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 
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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 + Ï€y = 4
(iv) 2x/3  y = 4
(ii) x = 6y
(iii) x + Ï€y = 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
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.
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 + Ï€y = 4
Substituting y = 0 in this equation, we get
x + Ï€(0) = 4
⇒ x = 4
So, (y, 0) is a solution of the give equation.
x + Ï€y = 4
Substituting y = 0 in this equation, we get
x + Ï€(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
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
⇒ 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
2x/3 = 5
⇒ x = 15/2
So, (15/2, 1) is a solution of the given equation.
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
(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
⇒ 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
⇒ 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
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
⇒ 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.
⇒ 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
⇒ 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
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)
(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
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
⇒ 3 + 8 = k
⇒ k = 5
5. Find the value of Î», if x = Î» 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 = Î» and y = 5/2 is a solution of the equation x + 4y  7 = 0
It is given that x = Î» and y = 5/2 is a solution of the equation x + 4y  7 = 0
6. If x = 2Î± + 1 and y = Î±  1 is a solution of the equation 2x  3y + 5 = 0, find the value of Î±.
Solution
We have
2x  3y + 5 = 0
It is given that x = 2Î± + 1 and y = Î±  1 is a solution of the equation 2x  3y + 5 = 0
It is given that x = 2Î± + 1 and y = Î±  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 + a^{2} = 0, find the value of a.
Solution
Given that
8x  ay + a^{2} = 0
It is given that x = 1 and y = 6 is a solution on the equation 8x  ay + a^{2} = 0
It is given that x = 1 and y = 6 is a solution on the equation 8x  ay + a^{2} = 0