# For what temperature value are the celsius value and the Fahrenheit value the same?

May 13, 2016

For this, we will need to find an expression for the celsius scale and set it equal to the fahrenheit scale.

Let ${T}_{\text{F}}$ be the temperature on the fahrenheit scale, and ${T}_{\text{C}}$ be the temperature on the celsius scale.

Recall that $\setminus m a t h b f \left({T}_{F} = \frac{9}{5} {T}_{C} + 32\right)$.

The trick is to start by assuming that ${T}_{F} = {T}_{C}$ (that is the premise of the problem) and finding ${T}_{C}$ such that that is the case.

If ${T}_{F} = {T}_{C}$, then:

${T}_{C} = \frac{9}{5} {T}_{C} + 32$

$\frac{5}{5} {T}_{C} = \frac{9}{5} {T}_{C} + 32$

$- 32 = \frac{4}{5} {T}_{C}$

$\textcolor{b l u e}{{T}_{C}} = - 32 \cdot \frac{5}{4}$

$= \textcolor{b l u e}{- {40}^{\circ} \text{C}}$

Now, let's test this claim by converting back to fahrenheit using the first equation we looked at.

$\textcolor{b l u e}{{T}_{F}} = \frac{9}{5} \cdot \left(- {40}^{\circ} \text{C}\right) + 32$

$= - 72 + 32$

$= \textcolor{b l u e}{- {40}^{\circ} \text{F}}$ color(green)(sqrt"")

Given that we were solving a linear equation, there exists one and only one solution. Thus, we have found all solutions.