# Ch 4 : Linear Equations

**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).

**Answer**

Let the cost of pen be y and the cost of notebook be x.

A/q,

Cost of a notebook = twice the pen = 2y.

∴2y = x

⇒ x – 2y = 0

This is a linear equation in two variables to represent this statement.

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 (ii) x – y/5 – 10 = 0 (iii) -2x + 3y = 6 (iv) x = 3y

(v) 2x = -5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x

**Answer**

(i) 2x + 3y = 9.35

⇒ 2x + 3y – 9.35 = 0

On comparing this equation with ax + by + c = 0, we get

a = 2x, b = 3 and c = -9.35

(ii) x – y/5 – 10 = 0

On comparing this equation with ax + by + c = 0, we get

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

(iii) -2x + 3y = 6

⇒ -2x + 3y – 6 = 0

On comparing this equation with ax + by + c = 0, we get

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

(iv) x = 3y

⇒ x – 3y = 0

On comparing this equation with ax + by + c = 0, we get

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

(v) 2x = -5y

⇒ 2x + 5y = 0

On comparing this equation with ax + by + c = 0, we get

a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0

⇒ 3x + 0y + 2 = 0

On comparing this equation with ax + by + c = 0, we get

a = 3, b = 0 and c = 2

(vii) y – 2 = 0

⇒ 0x + y – 2 = 0

On comparing this equation with ax + by + c = 0, we get

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

(viii) 5 = 2x

⇒ -2x + 0y + 5 = 0

On comparing this equation with ax + by + c = 0, we get

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

Page No: 70

**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

**Answer**

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

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

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

**Answer**

(i) 2x + y = 7

⇒ y = 7 – 2x

→ Put x = 0,

y = 7 – 2 × 0 ⇒ y = 7

(0, 7) is the solution.

→ Now, put x = 1

y = 7 – 2 × 1 ⇒ y = 5

(1, 5) is the solution.

→ Now, put x = 2

y = 7 – 2 × 2 ⇒ y = 3

(2, 3) is the solution.

→ Now, put x = -1

y = 7 – 2 × -1 ⇒ y = 9

(-1, 9) is the solution.

The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9

⇒ y = 9 – πx

→ Put x = 0,

y = 9 – π×0 ⇒ y = 9

(0, 9) is the solution.

→ Now, put x = 1

y = 9 – π×1 ⇒ y = 9-π

(1, 9-π) is the solution.

→ Now, put x = 2

y = 9 – π×2 ⇒ y = 9-2π

(2, 9-2π) is the solution.

→ Now, put x = -1

y = 9 – π× -1 ⇒ y = 9+π

(-1, 9+π) is the solution.

The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y

→ Put x = 0,

0 = 4y ⇒ y = 0

(0, 0) is the solution.

→ Now, put x = 1

1 = 4y ⇒ y = 1/4

(1, 1/4) is the solution.

→ Now, put x = 4

4 = 4y ⇒ y = 1

(4, 1) is the solution.

→ Now, put x = 8

8 = 4y ⇒ y = 2

(8, 2) is the solution.

The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

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)

**Answer**

(i) Put x = 0 and y = 2 in the equation x – 2y = 4.

0 – 2×2 = 4

⇒ -4 ≠ 4

∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x – 2y = 4.

2 – 2×0 = 4

⇒ 2 ≠ 4

∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x – 2y = 4.

4 – 2×0 = 4

⇒ 2 = 4

∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x – 2y = 4.

√2 – 2×4√2 = 4 ⇒ √2 – 8√2 = 4 ⇒ √2(1 – 8) = 4

⇒ -7√2 ≠ 4

∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x – 2y = 4.

1 – 2×1 = 4

⇒ -1 ≠ 4

∴ (1, 1) is not a solution of the given equation.

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

**Answer**

Given equation = 2x + 3y = k

x = 2, y = 1 is the solution of the given equation.

