$C=\frac{5 F-160}{9}$

$(i)$ 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 $35^{0}=\frac{5(F)-160}{9} \Rightarrow 315^{\circ}=5 F-160$

$\Rightarrow \quad 5 F=315+160=475$

$\therefore \quad F=\frac{475}{5}=95^{\circ}$

Hence, the temperature in Fahrenheit is $95^{\circ} F$.

$(iii)$ Putting $C =0^{\circ},$ we get

$0=\frac{5 F-160}{9} \Rightarrow 0=5 F-160$

$\Rightarrow \quad 5 F=160$

$\therefore \quad F=\frac{160}{5}=32^{\circ}$

Now, putting $F =0^{\circ},$ we get

$C=\frac{5 F-160}{9} \Rightarrow C=\frac{5(0)-160}{9}=\left(-\frac{160}{9}\right)^{0}$

If the temperature is $0^{\circ} C$, the temperature in Fahrenheit is $32^{\circ}$ and if the temperature is $0\, F,$ then the temperature in Celsius is $\left(-\frac{160}{9}\right)^{\circ} C$

$(iv)$ Putting $C = F$, in the given relation, we get

$F=\frac{5 F-160}{9} \Rightarrow 9 F=5 F-160$

$\Rightarrow \quad 4 F=-160$

$F=\frac{-160}{4}=-40^{\circ}$

Hence, the numerical value of the temperature which is same in both the scales is $-40 .$

The linear equation that converts Kelvin $(x)$ to Fahrenheit $(y)$ is given by the relation:

$y=\frac{9}{5}(x-273)+32$