NCERT Class 9 Solutions: Linear Equation in Two Variable (Chapter 4) Exercise 4.3 – Part-3

Q-4 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. 5 per kilometer. Taking the distance covered as x km and total fares as Rs. y, write a linear equation for this information, and draw its graph.

Solution:

Given,

Taxi fare for subsequent distance = Rs. 5

Let,

  • Total distance covered =x

  • Total fare =y

According to problem,

  • Taxi fare for first kilometer = Rs. 8

  • If the total distance is x, fare for rest of the (x1) kilometer ( at the rate of Rs. 5 per km) =5(x1)

  • Since the fare for first kilometer = Rs. 8, so, the total fare =5(x1)+8=y

    y=5(x1)+8

    y=5x5+8

    y=5x+3

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

Drawing the Graph

We can draw the graph by finding two points on this equation and drawing a line joining those points.

The equation is

y=5x+3 …………….equation (1)

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

y=5×0+3

y=0+3=3 So the solution is (0,3) this is first point (the y intercept)

Putting the value x=1 in equation (1)

y=5×1+3

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

Segment f Segment f: Segment [C3, C2] Point C2 Point C2: (A2, B2) Point C2 Point C2: (A2, B2) Point C2 Point C2: (A2, B2) Point C3 Point C3: (A3, B3) Point C3 Point C3: (A3, B3) Point C3 Point C3: (A3, B3) (0,3) text2 = "(0,3)" (1,8) text1 = "(1,8)" y=5x 3 text3 = "y=5x 3"

Graph for Equation Y = 5x + 3

x-axis and y-axis, distance covered as x km and total fare as Rs. Y. Equation is y = 5x + 3

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