RD Sharma Solutions for Class 9 Chapter 13 Linear Equation in Two Variables Exercise 13.2 helps students to solve linear equations in an easy way. Let ax+ by + c = 0, where a, b, c are real numbers, a, b are not equal to zero, and x, y are variable, then any pair of values of x and y which satisfies the equation is called a solution of it. Learn more about linear equations and their properties in detail with complete exercise-wise RD Sharma Solutions to prepare well for solving different complicated problems related to Linear Equations in Two Variables quickly.
RD Sharma Solutions for Class 9 Maths Chapter 13 Linear Equations in Two Variables Exercise 13.2
Access Answers to RD Sharma Solutions for Class 9 Maths Chapter 13 Linear Equations in Two Variables Exercise 13.2 Page Number 13.6
Question 1: Write two solutions for each of the following equations:
(i) 3x + 4y = 7
(ii) x = 6y
(iii) x + πy = 4
(iv) 2/3x – y = 4.
Solution:
(i) 3x + 4y =7 ….(1)
Step 1: Isolate the above equation in y.
Subtract 3x from both sides,
3x + 4y – 3x = 7 – 3x
4y = 7 – 3x
Divide each side by 4
y = 1/4 x (7 – 3x) ….(2)
Step 2: Find Solutions
Substituting x = 1 in (2)
y = 1/4 x (7 – 3) = 1/4 x 4 = 1
Thus x = 1 and y = 1 is the solution of 3x + 4y = 7
Again, Substituting x = 2 in (2)
y = 1/4 x (7 – 3 x 2) = 1/4 x 1 = 1/4
Thus x = 2 and y = 1/4 is the solution of 3x + 4y = 7
Therefore, (1, 1) and (2, 1/4) are two solutions of 3x + 4y = 7.
(ii) Given: x = 6y
Substituting x =0 in the given equation,
0 = 6y
or y = 0
Thus (0,0) is one solution
Again, substituting x=6
6 = 6y
or y = 1
Thus, (6, 1) is another solution.
Therefore, (0, 0) and (6, 1) are two solutions of x = 6y.
(iii) Given: x + πy = 4
Substituting x = 0 ⇒ 0 + πy = 4 ⇒ y = 4/ π
Substituting y = 0 ⇒ x + 0 = 4 ⇒ x = 4
Therefore, (0, 4/ π) and (4, 0) are two solutions of x + πy = 4.
(iv) Given: 2/3 x – y = 4
Substituting x = 0 ⇒ 0 – y = 4 ⇒ y = -4
Substituting x = 3 ⇒ 2/3 × 3 – y = 4 ⇒ 2 – y = 4 ⇒ y = -2
Therefore, (0, -4) and (3, -2) are two solutions of 2/3 x – y = 4.
Question 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: 5x – 2y = 10
Substituting x = 0 ⇒ 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 ⇒ 5x – 2 x 0 = 10 ⇒ 5x = 10 ⇒ x = 2
Thus x =2 and y = 0 is a solution of 5x – 2y = 10
(ii) Given, – 4x + 3y = 12
Substituting x = 0 ⇒ -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 ⇒ -4 x + 3 x 0 = 12 ⇒ – 4x = 12 ⇒ x = -3
Thus x = -3 and y = 0 is a solution of -4x + 3y = 12
(iii) Given, 2x + 3y = 24
Substituting x = 0 ⇒ 2 x 0 + 3y = 24 ⇒ 3y =24 ⇒ y = 8
Thus x = 0 and y = 8 is a solution of 2x+ 3y = 24
Substituting y = 0 ⇒ 2x + 3 x 0 = 24 ⇒ 2x = 24 ⇒ x =12
Thus x = 12 and y = 0 is a solution of 2x + 3y = 24
Question 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:
(i) Check for (3, 0)
Put x = 3 and y = 0 in equation 2x – y = 6
2(3) – (0) = 6
6 = 6
True statement.
⇒ (3,0) is a solution of 2x – y = 6.
(ii) Check for (0, 6)
Put x = 0 and y = 6 in 2x – y = 6
2 x 0 – 6 = 6
-6 = 6
False statement.
⇒ (0, 6) is not a solution of 2x – y = 6.
(iii) Check for (2, -2)
Put x = 0 and y = 6 in 2x – y = 6
2 x 2 – (-2) = 6
4 + 2 = 6
6 = 6
True statement.
⇒ (2,-2) is a solution of 2x – y = 6.
(iv) Check for (√3, 0)
Put x = √3 and y = 0 in 2x – y = 6
2 x √3 – 0 = 6
2 √3 = 6
False statement.
⇒(√3, 0) is not a solution of 2x – y = 6.
(v) Check for (1/2, -5)
Put x = 1/2 and y = -5 in 2x – y = 6
2 x (1/2) – (-5) = 6
1 + 5 = 6
6 = 6
True statement.
⇒ (1/2, -5) is a solution of 2x – y = 6.
Question 4: If x = -1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k.
Solution:
Given, 3 x + 4 y = k
(-1, 2) is the solution of 3x + 4y = k, so it satisfies the equation.
Substituting x = -1 and y = 2 in 3x + 4y = k, we get
3 (– 1 ) + 4( 2 ) = k
– 3 + 8 = k
k = 5
The value of k is 5.
Question 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, (-λ, 5/2) 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 + 10 – 7 = 0
λ = 3
Question 6: If x = 2 α + 1 and y = α -1 is a solution of the equation 2x – 3y + 5 = 0, find the value of α.
Solution:
Given, (2 α + 1, α – 1 ) is the solution of equation 2x – 3y + 5 = 0.
Substituting x = 2 α + 1 and y = α – 1 in 2x – 3y + 5 = 0, we get
2(2 α + 1) – 3(α – 1 ) + 5 = 0
4 α + 2 – 3 α + 3 + 5 = 0
α + 10 = 0
α = – 10
The value of α is -10.
Question 7: If x = 1 and y = 6 is a solution of the equation 8x – ay + a2 = 0, find the values of a.
Solution:
Given, (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 x 1 – a x 6 + a2 = 0
⇒ a2 – 6a + 8 = 0 (quadratic equation)
Using quadratic factorization
a2 – 4a – 2a + 8 = 0
a(a – 4) – 2 (a – 4) = 0
(a – 2) (a – 4)= 0
a = 2, 4
Values of a are 2 and 4.
RD Sharma Solutions for Class 9 Maths Chapter 13 Linear Equations in Two Variables Exercise 13.2
RD Sharma Solutions Class 9 Maths Chapter 13 Linear Equations in Two Variables Exercise 13.2 are based on finding a solution to a linear equation. Students can download RD Sharma solutions for Class 9, which are prepared by subject experts and start practising to score good marks in the annual exam.
