Do you remember the rules of fractions?
If you have some fraction, say, [itex]\frac{2}{3}[/itex] and you multiply it by 9, there are various ways to represent the same number. You could have:
[tex]=9\cdot \frac{2}{3}[/tex] (note, the dot just means multiply, as x could be confused with a variable)
[tex]=\frac{9\cdot 2}{3}[/tex]
[tex]=\frac{18}{3}[/tex]
[tex]=\frac{9}{3}\cdot 2[/tex]
[tex]=3\cdot 2[/tex]
[tex]=6[/tex]
Now notice the expression [itex]\frac{9}{3}\cdot 2=3\cdot 2[/itex]. We essentially cancel out the denominator this way, which is what you're trying to do in your problem.
So why do we multiply your question by 12 and not just any other number? Well, what we're looking for is something called the Lowest Common Denominator (LCD) of 3 and 4. What this means is that we want the first number that both 3 and 4 multiply into.
Multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, ...
Multiples of 4:
4, 8, 12, 16, 20, 24, ...
Notice the common multiples are 12 and 24. But we want the LCD which would be 12.
So now when you multiply [itex]\frac{2(x-1)}{3}[/itex] by 12, you get [tex]12\cdot \frac{2(x-1)}{3}=\frac{12}{3}\cdot 2(x-1)=4\cdot 2(x-1)=8(x-1)[/tex]
and if you multiply the other side by 12, you get [tex]12\left(\frac{x}{4}+1\right)=\frac{12x}{4}+12=3x+12[/tex]
Just remember the common rules of fractions:
[tex]a\cdot\frac{b}{c}=\frac{ab}{c}=\frac{a}{c}\cdot b=a\cdot b\cdot\frac{1}{c}[/tex]
etc.