Practical Applications of Linear Equations

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Linear equations arise in a lot of practical situations. For example, consider the two widely used temperature scales: Celsius and Fahrenheit. To convert from the Fahrenheit scale to the Celsius scale, the following relation is used:

\[F = \left( {\frac{9}{5}} \right)C + 32\]

Note carefully that this is a linear equation in the two variables F and C. Let us plot the graph for this linear equation using two particular solutions:

\[C = \]

\(0\) \(20\)
\[F = \] \(32\)



\[\left( {0,32}\right)\] \[\left( {20,68}\right)\]

The graph is plotted below:

Celsius to Fahrenheit linear graph

Using this graph, we can convert easily from Celsius to Fahrenheit and vice-versa. For example, we note that \(P\left( {30,86} \right)\) lies on the graph, which means that when \(C = 30\), then \(F = 86\), that is, \({30^0}C\) is \({86^0}F\). Similarly, we see that \(Q\left( {35,95} \right)\) lies on the graph, that is, \({35^0}C\) is \({95^0}F\).

We also note that the line intersects the y-axis at roughly \(\left( { - 17.7,0} \right)\), which means that \({0^0}F\) is approximately \( - {17.7^0}C\).

Is there a temperature whose numerical value is the same in both scales? To determine such a temperature, we need to determine a point on the line both of whose coordinates are the same, that is, a point for which

x-coordinate = y-coordinate

To do this, we draw the line \(y = x\) on the Celsius-Fahrenheit graph. Wherever the line \(y = x\) intersects the Celsius-Fahrenheit line, the point obtained will have the same x and y coordinates (can you see why?):

Intersecting Celsius to Fahrenheit linear graph

We see that the intersection point is \(\left( { - 40, - 40} \right)\). Thus, \( - {40^0}C\) when converted to Fahrenheit scale is \( - {40^0}F\). This temperature has the same numerical value on both the scales.

Example 1: In a certain city, there are two kinds of taxis available: black and yellow. Black taxis charge a down-payment of Rs. 20 and a per-km fare of Rs. 4. Yellow taxis charge a down-payment of Rs. 10 and per-km fare of Rs. 5.

I. Use \(f\) to denote the total fare (in rupees) and \(d\) to denote the distance covered (in km). Construct linear equations related \(f\) and \(d\) for the black and yellow taxis.

II. Draw graphs for these equations.

III. Under what conditions is a black taxi cheaper than a yellow taxi?


I. For the black taxi, we have \(f = 20 + 4d\), while for the yellow taxi, we have \(f = 10 + 5d\).

II. The two linear equations are plotted on the same axes below (the scale for the vertical axis is different than the horizontal axis):

Same axes two linear equations

Note that the two lines intersect at the point \(\left( {10,60} \right)\), that is, at \(d = 10,\,f = 60\).

III. For distances less than 10 km, the line for the black taxi lies above the yellow taxi (that is, fares for the black taxi will be higher). When \(d = 10\) km, the fares for the two taxis will be the same (Rs. 60). When \(d > 10\) km, the line for the black taxi goes below that of the yellow taxi. Thus, for distances greater than 10 km, a black taxi will be cheaper.

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Linear Equations
grade 9 | Answers Set 1
Linear Equations
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Linear Equations
grade 9 | Questions Set 1
Linear Equations
grade 9 | Answers Set 2