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

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 = 7Given: 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 = 9Given: 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 = 4yGiven: 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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