RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.
Which one of the following options is true, and why?
y = 3x + 5 has :
(i) a unique solution.
(ii) only two solutions.
(iii) infinitely many solutions.
Answer:
Given equation is y = 3x + 5.
When y = 0, 3x + 5 = 0 ⇒ x = – \(\frac{5}{3}\)
∴(- \(\frac{5}{3}\), 0) is one solution.
When x = 0, y = 0 + 5 = 5
∴ (0, 5) is another solution.
When x = 1 = 3 × 1 + 5 = 8
∴ (1, 8) is another solution.
Clearly, for different values of x, we get different values of y.
Thus, the chosen value of x together with the corresponding value of y constitutes another solution of the given equation. So, there is no end to different solutions of a linear equation in two variables.
∴ The given linear equation in two variables has infinitely many solutions. Thus, option (iii) is true.

Question 2.
Write, four solutions for each of the following equations :
(i) 2x + y – 7
Answer:
(i) The given equation can be written as y = 7 – 2x.
When x = 0, y = 7 – 2 × 0 = 7 – 0 = 7
When x = 1, y = 7 – 2 × 1 = 7 – 2 = 5
When x = 2, y = 7 – 2 × 2 = 7 – 4 = 3
When x = 3, y = 7 – 2 × 3 = 7 – 6 = 1
∴ The four solutions of the given equation are :
(0, 7), (1, 5), (2, 3) and (3, 1).

(ii) πx + y = 9
Answer:
The given equation can be written as y = 9 – πx.
When x = 0, y = 9 – 0 = 9
When x = 2, y = 9 – 2π
When x = 1, y = 9 – n
When x = 3, y = 9 – 3π
∴ The four solutions of the given equation are (0, 9), (1, 9 – π), (2, 9 – 2π) and (3, 9 – 3π).

(iii) x = 4y
Answer:
The given equation can be written as y = \(\frac{x}{4}\).
When x = 0, y = 0
When x = 2, y = \(\frac{1}{4}\) × 2 = \(\frac{1}{2}\)
When x = 1, y = \(\frac{1}{4}\) × 1 = \(\frac{1}{4}\)
When x = 3, y = \(\frac{1}{4}\) × 3 = \(\frac{3}{4}\)
∴ The four solutions of the given equation are (0, 0), (1, \(\frac{1}{4}\)), (2, \(\frac{1}{2}\)) and (3, \(\frac{3}{4}\)).

Question 3.
Cheek which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
Answer:
Putting x = 0, y = 2 in LHS of x – 2y = 4, we have:
LHS = 0 – 2 × 2 = -4 ≠ RHS
∴ (0, 2) is not its solution.

(ii) (2, 0)
Answer:
Putting x = 2, y = 0 in LHS of x – 2y = 4, we have:
LHS = 2 – 2 × 0 = 2 – 0 = 2 ≠ RHS
∴ (2, 0) is not its solution.

(iii) (4, 0)
Answer:
Putting x =4, y = 0 in the LHS of x – 2y = 4, we have:
LHS = 4 – 0 = 4 = RHS
∴ (4, 0) is its solution.

(iv) (√2, 4√2)
Answer:
Putting x = √2 y = 4√2 in the LHS of x – 2y = 4, we have:
LHS = √2 – 2 × 4√2 = √2 – 8√2 = -7√2 ≠ RHS
∴ (√2, 4√2) is not its solution.

(v) (1, 1)
Answer:
Putting x = 1, y = 1 in the LHS of x – 2y = 4, we have:
LHS = 1 – 2 × 1 = 1 – 2 = -1 ≠ RHS
∴ (1, 1) is not its solution.

Question 4.
Find the value of k, if x = 2,y = 1 is a solution of the equation 2x + 3y = k.
Answer:
if x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satis1 the equation.
∴ 2 × 2 + 3 × 1 = k ⇒ k = 4 + 3 = 7.