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• NCERT Solutions for Class 9 Maths

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

### (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 Ã— 0 + 5

= 0 + 5 = 5

Hence, (0, 5) is one solution

Now, let x = 1, Therefore y = 3 Ã— 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) Ï€x + 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 Ã— 0) + y = 7

y = 7

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

Let x = 1

Then,

2x+y = 7

(2Ã—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

(2Ã—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

(2Ã—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) Ï€x + y = 9

Given: Ï€x+y = 9

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

Let x = 0

Then,

Ï€x + y = 9

(Ï€Ã—0) + y = 9

y = 9

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

Let x = 1

Then,

Ï€x +y = 9

(Ï€Ã—1) + y = 9

Ï€ + y = 9

y = 9 – Ï€

Therefore, we get (x, y) = (1, 9 – Ï€)

Let x = 2

Then,

Ï€x +y = 9

(Ï€Ã—2) + y = 9

2Ï€ + y = 9

y = 9 – 2Ï€

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

Let x = 3

Then,

Ï€x +y = 9

(Ï€Ã—3) + y = 9

3Ï€ + y = 9

y = 9 – 3Ï€

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

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

(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 â€“ 2y = 4 and which are not:

(i) (0, 2)

(ii) (2, 0)

(iii) (4, 0)

(iv) (âˆš2 , 4âˆš2)

(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 â€“ 2y = 4

0 â€“ (2Ã—2) = 4

-4 â‰  4

L.H.S â‰  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 – (2Ã—0) = 4

2 – 0 = 4

2 â‰  4

L.H.S â‰  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 – 2Ã—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) (âˆš2, 4âˆš2)

Given: x – 2y = 4

As, x = âˆš2 and y = 4âˆš2

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

x – 2y = 4

âˆš2 – (2Ã—4âˆš2) = 4

âˆš2 – 8âˆš2 = 4

-7âˆš2 â‰  4

L.H.S â‰  R.H.S

Therefore, (âˆš2, 4âˆš2) 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 – (2Ã—1) = 4

1 – 2 = 4

-1 â‰  4

L.H.S â‰  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

(2Ã—2) + (3Ã—1) = k

4 + 3 = k

7 = k

k = 7

Therefore, the value of k is 7.

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