LHS = 2 × 0 - 6 = -6 ≠ RHS

So, (0, 6) is not a solution of the equation 2x - y = 6 

(iii) Substituting x = 2, y = -2 in 2x - y = 6, we get 

LHS = 2 × 2(-2) = 4 + 2 = 6 RHS 

(iv) Substituting x = √3  and y = 0 in 2x - y = 6, we get 

LHS = 2 × √3 - 0 = 2√3 ≠ RHS

So, (√3, 0) is not a solution of the equation 2x - y = 6 

(v) Substituting x = 1/2 and y = -5 in 2x - y = 6, we get 

LHS = 2× 1/2 - (-5) = 1 + 5 = 6 = RHS 

So, (1/2, -5) is a solution of the 2x - y = 6 

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

Solution

Given that 

3x + 4y = k 
It is given that x = -1 and y = 2 is a solution of the equation 3x + 4y = k

∴ 3 × (-1) ×4 ×2 = k 
⇒ -3 + 8 = k
⇒ k = 5 

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

Solution

Given that 

x + 4y - 7 = 0
It is given that x = -λ and y = 5/2 is a solution of the equation x + 4y - 7 = 0 

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

Solution

We have 

2x - 3y + 5 = 0 
It is given that x = 2α + 1 and y = α - 1 is a solution of the equation 2x - 3y + 5 = 0 

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

Solution

Given that 

8x - ay + a² = 0 
It is given that x = 1 and y = 6 is a solution on the equation 8x - ay + a² = 0