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

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

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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:

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

Question 2.

Write four solutions for each of the following equations:

(i) 2x + y = 7

(ii) πx + y = 9

(iii) x = 4y

Answer:

(i) 2x + y = 7

⇒ y = 7 – 2x

Put x = 0,

y = 7 – 2 × 0

⇒ y = 7

∴ (0, 7) is the solution.

Now, put x = 1

y = 7 – 2 × 1

⇒ y = 5

∴ (1, 5) is the solution.

Now, put x = 2

y = 7 – 2 × 2

⇒ y = 3

∴ (2, 3) is the solution.

Now, put x = -1

y = 7 – 2 × (-1)

⇒ y = 9

∴ (-1, 9) is the solution.

The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9

⇒ y = 9 – πx

Put x = 0,

y = 9 – π x 0

⇒ y = 9

∴ (0, 9) is the solution.

Now, put x = 1

y = 9 – π × 1

⇒ y = 9 – π

∴ (1, 9 – π) is the solution.

Now, put x = 2

y = 9 – π × 2

⇒ y = 9 – 2n

∴ (2, 9 – 2π) is the solution.

Now, put x = -1

y = 9 – π × -1

⇒ y = 9 + π

∴ (-1, 9 + 7t) is the solution.

The four solutions of the equation πx + y = 9 are (0, 9), (1, 9 – n), (2, 9 – 2n) and (-1, 9 + n).

(iii) x = 4y

Put x = 0,

0 = 4y

⇒ y=0

∴ (0, 0) is the solution.

Now, put x = 1

1 = 4y

⇒ y = 1/4

∴ (1, 1/4) is the solution.

Now, put x = 4

4 =4y

⇒ y=1

∴ (4,1) is the solution.

Now, put x = 8

8 = 4y .

⇒ y= 2

∴ (8, 2) is the solution.

The four solutions of the equation x = 4y are (0, 0), (1, 1/4), (4, 1) and (8, 2).

Question 3.

Check which of the following are solutions of the equation x – 2y = 4 and which are not:

(i) (0, 2)

(ii) (2, 0)

(iii) (4, 0)

(iv) (\(\sqrt{2}\), 4\(\sqrt{2}\))

(v) (1, 1)

Answer:

(i) Put x = 0 and y = 2 in the equation x – 2y = 4.

0 – 2 × 2 = 4

⇒ -4 ≠ 4

∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x – 2y = 4.

2 – 2 × 0 =4

⇒ 2 ≠ 4

∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x – 2y = 4.

4 – 2 × 0 =4

⇒ 4 = 4

∴ (4, 0) is a solution of the given equation.

(iv) Put \(\sqrt{2}\) and y= 4\(\sqrt{2}\) in the equation x – 2y = 4.

\(\sqrt{2}\) – 2 × 4\(\sqrt{2}\) = 4

⇒ \(\sqrt{2}\) – 8\(\sqrt{2}\) = 4

⇒ \(\sqrt{2}\) (1 – 8) = 4

⇒ – 7\(\sqrt{2}\) ≠ 4

∴ (\(\sqrt{2}\), 4\(\sqrt{2}\)) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x – 2y = 4.

1 – 2 × 1 =4

⇒ -1 ≠ 4

(1, 1) is not a solution of the given equation.

Question 4.

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

Answer:

Given equation is 2x + 3y = k.

x = 2, y = 1 is the solution of the given equation.

Putting the values of x and y in the equation, we get

2 × 2 + 3 × 1 = k

⇒ k = 4 + 3

⇒ k = 7