The temperature both Celsius & Fahrenheit scales showing same reading is

Let the temperature is t that gives same reading=$t^{0}C$

In celsius & Fahrenheit scale=$t^{0}F$

We know

$$\begin{gathered} i) F=\frac{9C}{5}+32 \\ t=\frac{9t}{5}+32 \\ 5t=9t+160 \\ -160=9t-5t \\ \Rightarrow 4t=-160 \\ \therefore t=-{40}^{0}C=-{40}^{0}F \\ ii) or consider \\ C=\frac{5}{9}(F-32) \\ t=\frac{5}{9}(t-32) \\ 9t=5t-160 \\ 4t=-160 \\ \therefore t=-{40}^{0}F \end{gathered}$$