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.

Answer:

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

Answer:

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

Answer:

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

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

Substituting x = 1 and y = 1, we get x – 2y = 1 -2(1) = – 1, whIch is not equal to 4. Hence, (1, 1) is not a solution of x – 2y = 4.

Question 4.

Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer:

x = 2, y = 1 is a solution of equation 2x + 3y = k. Hence, x = 2 and y = 1 satisfy the equation.

∴ 2(2) + 3(1) = k

∴ 4 + 3 = k

∴ 7 = k

∴ k = 7