The taxi fare in a city is as follows: For the first two kilometers, t

1.	The taxi fare in a city is as follows: For the first two kilometers, the fare is ₹ 14 and for the subsequent distance it is ₹ 6 per km. Taking the distance covered as x km and the total fare as ₹ y, write a linear equation and draw its graph.
Answer» ANSWER:x 0 1 2 y 3 8 13 Taxi fare for FIRST kilometer = Rs. 8 Taxi fare for subsequent distance = Rs. 5 Total distance covered =x Total fare =y Since the fare for first kilometer = Rs.8 According to problem, Fare for (x–1) kilometer = 5(x−1) So, the total fare y=5(x−1)+8 ⇒y=5(x−1)+8 ⇒y=5x–5+8 ⇒y=5x+3 Hence, y=5x+3 is the required linear equation. Now the equation is y=5x+3 ...(1) Now, putting the value x=0 in (1) y=5×0+3 y=0+3=3 So the solution is (0,3) Putting the value x=1 in (1) y=5×1+3 y=5+3=8. So the solution is (1,8) Putting the value x=2 in (1) y=5×2+3 y=10+3=13. So the solution is (2,13) Hope it helps you!!!

The taxi fare in a city is as follows: For the first two kilometers, the fare is ₹ 14 and for the subsequent distance it is ₹ 6 per km. Taking the distance covered as x km and the total fare as ₹ y, write a linear equation and draw its graph.

Answer»

ANSWER:x 0 1 2

y 3 8 13

Taxi fare for FIRST kilometer = Rs. 8

Taxi fare for subsequent distance = Rs. 5

Total distance covered =x

Total fare =y

Since the fare for first kilometer = Rs.8

According to problem,

Fare for (x–1) kilometer = 5(x−1)

So, the total fare y=5(x−1)+8

⇒y=5(x−1)+8

⇒y=5x–5+8

⇒y=5x+3

Hence, y=5x+3 is the required linear equation.

Now the equation is

y=5x+3 ...(1)

Now, putting the value x=0 in (1)

y=5×0+3

y=0+3=3 So the solution is (0,3)

Putting the value x=1 in (1)

y=5×1+3

y=5+3=8. So the solution is (1,8)

Putting the value x=2 in (1)

y=5×2+3

y=10+3=13. So the solution is (2,13)

Hope it helps you!!!