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

## BSEB Bihar Board Class 9th Maths Solutions 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

Solution:

Given equation is y = 3x + 5

For y = 0, 3x + 5 = 0 ⇒ x = – \(\frac { 5 }{ 3 }\)

∴ (- \(\frac { 5 }{ 3 }\), 0) is one solution.

For x = 0, y = 0 + 5 = 5

∴ (0, 5) is another solution.

For x = 1, y = 3 x 1 + 5 = 8

∴ (1, 8) is another solution.

Clearly, for different values of x, we get another value of y. Thus, the chosen value of x together with this value of y constitutes, another solution of the given solution. So, there is no end to different solutions of a linear equation in two variables.

∴ 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) The given equation can be written as y = 7 – 2x.

For x = 0, y = 7 – 2 x 0 = 7 – 0 = 7

For x = 1, y = 7 – 2 x 1 = 7 – 2 = 5

For x = 2, y = 7 – 2 x 2 = 7 – 4 = 3

For x = 3, y = 7 – 2 x 3 = 7 – 6 = 1

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

(ii) The given equation can be written as y = 9 – πx

For x = 0, y = 9 – 0 = 9

For x = 1, y = 9 – π

For x = 2, y = 9 – 2π

For x = 3, y = 9 – 3π

∴ The four solution of the given equation are (0, 9), (1, 9 – π), (2, 9 – π) and (3, 9 – 3π).

(iii) The given equation can be written as x = Ay

For x = 0, y = 0

For x = 1, y = 4 x 1 = 4

For x = 2, y = 4 x 2 = 8

For x = 3, y = 4 x 3 = 12

∴ The four solutions of the given equation are (0, 0), 0, 4), (2, 8) and (3, 12).

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)

Solution:

(i) Putting x = 0, y = 2 in L.H.S. of x – 2y = 4, we

have

L.H.S. = 0 – 2 x 2 = – 4 ≠ R.H.S.

∴ x = 0, y = 2 is not its solution.

(ii) Putting x = 2, y = 0 in L.H.S. of x – 2y = 4, we have

L.H.S. = 2 – 2 x 0 = 2 – 0 = 2 ≠ R.H.S.

∴ x = 2,y = 0 is not its solution.

(iii) putting x = 4, y = 0 in the L.H.S. of x – 2y = 4, we have

L.H.S. = 4 – 0 = 4 = R.H.S.

∴ x = 4, y = 0 is its solution.

(iv) Putting x = \(\sqrt{2}\) , y = 4\(\sqrt{2}\) in the L.H.S. of x – 2y = 4, we have

L.H.S. = \(\sqrt{2}\) – 2 x 4\(\sqrt{2}\) = \(\sqrt{2}\) – 8\(\sqrt{2}\)

= – 7\(\sqrt{2}\) ≠ R.H.S.

∴ (\(\sqrt{2}\), 4\(\sqrt{2}\)) is not its solution.

(v) Putting x = 1, y = 1 in the L.H.S. of x – 2y = 4, we have

L.H.S. = 1 – 2 x 1 = 1 – 2 = – 1 ≠ R.H.S.

∴ x = 1, y = 1 is not its solution.

Question 4.

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

Solution:

If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation.

∴ 2 x 2 + 3 x 1 = k ⇒ k = 4 + 3 = 7.