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
Page-70
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