Euclidean algorithm as a linear combination substitution and simplifying This problem is from my discrete mathematics textbook.
I'm trying to find $\gcd(420,66)$
I compute
$$\begin{align*}
420 &= 6 \times 66 + 24\\
66 &= 2\times 24 + 18\\
24 &= 1 \times 18 + 6\\
18 &= 3 \times 6 + 0
\end{align*}$$
then I rewrite the equation 
$$\begin{align*}
6 &= 24 - 1 \times 18\\
18 &= 66 - 2 \times 24\\
24 &= 420 - 6 \times 66\\
\end{align*}$$
Now I try to perform substitutions which give me 
$$\begin{align*}
6 &= 24 -1 \times 18\\
& = 24-1 (66 - 2 \times 24)\\
&= 3 \times 24 -66
\end{align*}$$
My question is how do you transition from 
$$    24-1  (66 - 2 \times 24)$$ 
to
$$3 \times 24 -66$$
I just can't wrap my head around this part. Maybe I'm way over thinking this step. 
Any help is appreciated thanks!

Distribute, reorder, associate:
$$\begin{align*}
24 - 1(66-2\times 24) &= 24 -1(66) -1(-2\times 24)\\
&= 24 - 66 +2\times 24\\
&= 24 + 2\times 24 - 66\\
&= (1+2)24 - 66\\
&= 3\times 24 - 66.
\end{align*}$$
