# Question #907e5

May 13, 2016

$- {40}^{o}$

#### Explanation:

Note the conversion formula:

$\frac{C}{5} = \frac{F - 32}{9.} \ldots \ldots \left(1\right)$

We are required to find such an $x$ such that $C = x = F$. Lets try putting this on relation $\left(1\right)$:

$\frac{x}{5} = \frac{x - 32}{9}$

$\implies 9 x = 5 x - 160$

$\implies 4 x = - 160$

$\implies x = - 40$

Hence, $- {40}^{o} C = - {40}^{o} F$