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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.
Solution
Here, total distance covered $= x,km$
Total taxi fare $=$ Rs. $y$
Fare for the $1^{st} \,km = Rs. $Rs. $ 8$
Remaining distance $=( x -1)\, km$
$\therefore $ Fare for $( x -1) \,km =$ Rs. $ 5 \times( x -1)\, km$
Total taxi fare $=$ Rs. $8+$ Rs. $ 5( x -1)$
$\therefore $ According to the condition,
$\Rightarrow $ $y=8+5(x-1)$
$\Rightarrow $ $y=8+5 x-5$
$\Rightarrow $ $y=5 x+3$
which is the required linear equation representing the given information.
Graph: We have $y =5 x +3$
$\therefore $ When $x=0$, $y=5(0)+3$
$\Rightarrow $ $y=3$
When $x=-\,1$, $y=5(-\,1)+3$
$\Rightarrow $ $y=-\,2$
When $x=-\,2$, $y=5(-\,2)+3$
$\Rightarrow $ $y=-\,7$
$\therefore $ 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, get a straight line $PQ$.
Thus, $PQ$ is the required graph of the linear equation $y =5 x +3$.
