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