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Class 9 NCERT Solutions- Chapter 4 Linear Equations in two variables – Exercise 4.2

  • Last Updated : 17 Dec, 2020
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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:

Given linear equation: y = 3x + 5

Let x = 0, Therefore y = 3 × 0 + 5 

= 0 + 5 = 5

Hence, (0, 5) is one solution



Now, let x = 1, Therefore y = 3 × 1 + 5

= 3 + 5 = 7

Hence, (1, 8) is another solution

Now, let y = 0, Therefore 0 = 3x + 5

x = 5/3

Hence, (5/3, 0) is one another solution.

This concludes that different values of x and y give the different values of y and x respectively.

As there is also no end to the different types of solution for the linear equation in two variables. Therefore, it can have infinitely many solutions.

Hence, option “(iii) infinitely many solutions” is the correct answer.

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 

Given: 2x + y = 7 

To find the four solutions we have to substitute different values for x.

Let x = 0

Then,



2x+y = 7

(2 × 0) + y = 7

y = 7

Therefore, we get (x, y) = (0, 7)

Let x = 1

Then,

2x+y = 7

(2×1)+y = 7

2+y = 7

y = 7-2

y = 5

Therefore, we get (x, y) = (1, 5)

Let x = 2

Then,

2x + y = 7

(2×2) + y = 7

4 + y = 7

y = 7 – 4

y = 3

Therefore, we get (x, y) = (2, 3)

Let x = 3

Then,

2x + y = 7

(2×3) + y = 7

6 + y = 7

y = 7 – 6

y = 1

Therefore, we get (x, y) = (3, 1)

Finally, the four solutions are (0, 7), (1,5), (2,3), (3, 1)

(ii) πx + y = 9

Given: πx+y = 9

To find the four solutions we have to substitute different values for x.

Let x = 0

Then,

πx + y = 9

(π×0) + y = 9

y = 9

Therefore, we get (x, y) = (0, 9)

Let x = 1

Then,



πx +y = 9

(π×1) + y = 9

π + y = 9

y = 9 – π

Therefore, we get (x, y) = (1, 9 – π)

Let x = 2

Then,

πx +y = 9

(π×2) + y = 9

2π + y = 9

y = 9 – 2π

Therefore, we get (x, y) = (1, 9 – 2π)

Let x = 3

Then,

πx +y = 9

(π×3) + y = 9

3π + y = 9

y = 9 – 3π

Therefore, we get (x, y) = (1, 9 – 3π)

Finally, the four solutions are (0, 9), (1, 9 – π), (2, 9 – 2π), (3, 9 – 3π)

(iii) x = 4y

Given: x = 4y

To find the four solutions we have to substitute different values for x.

Let x = 0

Then,

x = 4y

0 = 4y

4y= 0

y = 0/4

y = 0

Therefore, we get (x, y) = (0,0)

Let x = 1

Then,

x = 4y

1 = 4y

4y = 1

y = 1/4

Therefore, we get (x, y) = (1,1/4)

Let x = 2

Then,

x = 4y

2 = 4y

4y = 2

y = 2/4

Therefore, we get (x, y) = (2, 1/2)

Let x = 3

Then,

x = 4y

3 = 4y

4y = 3

y = 3/4

Therefore, we get (x, y) = (2, 3/4)

Finally, the four solutions are (0, 0), (1,1/4), (2, 1/2), (2, 3/4)

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)



Given: x – 2y = 4

As, x=0 and y=2

Hence, substituting the values of x and y in the equation, we get,

x – 2y = 4

0 – (2×2) = 4

-4 ≠ 4

L.H.S ≠ R.H.S

Therefore, (0, 2) is not a solution to the given equation x – 2y = 4.

(ii) (2, 0)

Given: x – 2y = 4

As, x = 2 and y = 0

Hence, substituting the values of x and y in the equation, we get,

x – 2y = 4

2 – (2×0) = 4

2 – 0 = 4

2 ≠ 4

L.H.S ≠ R.H.S

Therefore, (2, 0) is not a solution to the given equation x – 2y = 4.

(iii) (4, 0)

Given: x – 2y = 4

As, x= 4 and y=0

Hence, substituting the values of x and y in the equation, we get,

x – 2y = 4

4 – 2×0 = 4

4 – 0 = 4

4 = 4

L.H.S = R.H.S

Therefore, (4, 0) is a solution to the given equation x – 2y = 4.

(iv) (√2, 4√2)

Given: x – 2y = 4

As, x = √2 and y = 4√2

Hence, substituting the values of x and y in the equation, we get,

x – 2y = 4

√2 – (2×4√2) = 4

√2 – 8√2 = 4

-7√2 ≠ 4

L.H.S ≠ R.H.S

Therefore, (√2, 4√2) is not a solution to the given equation x – 2y = 4.

(v) (1, 1)

Given: x – 2y = 4

As, x= 1 and y= 1

Hence, substituting the values of x and y in the equation, we get,

x – 2y = 4

1 – (2×1) = 4

1 – 2 = 4

-1 ≠ 4

L.H.S ≠ R.H.S

Therefore, (1, 1) is not a solution to the given equation x – 2y = 4.

Question 4: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Solution:

Given: 2x + 3y = k



According to the question, x = 2 and y = 1 is solution of the given equation.

Hence, substituting the values of x and y in the equation 2x+3y = k, we get,

2x + 3y = k

(2×2) + (3×1) = k

4 + 3 = k

7 = k

k = 7

Therefore, the value of k is 7.

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