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### Exercise 4.2 Page: 70

1. Which one of the following options is true, and why?

y = 3x+5 has

1. A unique solution
2. Only two solutions
3. Infinitely many solutions

Solution:

Let us substitute different values for x in the linear equation y = 3x+5,

From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well.

Hence, (iii) infinitely many solutions is the only option true.

2. Write four solutions for each of the following equations:

(i) 2x+y = 7

Solution:

To find the four solutions of 2x+y =7 we substitute different values for x and y

Let x = 0

Then,

2x+y = 7

(20)+y = 7

y = 7

(0,7)

Let x = 1

Then,

2x+y = 7

(2×1)+y = 7

2+y = 7

y = 7-2

y = 5

(1,5)

Let y = 1

Then,

2x+y = 7

(2x)+1 = 7

2x = 7-1

2x = 6

x = 6/2

x = 3

(3,1)

Let x = 2

Then,

2x+y = 7

(2×2)+y = 7

4+y = 7

y =7-4

y = 3

(2,3)

The solutions are (0, 7), (1,5), (3,1), (2,3)

(ii) πx+y = 9

Solution:

To find the four solutions of πx+y = 9 we substitute different values for x and y

Let x = 0

Then,

πx+y = 9

(π0)+y = 9

y = 9

(0,9)

Let x = 1

Then,

πx +y = 9

(π×1)+y = 9

π+y = 9

y = 9-

(1, 9-)

Let y = 0

Then,

πx+y = 9

πx+0 = 9

πx = 9

x = 9/

(9/,0)

Let x = -1

Then,

πx + y = 9

(×-1) + y = 9

-+y = 9

y = 9+π

(-1,9+)

The solutions are (0,9), (1,9-), (9/,0), (-1,9+)

(iii) x = 4y

Solution:

To find the four solutions of x = 4y we substitute different values for x and y

Let x = 0

Then,

x = 4y

0 = 4y

4y= 0

y = 0/4

y = 0

(0,0)

Let x = 1

Then,

x = 4y

1 = 4y

4y = 1

y = 1/4

(1,1/4)

Let y = 4

Then,

x = 4y

x= 4×4

x = 16

(16,4)

Let y =

Then,

x = 4y

x = 4×1

x = 4

(4,1)

The solutions are (0,0), (1,1/4), (16,4), (4,1)

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)

Solutions:

(i) (0, 2)

(x,y) = (0,2)

Here, x=0 and y=2

Substituting the values of x and y in the equation x–2y = 4, we get,

x–2y = 4

⟹ 0 – (2×2) = 4

But, -4 4

(0, 2) is not a solution of the equation x–2y = 4

(ii) (2, 0)

(x,y) = (2, 0)

Here, x = 2 and y = 0

Substituting the values of x and y in the equation x -2y = 4, we get,

x -2y = 4

⟹ 2-(2×0) = 4

⟹ 2 -0 = 4

But, 2 ≠ 4

(2, 0) is not a solution of the equation x-2y = 4

(iii) (4, 0)

Solution:

(x,y) = (4, 0)

Here, x= 4 and y=0

Substituting the values of x and y in the equation x -2y = 4, we get,

x–2y = 4

⟹ 4 – 2×0 = 4

⟹ 4-0 = 4

⟹ 4 = 4

(4, 0) is a solution of the equation x–2y = 4

(iv) (√2,4√2)

Solution:

(x,y) = (√2,4√2)

Here, x = √2 and y = 4√2

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ √2-(2×4√2) = 4

√2-8√2 = 4

But, -7√2 ≠ 4

(√2,4√2) is not a solution of the equation x–2y = 4

(v) (1, 1)

Solution:

(x,y) = (1, 1)

Here, x= 1 and y= 1

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ 1 -(2×1) = 4

⟹ 1-2 = 4

But, -1 ≠ 4

(1, 1) is not a solution of the equation x–2y = 4

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

Solution:

The given equation is

2x+3y = k

According to the question, x = 2 and y = 1.

Now, Substituting the values of x and y in the equation2x+3y = k,

We get,

(2×2)+(3×1) = k

⟹ 4+3 = k

⟹ 7 = k

k = 7

The value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k, is 7.