Linear Equations in Two Variables (Exercise 4.2)

Linear Equations in Two Variables


Chapter 4


Exercise 4.2


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:

Infinitely many solutions, because a line has infinitely many points and each point is a solution of the linear equation.


EX 4.2 QUESTION 2.


Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:

(i) 2x + y = 7

x = 0, 2(0) + y = 7

⇒ y = 7                                     [∴ Solution is (0, 7)]
x =1, 2(1) + y = 7 ⇒ y = 7 – 2

⇒ y = 5                                      [ ∴ Solution is (1, 5)]

x = 2, 2(2) + y =7y = 7 – 4

⇒ y = 3                                       [∴ Solution is (2, 3)]

x = 3, 2(3) + y = 7y = 7 – 6

⇒ y = 1                                       [∴ Solution is (3, 1)]

∴ The solutions are (0, 7), (1,5), (3,1), (2,3).

(ii) πx + y = 9

x = 0, π(0) + y = 9 ⇒ y = 9 – 0

⇒ y = 9                                         [∴ Solution is (0, 9)]

x = 1, π(1) + y = 9

⇒ y = 9 – π                              [ ∴ Solution is (1, (9 – π))]
x = 2, π(2) + y = 9

⇒ y = 9 – 2π                            [∴ Solution is (2, (9 – 2π))]
x = -1,π(-1) + y = 9

⇒ y = 9 + π                              [∴ Solution is (-1, (9 + π))]

The solutions are (0,9), (1,9-), (9/,0), (-1,9+π)

(iii) x = 4y

x = 0, 4y = 1

⇒ y = 0                                     [∴ Solution is (0, 0)]
x = 1, 4y = 1

⇒ y = 1/4                                  [∴ Solution is (1,1/4)]

x = 2, 4y = 2

⇒ y = 1/2                                     [∴ Solution is (2, 1/2)]
x = 3, 4y = 3

⇒ y = 4/3                                   [∴ Solution is (3, 4/3)]


EX 4.2 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) (√2, 4√2)
(v) (1, 1)
Solution:
(i) (0,2) ,
x = 0 and y = 2

x – 2y = 4,
L.H.S. = 0 – 2(2) = -4.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ x =0, y =2 is not a solution.

(ii) (2, 0)
x = 2 and y = 0

x – 2y = 4,
L.H:S. 2 – 2(0) = 2 – 0 = 2.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (2,0) is not a solution.

(iii) (4, 0)
x = 4 and y = o

x – 2y = 4,
L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S.
∴ L.H.S. = R.H.S.
∴ (4, 0) is a solution.

(iv) (√2, 4√2)
x = √2 and y = 4√2

x – 2y = 4, we get
L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (√2 , 4√2) is not a solution.

(v) (1, 1)

x = 1 and y = 1 in

x – 2y = 4,
LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4
∴ LH.S. ≠ R.H.S.
∴ (1, 1) is not a solution.


EX 4.2 QUESTION 4.


Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.
Solution:
2x + 3y = k
x = 2 and y = 1

2x+3y = k,
2(2) + 3(1) ⇒ k = 4 + 3 – k

⇒ 7 = k
Thus, the required value of k is 7.