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:
$F = \left(\frac{9}{5}\right) C + 32$
$(i)$ Draw the graph of the linear equation above using Celsius for $x$-axis and Fahrenheit for $y$-axis.
$(ii)$ If the temperature is $30\,^oC$,what is the temperature in Fahrenheit?
$(iii)$ If the temperature is $95\,^oF$,what is the temperature in Celsius?
$(iv)$ If the temperature is $0\,^oC$,what is the temperature in Fahrenheit and if the temperature is $0\,^oF$,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.

(N/A) $(i)$ We have the equation $F = \left(\frac{9}{5}\right) C + 32$.
When $C = 0$,$F = \left(\frac{9}{5}\right) \times 0 + 32 = 32$.
When $C = -15$,$F = \frac{9}{5}(-15) + 32 = -27 + 32 = 5$.
When $C = -10$,$F = \frac{9}{5}(-10) + 32 = 9(-2) + 32 = 14$.
Table of values:

$C$	$0$	$-15$	$-10$
$F$	$32$	$5$	$14$

Plotting the points $(0, 32)$,$(-15, 5)$,and $(-10, 14)$ on a graph and joining them gives a straight line.
$(ii)$ For $C = 30$,$F = \frac{9}{5}(30) + 32 = 9(6) + 32 = 54 + 32 = 86\,^oF$.
$(iii)$ For $F = 95$,$95 = \frac{9}{5}C + 32 \implies 63 = \frac{9}{5}C \implies C = 63 \times \frac{5}{9} = 35\,^oC$.
$(iv)$ For $C = 0$,$F = 32\,^oF$. For $F = 0$,$0 = \frac{9}{5}C + 32 \implies -32 = \frac{9}{5}C \implies C = -\frac{160}{9} \approx -17.8\,^oC$.
$(v)$ Let $F = C = x$. Then $x = \frac{9}{5}x + 32 \implies x - \frac{9}{5}x = 32 \implies -\frac{4}{5}x = 32 \implies x = 32 \times (-\frac{5}{4}) = -40$. Thus,$-40\,^oC = -40\,^oF$.