CBSE Class 9 - CH 4.2 Linear equations in two variables (NCERT Ex 4.2)

CBSE Class 9 - CH 4.2 Linear equations in two variables (NCERT Ex 4.2)

LINEAR EQUATIONS IN TWO VARIABLES

Exercise 4.2

Q1: 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:

y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. As for every value of
x, there will be a value of y satisfying the above equation and vice-versa.

Hence, the correct answer is (iii).


Q2: Write four solutions for each of the following equations:

(i) 2x + y = 7 (ii) ∏x + y = 9 (iii) x = 4y
Answer:

(i) 2x + y = 7
    For x = 0,
    2(0) + y = 7
⇒ y = 7

    Therefore, (0, 7) is a solution of this equation.
    For x = 1,
    2(1) + y = 7
⇒ y = 5

    Therefore, (1, 5) is a solution of this equation.

    For x = −1,
    2(−1) + y = 7
⇒ y = 9
     Therefore, (−1, 9) is a solution of this equation.

     For x = 2,
     2(2) + y = 7
⇒ y = 3
     Therefore, (2, 3) is a solution of this equation.


(ii) ∏x + y = 9

    For x = 0,
    ∏(0) + y = 9

⇒ y = 9
    Therefore, (0, 9) is a solution of this equation.

    For x = 1,
    ∏(1) + y = 9
⇒y = 9 − ∏
    Therefore, (1, 9 − ∏) is a solution of this equation.

    For x = 2,
    ∏(2) + y = 9
⇒ y = 9 − 2∏
    Therefore, (2, 9 −2∏) is a solution of this equation.

     For x = −1,
     ∏(−1) + y = 9
⇒ y = 9 + ∏
⇒ (−1, 9 + ∏) is a solution of this equation.


(iii) x = 4y      For x = 0,
      0 = 4y
⇒  y = 0

Therefore, (0, 0) is a solution of this equation.

     For y = 1,
     x = 4(1) = 4
Therefore, (4, 1) is a solution of this equation.

    For y = −1,
    x = 4(−1)
⇒ x = −4
Therefore, (−4, −1) is a solution of this equation.
   

    For x = 2,
    2 = 4y
    y = 2/4 = 1/2

Therefore,(2 , 1/2) is a solution of this equation.


Q3: 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)


Answer:

(i) (0, 2)

Putting x = 0 and y = 2 in the L.H.S of the given equation,
x − 2y = 0 − 2 ☓ 2 = − 4 ≠ 4
L.H.S ≠ R.H.S
Therefore, (0, 2) is not a solution of this equation.


(ii) (2, 0)

Putting x = 2 and y = 0 in the L.H.S of the given equation,
x − 2y = 2 − 2 ☓ 0 = 2 ≠ 4
L.H.S ≠ R.H.S
Therefore, (2, 0) is not a solution of this equation.


(iii) (4, 0)

Putting x = 4 and y = 0 in the L.H.S of the given equation,
x − 2y = 4 − 2(0) = 4 = R.H.S
Therefore, (4, 0) is a solution of this equation.


(iv) (√ 2 , 4√ 2 )

Putting x = √ 2 and y = 4√ 2 in the L.H.S of the given equation,
x - 2y = √ 2 - 2(4√ 2 )
= √ 2 - 8√ 2 = -7√ 2 ≠ 4
L.H.S ≠ R.H.S
Therefore,(√ 2 , 4√ 2 ) is not a solution of this equation.


(v) (1, 1)
Putting x = 1 and y = 1 in the L.H.S of the given equation,
x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4
L.H.S ≠ R.H.S
Therefore, (1, 1) is not a solution of this equation.



Q4: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer: 

Putting x = 2 and y = 1 in the given equation,
2x + 3y = k
⇒ 2(2) + 3(1) = k
⇒ 4 + 3 = k
⇒ k = 7
Therefore, the value of k is 7.