The linear equation that converts Fahrenheit $(F)$ to Celsius $(C)$ is given by the relation:
$C = \frac{5F - 160}{9}$
$(i)$ If the temperature is $86^{\circ} F$,what is the temperature in Celsius?
$(ii)$ If the temperature is $35^{\circ} C$,what is the temperature in Fahrenheit?
$(iii)$ If the temperature is $0^{\circ} C$,what is the temperature in Fahrenheit? And if the temperature is $0^{\circ} F$,what is the temperature in Celsius?
$(iv)$ What is the numerical value of the temperature which is the same in both the scales?

(N/A) $C = \frac{5F - 160}{9}$
$(i)$ Putting $F = 86^{\circ}$,we get $C = \frac{5(86) - 160}{9} = \frac{430 - 160}{9} = \frac{270}{9} = 30^{\circ}$.
Hence,the temperature in Celsius is $30^{\circ} C$.
$(ii)$ Putting $C = 35^{\circ}$,we get $35 = \frac{5F - 160}{9} \Rightarrow 315 = 5F - 160$.
$\Rightarrow 5F = 315 + 160 = 475$.
$\therefore F = \frac{475}{5} = 95^{\circ}$.
Hence,the temperature in Fahrenheit is $95^{\circ} F$.
$(iii)$ Putting $C = 0^{\circ}$,we get $0 = \frac{5F - 160}{9} \Rightarrow 0 = 5F - 160$.
$\Rightarrow 5F = 160 \Rightarrow F = \frac{160}{5} = 32^{\circ}$.
Now,putting $F = 0^{\circ}$,we get $C = \frac{5(0) - 160}{9} = -\frac{160}{9} \approx -17.78^{\circ} C$.
If the temperature is $0^{\circ} C$,the temperature in Fahrenheit is $32^{\circ} F$,and if the temperature is $0^{\circ} F$,the temperature in Celsius is $-\frac{160}{9}^{\circ} C$.
$(iv)$ Putting $C = F$ in the given relation,we get $F = \frac{5F - 160}{9} \Rightarrow 9F = 5F - 160$.
$\Rightarrow 4F = -160 \Rightarrow F = -40^{\circ}$.
Hence,the numerical value of the temperature which is the same in both scales is $-40$.