Here, total distance covered = x km and total taxi fare = Rs. yFare for 1km = Rs. 8Remaining distance = (x – 1) km∴ Fare for (x – 1)km = Rs.5 x(x – 1)Total taxi fare = Rs. 8 + Rs. 5(x – 1)According to question,y = 8 + 5(x – 1) = y = 8 + 5x – 5⇒ y = 5x + 3,which is the required linear equation representing the given information.Graph: We have y = 5x + 3Wben x = 0, then y = 5(0) + 3 ⇒ y = 3x = -1, then y = 5(-1) + 3 ⇒ y = -2x = -2, then y = 5(-2) + 3 ⇒ y = -7∴ We get the following table:Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.Thus, the line PQ is the required graph of the linear equation y = 5x + 3.