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

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 
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