The taxi fare in a city is as follows:
For the first kilometre,the fare is Rs. $8$ and for the subsequent distance it is Rs. $5$ 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.

(N/A) Here,total distance covered $= x \, km$
Total taxi fare $= y$
Fare for the $1^{st} \, km = Rs. \, 8$
Remaining distance $= (x - 1) \, km$
$\therefore$ Fare for $(x - 1) \, km = Rs. \, 5(x - 1)$
Total taxi fare $= Rs. \, 8 + 5(x - 1)$
$\therefore$ According to the condition,
$y = 8 + 5(x - 1)$
$y = 8 + 5x - 5$
$y = 5x + 3$
This is the required linear equation representing the given information.
Graph: We have $y = 5x + 3$
When $x = 0, y = 5(0) + 3 = 3$
When $x = -1, y = 5(-1) + 3 = -2$
When $x = -2, y = 5(-2) + 3 = -7$
We get the following table:

$x$	$0$	$-1$	$-2$
$y$	$3$	$-2$	$-7$

Now,plotting the ordered pairs $(0, 3), (-1, -2)$ and $(-2, -7)$ on a graph paper and joining them,we get a straight line $PQ$. Thus,$PQ$ is the required graph of the linear equation $y = 5x + 3$.