# Ex.4.3 Q4 Linear Equations in Two Variables Solution - NCERT Maths Class 9

## Question

The taxi fare in a city is as follows: For the first kilometer, the fare is \(₹\, 8\) and for the subsequent distance it is \(₹\, 5 \,\rm{per \,km.}\) Taking the distance covered as \(x\) \(\rm{km}\) and total fare as \(₹\) \(y\), write a linear equation for this information, and draw its graph.

## Text Solution

**What is known? **

Distance covered is \(x\) \(\rm{km}\) and total fare is \(₹\). \(y\). Also fare of first kilometer is \(₹\, 8\) and for subsequent distance is \(₹\, 5\)

**What is Unknown?**

Linear equation for given condition and also graph of it.

**Reasoning: **

We know fare for first \(\rm{km}\) and add the fare of subsequent distance and the result will form linear equation. After that by finding the solution for linear equation we can draw graph of it.

**Steps:**

**Given,**

Let the Total distance covered \(= x\,\rm{km}\)

Let the Total fare covered \(= ₹\) \(y\)

Therefore, subsequent distance \(= (x -1) \rm{km}\)

Fare for the subsequent distance \(= ₹ 5 (x − 1)\)

Fare for \(1\)st kilometer \(= ₹\, 8\)

The linear equation for the above information is given by,

Total fare

\[\begin{align} &(y) = [ 8 + ( x - 1 ) 5 ] \\ &y = 8 + 5 x - 5 \\ &y = 5x + 3\\ &5 x - y + 3 = 0 \end{align}\]

Re-write the equation by transposing as \(y = 5x + 3 \qquad \dots \dots \text{ Equation } (1)\)

By substituting different values of \(x\) in the Equation (1) we get different values for \(y\)

- When \(x = 0\),

\(y = 5 × 0 + 3 = 0 + 3 = 3\)

- When \(x = 1\),

\(y = 5 × (1) + 3 = 5 + 3 = 8\)

- When \(x = 2\),

\(y = 5 × (2) + 3 = 10 + 3 = 13 \)

- When \(x = -1\),

\(y = 5 × (–1) + 3 =\, –5 + 3 =\, -2 \)

- When \(x = -2\),

\(y = 5 × (–2) + 3 =\, –10 + 3 =\, -7\)

Thus, we have the following table with all the obtained solutions:

\(x\) | \(0\) | \(1\) | \(2\) | \(-1\) | \(-2\) |

\(y\) | \(3\) | \(8\) | \(13\) | \(-2\) | \(-7\) |

By Plotting the points \((0, 3), (1, 8), (2, 13),\) \((-1, -2), (-2, -7)\) on the graph paper and drawing a line joining them, we, obtain the required graph

The graph of the line represented by the given equation as shown:

Here, the variable \(x\) and \(y\) are representing the distance covered and the fare paid for that distance respectively and these quantities may not be negative.

Hence, only those values of \(x\) and \(y\) which are lying in the \(1\rm st\) quadrant will be considered.