How to cancel out a negative in a denominator? 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?

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