NCERT solution class 9 chapter 4 Linear Equations in Two Variables exercise 4.2 mathematics

EXERCISE 4.2


Page No 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:

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).


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

For x = 0,

2(0) + = 7

⇒ = 7

Therefore, (0, 7) is a solution of this equation.

For = 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 = 2,

2(2) + y = 7

y = 3

Therefore, (2, 3) is a solution of this equation.

(ii) πx + y = 9

For x = 0,

π(0) + = 9

⇒ = 9

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

For x = 1,

π(1) + = 9

= 9 − π

Therefore, (1, 9 − π) is a solution of this equation.

For x = 2,

π(2) + = 9

⇒ = 9 − 2π

Therefore, (2, 9 −2π) is a solution of this equation.

For x = −1,

π(−1) + = 9

⇒ = 9 + π

⇒ (−1, 9 + π) is a solution of this equation.

(iii) x = 4y

For x = 0,

0 = 4y

⇒ = 0

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

For = 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

Therefore, is a solution of this equation.


Question 3:

Check which of the following are solutions of the equation x − 2= 4 and which are not:

(i) (0, 2 (ii) (2, 0) (iii) (4, 0)

(iv)  (v) (1, 1)

Answer:

(i) (0, 2)

Putting x = 0 and y = 2 in the L.H.S of the given equation,

− 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,

− 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,

− 2= 4 − 2(0)

= 4 = R.H.S

Therefore, (4, 0) is a solution of this equation.

(iv) 

Putting

 and in the L.H.S of the given equation,

L.H.S ≠ R.H.S

Therefore, 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,

− 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4

L.H.S ≠ R.H.S

Therefore, (1, 1) is not a solution of this equation.

Question 4:

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

Answer:

Putting x = 2 and y = 1 in the given equation,

2+ 3y = k

⇒ 2(2) + 3(1) = k

⇒ 4 + 3 = k

⇒ k = 7

Therefore, the value of k is 7.


 

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