An auto rickshaw fare in the cityis charyedR 1 for the first kilometer

1.	An auto rickshaw fare in the cityis charyedR 1 for the first kilometer and 4 per kilometertor subsequent distance covered Write the linear equation to express the above statement.Draw the graph of the linear equation.
Answer» Solution :- Let the total distance covered be x km and, the total fare be Rs. y Since, Total fare = fare for the first kilometer + fare for the remaining kilometer of distance ...........................(1) We know that the fare for first kilometer is Rs. 10 Fare for the remaining kilometers i.e. (x - 1) = 4(x - 1) Put values in (1) ⇒ y = 10(1) + 4(x - 1) ⇒ y = 10 + 4x - 4 ⇒ y = 4x + 6 When x = 0, we have, y = 4× (0) + 6, So, y = 6 When x = - 1, we have, y = 4× (- 1) + 6, = - 4 + 6 So, y = 2When x = - 2, we have, y = 4× (- 2) + 6 = - 8 + 6 So, y = - 2 x = 0 ; y = 6 x = - 1 ; y = 2 x = - 2 ; y = - 2 Now, you can draw the graph easily.

An auto rickshaw fare in the cityis charyedR 1 for the first kilometer and 4 per kilometertor subsequent distance covered Write the linear equation to express the above statement.Draw the graph of the linear equation.

Answer»

Solution :-

Let the total distance covered be x km

and, the total fare be Rs. y

Since, Total fare = fare for the first kilometer + fare for the remaining kilometer of distance ...........................(1)

We know that the fare for first kilometer is Rs. 10

Fare for the remaining kilometers i.e. (x - 1)

= 4(x - 1)

Put values in (1)

⇒ y = 10(1) + 4(x - 1)

⇒ y = 10 + 4x - 4

⇒ y = 4x + 6

When x = 0, we have, y = 4× (0) + 6, So, y = 6

When x = - 1, we have, y = 4× (- 1) + 6, = - 4 + 6

So, y = 2When x = - 2, we have, y = 4× (- 2) + 6 = - 8 + 6

So, y = - 2

x = 0 ; y = 6

x = - 1 ; y = 2

x = - 2 ; y = - 2

Now, you can draw the graph easily.