Question: I have this statement: 

> 
> If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
> 
> 
> 

My attempt was:

Move the function $3$ units to the left

$f(2x) = 4(x+3) -2 = 4x+10$

Divide $x$ by $2$

$f(x) = 2x+10$

Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}\_{f(2x-3)}$

But according to the guide the correct answer is $2x+4$ and i don't know why. Thans in advance.

Answer: 
Here is where you went wrong:

If $f(x)= $ [some expression involving $x$], then you can indeed say that $f(x+3)$ = [that expression with $x+3$ filled in for $x$

And, if you have $f(x-3)=$ [some expression involving $x$], then you can indeed say that $f(x)=$ [that expression with $x+3$ filled in for $x$]

So far so good.

But, all you have is that $f(\color{red}{2}x-3)=...$. If you want to 'shift' this, you will have to do this relative to $2x$ ... your mistake was to still treat this as if it was just $x$

What to do? Well, treat the $2x$ as a variable, i.e. Say $y=2x$, and rewrite the formula in terms of this $y$. So, in your case, we can say:

$4x-2=2y-2$

OK, and now we can do our normal shift:

$f(y-3)=2y-2$

And thus:

$f(y)=2(y+3)-2=2y+4$

Substituting $y=2x$ back:

$f(2x)=4x+4$

And hence

$f(x)=2x+4$
Put $y=2x-3$. Then $x=(y+3)/2$. In particular
$$
f(y)=4x-2=4\left(\frac{y+3}{2}\right)-2=2(y+3)-2=2(y+2)
$$
as desired.