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In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
\[\mathrm{F}=\left(\frac{9}{5}\right) \mathrm{C}+32
\]
(i) Draw the graph of the linear equation above using Celsius for t-axis and Fahrenheit for $ y $-axis.
(ii) If the temperature is $ 30^{\circ} \mathrm{C} $, what is the temperature in Fahrenheit?
(iii) If the temperature is $ 95^{\circ} \mathrm{F} $, what is the temperature in Celsius?
(iv) If the temperature is $ 0^{\circ} \mathrm{C} $, what is the temperature in Fahrenheit and if the temperature is $ 0^{\circ} \mathrm{F} $, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Given:
The linear equation that converts Fahrenheit to Celsius is
$F=(\frac{9}{5})C+32$.
To do:
We have to find the solutions of given questions.
Solution:
Given,
$F=(\frac{9}{5})C+32$
(i) To draw the linear equation on a graph by taking Celsius on x-axis and Fahrenheit on y-axis.
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 $F=(\frac{9}{5})C+32$.
Let us substitute $C=0$ in equation $F=(\frac{9}{5})C+32$
For $C=0$
We get,
$F=(\frac{9}{5})0+32$
$F=32$
For $C=-10$
We get,
$F=(\frac{9}{5})-10+32$
$F=9(-2)+32$
$F=-18+32$
$F=14$
Therefore,
$(0, 32)$ and $(-10, 14)$ are two solutions of the equation $F=(\frac{9}{5})C+32$
Hence,
The graph of the linear equation $F=(\frac{9}{5})C+32$ in two variables is,
(ii) The temperature is $30^o\ C$,
This implies,
$C=30$
By substituting $C$ in the linear equation $F=(\frac{9}{5})C+32$
We get,
$F=(\frac{9}{5})30+32$
$F=9(6)+32$
$F=54+32$
$F=86$
Therefore, the temperature in Fahrenheit is $86^o$.
(iii) The temperature is $95^oF$
This implies,
$F = 95$
By substituting $F$ in the linear equation $F=(\frac{9}{5})C+32$
We get,
$95=(\frac{9}{5})C+32$
$(\frac{9}{5})C=95-32$
$(\frac{9}{5})C=63$
$C=\frac{63\times5}{9}$
$C=35$
Therefore, the temperature in Celsius is $35^o$.
(iv) The temperature is $0^o\ C$,
This implies,
$C=0$
By substituting $C$ in the linear equation $F=(\frac{9}{5})C+32$
We get,
$F=(\frac{9}{5})0+32$
$F=0+32$
$F=32$
Therefore, the temperature in Fahrenheit is $32^o$.
The temperature is $0^oF$
This implies,
$F = 0$
By substituting $F$ in the linear equation $F=(\frac{9}{5})C+32$
We get,
$0=(\frac{9}{5})C+32$
$(\frac{9}{5})C=-32$
$C=\frac{-32\times5}{9}$
$C=-17.777$
$C=-17.8$
Therefore, the temperature in Celsius is $-17.8^o$.
(v) The temperature is numerically the same in both Fahrenheit and celsius
This implies,
$F=C$
$C=(\frac{9}{5})C+32$
$C-(\frac{9}{5})C=32$
$\frac{(5-9)C}{5}=32$
$(\frac{-4}{5})C=32$
$(\frac{-4}{5})C=\frac{-32\times5}{4}$
$C=-40$
Therefore,
The temperature which is numerically the same in both Fahrenheit and Celsius is $-40^o$.
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