the taxi fare in a city is as follows:for the first kilometre ,the fare is rupee 8 and for the subsequent distance it is 5 rupees per km. Taking the distance covered as x km and total fare as y rupee, write a linear equation for this information,and draw its graph.
Answers
Answer:
Given, Total distance covered= x km=1+(x-1)km
Fare For first kilometre= ₹8
Fare for subsequent distance= ₹5 per km
Fare for next(x-1)km= 5(x-1)
A.T.Q
Total fare= y
8+5(x-1)=y
8+5x-5=y
5x-y+3=0
Which is the required linear equation.
It can also be written as y= 5x+3
When X = 0 ,then Y = 3,When x=1, then y= 5+3=8
When x= 2, then y= 10+3=13 ANS
:
Mark Me As Brainliest And Thank Me If The Answer Is Useful.
-- :
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)