# PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 9 Maths 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.
Option (iii) is true. Since y = 3x + 5 is a linear equation in two variables, it has infinitely many solutions. e.g., (1, 8), (2, 11), (3, 14), (4, 17), (0, 5), (- 1, 2) are all solutions of the given equation y = 3x + 5.

Question 2.
Write four solutions for each of the following equations:
(i) 2x + y = 7
2x + y = 7
∴ y = 7 – 2x
Taking x = 0, 1, 2, 3, we get the values of y as 7, 5, 3 and 1 respectively. Thus, (0, 7), (1, 5), (2, 3) and (3, 1) are four solutions of the given equation 2x + y = 7. We can give other answers as well because the given linear equation in two variables has infinitely many solutions.

(ii) πx + y = 9
πx + y = 9
∴ y = 9 – πx
For x = 0, y = 9.
For x = 1, y = 9 – π.
For x = – 1, y = 9 + π.
For x = , y = 8.
Thus, (0, 9), (1, 9 – π), (- 1, 9 + π) and ($$\frac{1}{\pi}$$, 8) are four of the infinitely many solutions of the given equation πx + y = 9.

(iii) x = 4y
x = 4y
For y = 0, x = 0.
For y = 1, x = 4.
For y = – 1, x = – 4.
For y = 2, x = 8.
Thus, (0, 0), (4, 1), (- 4, – 1) and (8, 2) are four of the infinitely many solutions of the given equation x = 4y.

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
Substituting x = 0 and y = 2, we get x – 2y = 0 – 2(2) = – 4, which is not equal to 4. Hence, (0, 2) is not a solution of x – 2y = 4.

(ii) (2, 0)
Substituting x = 2 and y = 0, we get x – 2y = 2 – 2 (0) = 2, which is not equal to 4. Hence, (2, 0) is not a solution of x – 2y = 4.

(iii) (4, 0)
Substituting x = 4 and y = 0, we get x – 2y = 4 – 2 (0) = 4. Hence, (4, 0) is a solution of x – 2y = 4.

(iv) (√2, 4√2)
Substituting x = √2 and y = 4√2, we get x – 2y = √2 – 2(4√2) = – 7√2, which is not equal to 4. Hence, (√2, 4√2) is not a solution of x – 2y = 4.

(v) (1, 1)