If $9 ≥ 4x + 1$, which inequality represents the possible range of values of $12x + 3?$ If $9 ≥ 4x + 1$, which inequality represents the possible range of values of $12x + 3$?
I've been trying to do SAT prep, and I came across this question. It allowed me to show an explanation and it still didn't make any sense to me. 
"If we look closely, we see that $12x+3=3(4x+1)$." 
What did they do to get $12x+3=3(4x+1)?$
The answer choices:
A. $12x+3≥17$
B. $12x+3≤17$
C. $12x+3≥27$
D. $12x+3≤27$
Thank you for your time. 

$\begin{array}{llr}12x+3 &= 1\cdot (12x + 3)&\text{due to properties of 1}\\
&=3\cdot\frac{1}{3}\cdot(12x+3)&\text{by replacing 1 with}~3\cdot\frac{1}{3}\\
&=3\cdot(\frac{1}{3}\cdot 12x + \frac{1}{3}\cdot 3)&\text{by distributivity of multiplication over addition}\\
&=3\cdot(4x+1)&\text{by evaluating}~\frac{1}{3}\cdot 12~\text{and}~\frac{1}{3}\cdot 3\end{array}$
Going in reverse is faster:
$3(4x+1)=3\cdot 4x + 3\cdot 1 = 12x+3$, again by the distributivity property of multiplication over addition.
In general $a(b+c)=ab+ac$
