Question

19. The taxi fare in a city is as follows: For the first kilometer, the fare is Rs 8 and for the subsequent
distance it is Rs 3 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. (Read statement carefully).​

Answer 1

Step-by-step explanation:

Given,Total distance covered= x km=1+(x-1)km

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1)

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1)

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation.

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation.

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation. It can also be written as y= 5x+3

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation. It can also be written as y= 5x+3When X = 0 ,then Y = 3,When x=1, then y= 5+3=8

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation. It can also be written as y= 5x+3When X = 0 ,then Y = 3,When x=1, then y= 5+3=8When x= 2, then y= 10+3=13

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation. It can also be written as y= 5x+3When X = 0 ,then Y = 3,When x=1, then y= 5+3=8When x= 2, then y= 10+3=13

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation. It can also be written as y= 5x+3When X = 0 ,then Y = 3,When x=1, then y= 5+3=8When x= 2, then y= 10+3=13 [Table & graph are on the attachment]

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation. It can also be written as y= 5x+3When X = 0 ,then Y = 3,When x=1, then y= 5+3=8When x= 2, then y= 10+3=13 [Table & graph are on the attachment]

Given,Total distance covered= x km=1+(x-1)kmFare For first kilometre= ₹8Fare for subsequent distance= ₹5 per km Fare for next(x-1)km= 5(x-1) A.T.Q Total fare= y8+5(x-1)=y8+5x-5=y5x-y+3=0Which is the required linear equation. It can also be written as y= 5x+3When X = 0 ,then Y = 3,When x=1, then y= 5+3=8When x= 2, then y= 10+3=13 [Table & graph are on the attachment] Now plot the points A(0,3), B(1,8), C( 2,13) on the graph paper and join them to form a line BC, which represents the required graph of linear equation.