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. 
 A: Note that
$$
12x+3=3(4x+1)
$$
so $12x+3\le 27$
A: You can multiply both sides of an inequality by the same positive number and the inequality will still hold.
The "look-closely" part points out that you can multiply $4x+1$ by $3$ to get $12x + 3$.  Just distribute the $3$ through each term:
$$3 \cdot (4x+1) = 3 \cdot 4x + 3 \cdot 1 = 12x + 3.$$
Hence, to keep the inequality the same, you multiply the other side by $3$:
$$9 \geq 4x + 1$$
$$3 \cdot 9 \geq 3 \cdot (4x+1)$$
$$27 \geq 12x + 3$$
$$12x + 3 \leq 27$$
The last step just flips the sides to make it match the form of the choices.
(If we were multiplying by a negative number, the direction of the inequality would switch.)
A: If you want to do it the easy but harder way
$9 \ge 4x + 1$
$8 \ge 4x$
$2 \ge x$
$24 \ge 12 x$
$27 \ge 12x + 3$
But if you want to do it the hard by easier way
$9 \ge 4x + 1$
$3*9 \ge 3(4x + 1)$
$27 \ge 12x + 3$
A: $\begin{array}{llr}12x+3 &= 1\cdot (12x + 3)&\text{due to properties of 1}\\
&=3\cdot\frac{1}{3}\cdot(12x+3)&\text{by replacing 1 with}~3\cdot\frac{1}{3}\\
&=3\cdot(\frac{1}{3}\cdot 12x + \frac{1}{3}\cdot 3)&\text{by distributivity of multiplication over addition}\\
&=3\cdot(4x+1)&\text{by evaluating}~\frac{1}{3}\cdot 12~\text{and}~\frac{1}{3}\cdot 3\end{array}$
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 $a(b+c)=ab+ac$
