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 $ \mathrm{km} $. Taking the distance covered as $ x \mathrm{~km} $ and total fare as Rs $ y $, write a linear equation for this information, and draw its graph.
Given:
The taxi fare in a city is $\ Rs. 8$ for the first kilometre and for the subsequent distance it is $Rs.5\ per\ kilometre$.
To do:
We have to write the linear equation by taking the distance covered as \( x \mathrm{~km} \) and total fare as \( ₹ y \) and draw its graph
Solution:
Let the total distance covered be $x\ km$.
Fare for the first kilometre $=1\times 8=Rs.\ 8$
Fare for the subsequent distance$=Rs.\ ( x-1)\times5$
According to the question,
$8+( x-1)5=y$
$\Rightarrow 8+5x-5=y$
$\Rightarrow 5x-y+3=0$
The linear equation representing the given information is $5x-y+3=0$.
We know that,
To draw a graph of a linear equation in two variables, we need at least two solutions to the given equation.
To find the solutions to the given equation $5x-y+3=0$.
This implies,
$5x-y=-3$
Let us substitute $x=0$ and $y=0$ in equation $5x-y=-3$
For $x=0$
We get,
$5(0)-y=-3$
$0-y=-3$
$y=3$
For $y=0$
We get,
$5x-0=-3$
$5x=-3$
$x=\frac{-3}{5}$
Therefore,
$(0, 3)$ and $(\frac{-3}{5}, 0)$ are two solutions of the equation$5x-y=-3$.
Hence,
The graph of the linear equation $5x-y=-3$ in two variables is,
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