R.D. Sharma Solutions Class 9th: Ch 13 Linear Equations in Two Variables Exercise 13.2

Chapter 13 Linear Equations in Two Variables R.D. Sharma Solutions for Class 9th Exercise 13.2

Exercise 13.2

1. Write two solutions for each of the following equations:
(i) 5x–2y = 7
(ii) x = 6y
(iii) x + πy = 4
(iv) 2/3x – y = 4.

Solution








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 x 0 = 10
5x = 10
x = 10/2
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;
- 4 x + 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;
× 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

3. Check which of the following are solutions of the equations 2x - y = 6 and which are not:


Solution


We are given, 2x – y = 6

(i) In the equation 2 x – 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 x 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
LHS = 2x-y and RHS = 6
Substituting x = √3 and y = 0 in 2x - y, we get
LHS = 2×√3-0 = 2√3
LHS ≠ RHS
(√3, 0) is not the solution of 2x - y = 6.

(v) In the equation 2x - y = 6, we have
LHS = 2x-y and RHS = 6
Substituting x = 1/2 and y = -5 in 2x-y, we get
LHS = 2× 1/2 and y = -5 in 2x-y, we get
LHS = RHS
(1/2, -5) is the solution of 2x-y = 6.

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


Solution


We are given, 3 x + 4 y = 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

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


  Solution



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 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 x 2a + (1- 3) x a – 1 + 5 = 0
⇒ 4a + 2 – 3a + 3 + 5 = 0
⇒ a + 10 = 0
⇒ a = – 10

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


Solution


Given,
8x - ay + a2 = 0
(1, 6) is the 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
Therefore, a = 2, 4

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