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

Gujarat Board GSEB Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.
Which of the following options is true and why?
y = 3x + 5
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions
Solution:
Option (iii) is true y = 3x + 5 has infinitely many solutions.
Reason: for every real value of x, there will be a corresponding value of y and vice versa.

Question 2.
Write four solutions for each of the following equations.
1. 2x + y = 7
2. πx + y = 9
3. x = 4y
Solution:
1. 2x + y = 7
⇒ y = 7 – 2x
Taking values of x = 0, 1, 2, 3
Putting x = 0, y = 7 – 2 x 0 = 7
Putting x = 1, y = 7 – 2 x 1 = 7 – 2 = 5
Putting x = 2, y = 7 – 2x
Putting x = 3, y = 7 – 2 x 3 = 7 – 6 = 1
Therefore four solutions are
(0, 7), (1, 5), (2, 5) and (3, 1)

2. πx + y = 9
⇒ y = 9 – πx
Now putting the values of x = 0, 1, 2, 3
Putting x = 0, y = 9 – π x 0
y = 9
Putting x = 1, y = 9 – π
Putting x = 2, y = 9 – πx = 9 – π x 2
y = 9 – 2π
Putting x = 3, y = 9 – πx
y = 9 – π x 3
⇒ y = 9 – 3π
Therefore four solutions are
(0, 9), (1, 9 – π ), (2, 9 – 2π) and (3, 9 – 3π)

3. x = 4y
Putting value ofy in given eqn, we get value of x.
Let’s take values of y = 0, 1, 2, 3
Puttmg y = 0, x = 0
Putting y = 1, x = 4 x 1
x = 4
Putting y = 2, x = 4 x 2
x = 8
Putting y = 3, x = 4 x 3
x = 12
Therefore four solutions are
(0, 0), (4, 1), (8, 2) and (12, 3)

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not?
1. (0, 2)
2. (2, 0)
3. (4, 0)
4. \((\sqrt{2}, 4 \sqrt{2})\)
5. (1,1)
Solution:
1. x – 2y = 4
(0, 2) will be the solution of given equation
if they will satisfy the given equation.
x – 2y = 4
Puttingx = 0 and y = 2 in LHS,
0 – 2 x 2 = -4
⇒ LHS = RHS
∴ (0,2) is not the solution of given equation.

2.  x – 2y = 4
(2, 0) will be the solution of given equation
if they satisfy the equation.
Putting x = 2 and y = 0 in LHS,
⇒ 2 – 2 x 0 = 2
LHS = RHS
∴ (2,0) is not the solution of given equation.

3. x – 2y = 4
(4, 0) will be the solution of given equation
if they satisfy the equation.
Putting x = 4 and y = 0 in LHS,
4- 2 x 0 = 4
LHS = RHS
∴ (4, 0) is the solution of given equation.

4.  x – 2y = 4
\((\sqrt{2}, 4 \sqrt{2})\) will be the solution of given equation if they satisfy the equation.

Putting x = \(\sqrt{2}\) and y = 4\(\sqrt{2}\) in LHS
\(\sqrt{2}\) – 2 x 4\(\sqrt{2}\)
\(\sqrt{2}\) – 8\(\sqrt{2}\) = -7\(\sqrt{2}\)
⇒ LHS = RHS
\((\sqrt{2}, 4 \sqrt{2})\) is not the solution of given equation.

5. x – 2y = 4
(1, 1) will be the solution of given equation
if they satisfy the equation.
Putting x = 1 and y = 1 in LHS,
1 – 2 x 1= -1
⇒ LHS = RHS
∴ (1, 1) is not the solution of given equation.

Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
Given 2x + 3y = k
x = 2 and y = 1 will be the solution of given equation if this point satisfies the equation.
2x + 3y = k
2 x 2 + 3 x 1 = k
4 + 3 = k
k = 7