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Linear Equations in Two Variables

    Exercise 4.2

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9th class > Linear Equations in Two Variables > Exercise 4.2

Exercise 4.2

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

x012345
y

From the table, it is clear that 'x' can have 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.
  1. Write four solutions for each of the following equations:

(i) 2x + y = 7

Let's substituting x = 0, 1, 2, 3

Thus,

2x + y = 7 ⇒ 2 × + y = 7 ⇒ y =

First solution: (0,7)

Putting x = 1: 2x + y = 7

⇒ 2 × + y = 7 ⇒ y =

Second solution: (1,5)

Putting x = 2: 2x + y = 7 ⇒ 2× + y = 7 ⇒ y =

Third solution: (2,3)

Putting x = 3: 2x + y = 7 ⇒ 2× + y = 7 ⇒ y =

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

(ii) πx + y = 9

Putting x = 0,1,2 and 3.

Thus,

Putting x = 0: πx+y = 9 ⇒ π × + y = 9 ⇒ y =

First solution: (0,9)

Putting x = 1: πx+y = 9 ⇒ π × + y = 9 ⇒ y =

Second solution: (1, 9-π)

Putting x = 2: πx+y = 9 ⇒ π × + y = 9 ⇒ y =

Third solution: (2, 9-2π)

Putting x = 3: πx+y = 9 ⇒ π × + y = 9 ⇒ y =

The solutions are (0,9), (1,9-π), (2,9-2π), (3,9-3π)

(iii) x = 4y

Putting x = 0,1,2 and 4.

Putting x = 0: x = 4y ⇒ = 4y ⇒ y =

First solution: (0, 0)

Putting x = 1: x = 4y ⇒ = 4y ⇒ y =

Second solution: (1,1/4)

Putting x = 2: x = 4y ⇒ = 4y ⇒ y =

Third solution: (2,1/2)

Putting x = 4: x = 4y ⇒ = 4y ⇒ y =

The solutions are (0,0), (1,1/4), (2,1/2), (4,1).
  1. Check which of the following are solutions of the equation x – 2y = 4 and which are not:

(i) (0, 2)

Here, x = and y =

Substituting the values of x and y in the equation x−2y=4:

x−2y=4 ⟹ – = 4

⟹ -4 4

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

(ii) (2, 0)

Here, x = and y =

Substituting the values of x and y in the equation x−2y=4:

x -2y = 4 ⟹ 2−2×0=4 ⟹ - = 4

⟹ 2 4

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

(iii) (4, 0)

Here, x = and y =

Substituting the values of x and y in the equation x−2y=4:

x−2y=4 ⟹ 4 – 2 × 0 = 4 ⟹ - = 4

⟹ 4 4

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

(iv) (2, 42)

Here, x = and y =

Substituting the values of x and y in the equation x−2y=4:

x−2y=4 ⟹ 2−2×42=4 ⟹ - = 4

But, -7√2 4

(2,42) is not a solution of the equation x−2y=4.

(v) (1,1)

Here, x = and y =

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

x−2y=4 ⟹ 1−2×1=4 ⟹ 1 - = 4

But, -1 4

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

Q4

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

Sol

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 equation: 2x+3y=k

⟹ 2 × + 3 × = k

⟹ + = k

⟹ k =

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