A/q,

Putting the value of x and y in the equation, we get

2×2 + 3×1 = k

⇒ k = 4 + 3

⇒ k = 7

Page No: 74

**Exercise 4.3**

1. Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4 (ii) x – y = 2 (iii) y = 3x (iv) 3 = 2x + y

**Answer**

(i) x + y = 4

Put x = 0 then y = 4

Put x = 4 then y = 0

x | 0 | 4 |

y | 4 | 0 |

(ii) x – y = 2

Put x = 0 then y = -2

Put x = 2 then y = 0

x | 0 | 2 |

y | -2 | 0 |

(iii) y = 3x

Put x = 0 then y = 0

Put x = 1 then y = 3

x | 0 | 1 |

y | 0 | 3 |

(iv) 3 = 2x + y

Put x = 0 then y = 3

Put x = 1 then y = 1

x | 0 | 1 |

y | 3 | 1 |

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

**Answer**

Here, x = 2 and y =14.

Thus, x + y = 1

also, y = 7x ⇒ y – 7x = 0

∴ The equations of two lines passing through (2, 14) are

x + y = 1 and y – 7x = 0.

There will be infinite such lines because infinite number of lines can pass through a given point.

3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

**Answer**

The point (3, 4) lies on the graph of the equation.

∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get

3×4 = a×3 + 7

⇒ 12 = 3a + 7

⇒ 3a = 12 – 7

⇒ a = 5/3

4. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

**Answer**

Total fare = y

Total distance covered = x

Fair for the subsequent distance after 1st kilometre = Rs 5

Fair for 1st kilometre = Rs 8

A/q

y = 8 + 5(x-1)

⇒ y = 8 + 5x – 5

⇒ y = 5x + 3

x | 0 | -3/5 |

y | 3 | 0 |

5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.

__For Fig. 4. 6 __ __For Fig. 4.7__

(i) y = x (i) y = x + 2

(ii) x + y = 0 (ii) y = x – 2

(iii) y = 2x (iii) y = –x + 2

(iv) 2 + 3y = 7x (iv) x + 2y = 6

**Answer**

In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).

∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.

In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).

∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points.

Page No: 75

6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is

(i) 2 units (ii) 0 unit

**Answer**

Let the distance traveled by the body be x and y be the work done by the force.

y ∝ x (Given)

⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)

A/q,

(i) When x = 2 units then y = 10 units

(ii) When x = 0 unit then y = 0 unit

x | 2 | 0 |

y | 10 | 0 |

7. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

**Answer**

Let the contribution amount by Yamini be x and contribution amount by Fatima be y.

A/q,

x + y = 100

When x = 0 then y = 100

When x = 50 then y = 50

When x = 100 then y = 0

x | 0 | 50 | 100 |

y | 100 | 50 | 0 |

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 95°F, what is the temperature in Celsius?

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

**Answer**

C | 0 | -10 |

F | 32 | 14 |

(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get

F = (9/5)×30 + 32

⇒ F = 54 + 32

⇒ F = 86

(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get

95 = (9/5)C + 32

⇒ (9/5)C = 95 – 32

⇒ C = 63 × 5/9

⇒ C = 35

(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get

0 = (9/5)C + 32

⇒ (9/5)C = -32

⇒ C = -32 × 5/9

⇒ C = -160/9

Putting the value of C = 0 in F = (9/5)C + 32, we get

F = (9/5)× 0 + 32

⇒ F = 32

(v) Here, we have to find when F = C.

Therefore, Putting F = C in F = (9/5)C + 32, we get

F = (9/5)F + 32

⇒ F – 9/5 F = 32

⇒ -4/5 F = 32

⇒ F = -40

Therefore at -40, both Fahrenheit and Celsius numerically the same.

Page No: 77

**Exercise 4.4**

1. Give the geometric representations of y = 3 as an equation

(i) in one variable

(ii) in two variables

**Answer**

(i) in one variable, it is represented as

y = 3

2. Give the geometric representations of 2x + 9 = 0 as an equation

(i) in one variable

(ii) in two variables

**Answer**

(i) in one variable, it is represented as

x = -9/2