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

## Haryana Board 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Exercise 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.

Solution:

(iii) is true, because y = 3x + 5 is a linear equation in two variables. We know that a linear equation in two variables has 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.

Solution:

(i) We have,

2x + y = 7 …(i)

Substituting x = 0 in the equation (i), we get

2 × 0 + y = 7

⇒ 0 + y = 7

⇒ y = 7

∴ (0, 7) is a solution of the given equation.

Substituting x = 1 in the equation (i), we get

2 × 1 + y = 7

⇒ 2 + y = 7

⇒ y = 7 – 2 = 5

∴ (1, 5) is a solution of the given equation.

Substituting x = 2 in equation (i), we get

2 × 2 + y = 7

⇒ 4 + y = 7

⇒ y = 7 -4 = 3

∴ (2, 3) is a solution of the given equation.

Substituting x = 3 in the equation (i), we get

2 × 3 + y = 7

⇒ 6 + y = 7

⇒ y = 7 – 6 = 1

∴ (3, 1) is a solution of the given equation.

Hence, four solutions of the given equation are (0, 7), (1, 5), (2, 3) and (3, 1).

(ii) We have πx + y = 9 ……..(i)

Substituting x = 0 in the equation (i), we get

π × 0 + y = 9

⇒ 0 + y = 9

⇒ y = 9 .

∴ (0, 9) is a solution of the given equation.

Substituting x = 1 in the equation (i), we get

π × 1 + y = 9

⇒ π + y = 9

⇒ y = 9 – π

∴ (1, 9 – π) is a solution of the given equation.

Substituting x = \(\frac {9}{π}\) in the equation (i), we get

π × \(\frac {9}{π}\) + y = 9

⇒ 9 + y = 9

⇒ y = 9 – 9 = 0

∴ (\(\frac {9}{π}\), 0) is a solution of the given equation.

Substituting x = – 1 in the equation (i), we get

π × (-1) + y = 9

⇒ – π + y = 9

⇒ y = 9 + π

∴ (-1, 9 + π) is a solution of the given equation

Hence, four solutions of the given equation are (0, 9), (1, 9 – π), (\(\frac {9}{π}\), 0) and (- 1, 9 + π).

(iii) We have, x = 4y …(i)

Substituting x = 0 in the equation (i), we get

0 = 4y

⇒ \(\frac {0}{4}\) = y

⇒ y = 0

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

Substituting x = 4 in the equation (i), we get

4 = 4y

⇒ \(\frac {4}{4}\) = y

⇒ y = 1

∴ (4, 1) is a solution of the given equation.

Substituting x = – 4 in the equation (i), we get

– 4 = 4y

⇒ \(\frac {-4}{4}\) = y

⇒ – 1 = y

⇒ y = – 1

∴ (-4, -1) is a solution of the given equation.

Substituting x = 2 in the equation (i), we get

2 = 4y

⇒ \(\frac {2}{4}\) = y

⇒ y = \(\frac {1}{2}\)

∴ (2, \(\frac {1}{2}\)) is a solution of the given equation.

Hence, four solutions of the given equation are (0, 0), (4, 1) (-4, -1) and (2, \(\frac {1}{2}\))

Question 3.

Check which of the following are solutions of the equation 1 – 2y = 4 and which are not :

(i) (0, 2)

(ii) (2, 0)

(iii) (4, 0)

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

(v) (1, 1).

Solution:

We have, x – 2y = 4 …(i)

(i) Substituting x = 0, y = 2 in the L.H.S. of the equation (i), we get

L.H.S. = 0 – 2 × 2 = – 4

≠ R.H.S.

∵ L.H.S. ≠ R.H.S.

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

(ii) Substituting x = 2, y = 0 in the L.H.S. of the equation (i), we get

L.H.S. = 2 – 2 × 0

= 2 – 0 = 2

≠ R.H.S.

∵ L.H.S. ≠ R.H.S.

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

(iii) Substituting x = 4, y = 0 in the L.H.S. of equation (i), we get

L.H.S. = 4 – 2 × 0

= 4 – 0 = 4

= R.H.S.

∵ L.H.S. = R.H.S.

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

(iv) Substituting x = \(\sqrt{2}\), y = 4\(\sqrt{2}\) in the L.H.S. of equation (i), we get

L.H.S. = 1\(\sqrt{2}\) – 2 × 4\(\sqrt{2}\)

= \(\sqrt{2}\) – 8\(\sqrt{2}\) = – 7\(\sqrt{2}\)

≠ R.H.S.

∵ L.H.S. ≠ R.H.S.

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

(v) Substituting x = 1, y = 1 in the L.H.S. of equation (i), we get

L.H.S. = 1 – 2 × 1

= 1 – 2 = – 1

≠ R.H.S.

∵ L.H.S. ≠ R.H.S.

∴ (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.

Solution:

Since x = 2, y = 1 is a solution of the equation

2x + 3y = k

Therefore, x = 2, y = 1 will satisfy the given equation.

⇒ 2 × 2 + 3 × 1 = k

⇒ 4 + 3 = k

⇒ 7 = k

Hence, k = 7