℉ = \(\frac{9}{5}\)(ºC) + 32
Or ℉ = \(\big(\)ºC x \(\frac{9}{5}\)\(\big)\)+ 32 t
℃ = (℉ - 32) x \(\frac{5}{9}\)
To find the temperature when both are equal, we use an old algebra trick and just set ℉ = ℃ and solve one of the equations.
℃ = \(\big(\)ºC x \(\frac{9}{5}\)\(\big)\)+ 32
Or, ℃ - \(\big(\)℃ x \(\frac{9}{5}\)\(\big)\) = 32
Or, \(-\frac{4}{5}\) x ℃ = 32
Or, ℃ = \(-\frac{32\times5}{4}\) = -40
℉ = \(\big(^oF\times\frac{9}{5}\big)\)+ 32
Or, ℉ - \(\big(^oF\times\frac{9}{5}\big)\) = 32
Or, \(-\frac{4}{5}\) x ℉ = 32
℉ = \(-\frac{-32\times5}{4}\) = -40
So, at – 40-degree temperature, both the Celsius and Fahrenheit scales read the same value.
