Question

the taxi fare in a city is as follows.for the first km the fare is rs. 8 and for the subsequent km is is rs 5 take fare as y and distance as x form a linear equation in two variables

Answer 1

$Total \: distance \: covered\: =\: x\: km$

$Fare\: for\: {1}^{st} \: kilometre \: = Rs \: 8$

$Fare \: for \: the \: rest \: of \: the \: distance = rs\: 5 \times reamaining \: distance$

$= rs \: 5 \times (x - 1)$

$. ^{.} . \: total \: fare \: = \: fare \: for \: 1^{st} \: km \: + \: fare \: for \: rest \: km$

$y \: = \: 8 + (x - 1)5$

$y \: = \: 8 + 5x - 5$

$y \: = \: 5x + 3$

(The graph is given above)

Answer 2

⇒ Let the total fare be = y Rs

⇒ Distance covered be = x km

⇒ For first subsequent distance (x - 1) km fare = 5(x - 1)

⇒ y = 8 + 5 (x - 1)

⇒ y = 8 + 5x - 5

⇒ y = 5x + 3

⇒ The coordinates of the following linear equation

1) x = 0, y = 3 (0,3)

2) x = 1, y = 8 (1,8)

3) x = 2, y = 13 (2,13)

4) x = 3, y = 18 (3,18)