Jharkhand Board JAC Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

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Question 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 |

Question 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 = 16

also, y = 7x (∵ x = 2, y = 14)

⇒ y – 7x = 0

∴ The equations of two lines passing through (2, 14) are x + y = 16 and y – 7x = 0.

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

Question 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 3y = ax + 7.

∴ 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 = \(\frac{5}{3}\)

Question 4.

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

Answer:

Total fare = ₹y

Total distance covered = x kilometre

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

Fair for 1st kilometre = ₹8

y = 8 + 5(x – 1)

⇒ y = 8 + 5x – 5

⇒ y = 5x + 3

x | 0 | \(\frac{-3}{5}\) |

y | 3 | 0 |

Question 5.

From the choices given below, choose the equation whose graphs are given in Fig (a) and Fig (b).

For Fig (a)

(i) y = x

(ii) x + y = 0

(iii) y = 2x

(iv) 2 + 3y = 7x

For Fig (b)

(i) y = x + 2

(ii) y = x – 2

(iii) y = -x + 2

(iv) x + 2y = 6

Answer:

In fig. (a), points are (0, 0), (-1, 1) and (1,-1).

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

In fig. (b), points are (-1, 3), (0, 2) and (2, 0).

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

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Question 6.

If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body,

(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

When x = \(\frac{1}{4}\), y = \(\frac{5}{4}\)

When x = \(\frac{1}{2}\) , y = \(\frac{5}{2}\)

x | \(\frac{1}{4}\) | \(\frac{1}{2}\) |

y | \(\frac{5}{4}\) | \(\frac{5}{2}\) |

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

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

Question 7.

Yamini and Fatima, two students of Class IX of a school, together contributed ? 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 ₹ x and ₹ y.) Draw the graph of the same.

Answer:

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

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 |

Question 8.

In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

F = (\(\frac {9}{5}\))C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(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:

(i) F = (\(\frac {9}{5}\))C + 32

When C = 0 then F = 32

also, when C = -10 then F = 14

(ii) Putting the value of C = 30° in

F = (\(\frac {9}{5}\))C + 32, we get

F = \(\frac {9}{5}\) × 30 + 32

⇒ F = 54 +32

⇒ F = 86

(iii) Putting the value of F = 95 in

F = (\(\frac {9}{5}\))C + 32, we get

95 = (\(\frac {9}{5}\))C + 32

⇒ \(\frac {9}{5}\)C = 95 – 32

⇒ C = 63 × \(\frac {5}{9}\) = 35

(iv) Putting the value of F = 0 in

F = (\(\frac {9}{5}\))C + 32, we get

0 = (\(\frac {9}{5}\))C + 32

⇒ \(\frac {9}{5}\)C = – 32

⇒ C = – 32 × \(\frac {5}{9}\)

⇒ C = \(\frac{-160}{9}\)

Putting the value of C = 0 in

F = (\(\frac {9}{5}\))C + 32, we get

F = (\(\frac {9}{5}\)) × 0 + 32

⇒ F = 32

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

Therefore, putting F = C in F = \(\frac {9}{5}\)C + 32, we get

F = \(\frac {9}{5}\)F + 32

⇒ F – \(\frac {9}{5}\)F = 32

⇒ –\(\frac {4}{5}\)F = 32

⇒ F = -40

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