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

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.