**Q.8.** 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:

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

- Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
- If the temperature is 30°C, what is the temperature in Fahrenheit?
- If the temperature is 95°F, what is the temperature in Celsius?
- If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
- Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

**Solution:**

(i) F = $\left(\frac{9}{5}\right)$C + 32

The solutions can be tabulated as follows,

C |
0 |
-10 |

F |
32 |
14 |

Taking 1 unit = 10 on both x and y-axis

(ii) Putting the value of C = 30 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get,

F = $\left(\frac{9}{5}\right)$×30 + 32

= 54 + 32 = 86

(iii) Putting the value of F = 95 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get,

95 = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32

⇒C = (95 - 32)× $\left(\frac{5}{9}\right)$

= 63 × $\left(\frac{5}{9}\right)$

⇒ C = 35

(iv) Putting the value of F = 0 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get

0 = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32

⇒ C = - 32× $\left(\frac{5}{9}\right)$

= 63 × $\left(\frac{5}{9}\right)$

⇒ C = - $\frac{160}{9}$

Putting the value of C = 0 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get,

F = $\left(\frac{9}{5}\right)\mathrm{}$× 0 + 32 = 32

(v) Putting F = C in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get,

F = $\left(\frac{9}{5}\right)\mathrm{}$F + 32

⇒ F - $\left(\frac{9}{5}\right)\mathrm{}$F = 32

⇒ - $\left(\frac{9}{5}\right)\mathrm{}$F = 32

⇒ F = - 40

Therefore both Fahrenheit and Celsius numerically the same at -40 ^{o}C^{ }= -40 ^{o}F.