# Question a8045

May 24, 2017

-40°#

#### Explanation:

We know $F = \frac{9}{5} C + 32$
We need to find a point in which $F = C$
So, we let $F = C$ and rewrite the equation to look like this:
$C = \frac{9}{5} C + 32$

From here, we just solve for C. Now, in a many cases, when we substitute $F$ with $C$, there is no solution and we get something that looks like $1 = 2$. This means that there is no point in which $F = C$

Let's solve this.
$C = \frac{9}{5} C + 32$
$C - \frac{9}{5} C = 32$

Factorize $C$ out.

$C \left(1 - \frac{9}{5}\right) = 32$
$- \frac{4}{5} C = 32$
$C = 32 \cdot - \frac{5}{4} = - 40$

$\therefore$ When $C = - 40$ $F$ is also $- 40$

We can check:
$F = \frac{9}{5} C + 32$
$- 40 = \frac{9}{5} \cdot - 40 + 32$
$- 40 = - 40$

$Q E D$

May 24, 2017

$C = - 40$

#### Explanation:

$F = \frac{9}{5} C + 32$

Let $F = C$ so $C = \frac{9}{5} C + 32$

Bring all $C$ terms to one side

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

Divide both sides by $- \frac{4}{5}$ to get $C$ on its own

$C = - 40$