Linear Equations in Two Variables - NCERT Solutions

Exercise 4.1
Exercise 4.2
Examples
Summary

Exercise 4.1

1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y).

Let the cost of a notebook = ₹ x

Let the cost of a pen = ₹ y

According to the problem:

Cost of notebook = 2 × Cost of pen

So, x = 2y

This can be written as: x - 2y = 0

2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) 2x + 3y = 9.35

2x + 3y - 9.35 = 0

Here, a = 2, b = 3, c = -9.35

(ii) x - y/5 - 10 = 0

This is already in the required form

Here, a = 1, b = -1/5, c = -10

(iii) -2x + 3y = 6

-2x + 3y - 6 = 0

Here, a = -2, b = 3, c = -6

(iv) x = 3y

x - 3y = 0

Here, a = 1, b = -3, c = 0

(v) 2x = -5y

2x + 5y = 0

Here, a = 2, b = 5, c = 0

(vi) 3x + 2 = 0

3x + 0.y + 2 = 0

Here, a = 3, b = 0, c = 2

(vii) y - 2 = 0

0.x + 1.y - 2 = 0

Here, a = 0, b = 1, c = -2

(viii) 5 = 2x

-2x + 0.y + 5 = 0

Here, a = -2, b = 0, c = 5

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

The equation y = 3x + 5 is a linear equation in two variables.

A linear equation in two variables has infinitely many solutions.

For example, when x = 0, y = 5; when x = 1, y = 8; when x = 2, y = 11, and so on.

So, the correct option is (iii) infinitely many solutions.

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

(i) 2x + y = 7

When x = 0, 2(0) + y = 7 ⇒ y = 7 → (0, 7)

When x = 1, 2(1) + y = 7 ⇒ y = 5 → (1, 5)

When x = 2, 2(2) + y = 7 ⇒ y = 3 → (2, 3)

When x = 3, 2(3) + y = 7 ⇒ y = 1 → (3, 1)

(ii) πx + y = 9

When x = 0, π(0) + y = 9 ⇒ y = 9 → (0, 9)

When x = 1, π(1) + y = 9 ⇒ y = 9 - π → (1, 9 - π)

When x = 2, π(2) + y = 9 ⇒ y = 9 - 2π → (2, 9 - 2π)

When x = 3, π(3) + y = 9 ⇒ y = 9 - 3π → (3, 9 - 3π)

(iii) x = 4y

When y = 0, x = 4(0) = 0 → (0, 0)

When y = 1, x = 4(1) = 4 → (4, 1)

When y = 2, x = 4(2) = 8 → (8, 2)

When y = 3, x = 4(3) = 12 → (12, 3)

3. Check which of the following are solutions of the equation x - 2y = 4 and which are not:

(i) (0, 2)

Put x = 0, y = 2: 0 - 2(2) = -4 ≠ 4

So, (0, 2) is not a solution

(ii) (2, 0)

Put x = 2, y = 0: 2 - 2(0) = 2 ≠ 4

So, (2, 0) is not a solution

(iii) (4, 0)

Put x = 4, y = 0: 4 - 2(0) = 4

So, (4, 0) is a solution

(iv) (√2, 4√2)

Put x = √2, y = 4√2: √2 - 2(4√2) = √2 - 8√2 = -7√2 ≠ 4

So, (√2, 4√2) is not a solution

(v) (1, 1)

Put x = 1, y = 1: 1 - 2(1) = -1 ≠ 4

So, (1, 1) is not a solution

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

If (2, 1) is a solution of 2x + 3y = k, then it must satisfy the equation.

Put x = 2, y = 1 in the equation:

2(2) + 3(1) = k

4 + 3 = k

k = 7

Important Examples

Example 1: Write each of the following equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) 2x + 3y = 4.37

2x + 3y - 4.37 = 0

Here a = 2, b = 3, c = -4.37

(ii) x - 4 = √3 y

x - √3 y - 4 = 0

Here a = 1, b = -√3, c = -4

(iii) 4 = 5x - 3y

5x - 3y - 4 = 0

Here a = 5, b = -3, c = -4

It can also be written as -5x + 3y + 4 = 0, where a = -5, b = 3, c = 4

(iv) 2x = y

2x - y + 0 = 0

Here a = 2, b = -1, c = 0

Example 2: Write each of the following as an equation in two variables:

(i) x = -5

1.x + 0.y = -5 or 1.x + 0.y + 5 = 0

(ii) y = 2

0.x + 1.y = 2 or 0.x + 1.y - 2 = 0

(iii) 2x = 3

2x + 0.y - 3 = 0

(iv) 5y = 2

0.x + 5y - 2 = 0

Example 3: Find four different solutions of the equation x + 2y = 6.

When x = 2, 2 + 2y = 6 ⇒ 2y = 4 ⇒ y = 2 → (2, 2)

When x = 0, 0 + 2y = 6 ⇒ 2y = 6 ⇒ y = 3 → (0, 3)

When y = 0, x + 2(0) = 6 ⇒ x = 6 → (6, 0)

When y = 1, x + 2(1) = 6 ⇒ x = 4 → (4, 1)

So four solutions are: (2, 2), (0, 3), (6, 0), (4, 1)

Example 4: Find two solutions for each of the following equations:

(i) 4x + 3y = 12

When x = 0, 3y = 12 ⇒ y = 4 → (0, 4)

When y = 0, 4x = 12 ⇒ x = 3 → (3, 0)

(ii) 2x + 5y = 0

When x = 0, 5y = 0 ⇒ y = 0 → (0, 0)

When x = 1, 2(1) + 5y = 0 ⇒ 5y = -2 ⇒ y = -2/5 → (1, -2/5)

(iii) 3y + 4 = 0

This can be written as 0.x + 3y + 4 = 0

y = -4/3 for any value of x

So two solutions are: (0, -4/3) and (1, -4/3)

Chapter Summary

Key Points:

  • An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.
  • A linear equation in two variables has infinitely many solutions.
  • Every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation.
Important Concepts:

Solution of a Linear Equation:

A solution of a linear equation in two variables means a pair of values, one for x and one for y, which satisfy the given equation.

For example, for the equation 2x + 3y = 12:

  • (3, 2) is a solution because 2(3) + 3(2) = 6 + 6 = 12
  • (0, 4) is a solution because 2(0) + 3(4) = 0 + 12 = 12
  • (1, 4) is not a solution because 2(1) + 3(4) = 2 + 12 = 14 ≠ 12

Finding Solutions:

To find solutions of a linear equation in two variables:

  1. Choose any value for x (or y)
  2. Substitute this value in the equation
  3. Solve for the other variable

For example, for x + 2y = 6:

If we choose x = 2, then 2 + 2y = 6 ⇒ 2y = 4 ⇒ y = 2

So (2, 2) is a solution.