The taxi fare in a city is as follows: For the first kilometre, the fare is $\mathrm{F} 8$ 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.

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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,

Updated on 10-Oct-2022 13:40:07