How to cancel out a negative in a denominator?

Question

The question was to make $y$ the subject in $x=5-3y$ (i.e. solve for $y$). My working was this: $$\begin{align*} x&=5-3y \\ -3y&=x-5 \\ y&=(x-5)/(-3) \end{align*}$$

But this was apparently wrong and the correct answer was $(5-x)/3$. Does anyone know how to get that and how to cancel out the $-3$ I had in the denominator of the equation?

Answer 1

$$ \require{cancel} \begin{align*} y \;\;&=\;\; \frac{(x-5)}{(-3)} \\\\&=\;\; \frac{-1(-x+5)}{-1(3)} \\\\&=\;\; \frac{-1}{-1}\times\frac{(-x+5)}{3} \\\\&=\;\; \cancel{\frac{-1}{-1}}\times\frac{(-x+5)}{3} \\\\&=\;\; \frac{(-x+5)}{3} \\\\&=\;\; \frac{(5-x)}{3} \end{align*} $$

Answer 2

Multiplying both the numerator and denominator with -1 will do the trick here.