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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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