How to solve an equation involving the whole part of a number? 
If $[x]$ represents the floor of $x$, solve the equation:
$$\left[\frac{x+1}3\right]=\frac{x-1}2$$
Choose the correct answer:
a) $(2,7)\cup(9,15)$  
b) $(-3,2)\cup[3,4)\cup(6,14)$  
c) $(-1,1]\cup[2,3)\cup(5,8)$  
d) $(-5,-3)\cup(1,3]\cup[5,7)$  
e) $[1,\frac32]\cup(2,4)\cup[5,7)$  
f) $[0,2]\cup[4,7]\cup(9,+\infty)$

Can someone please explain how I can solve this type of equation? Thank you very much!
P.S. I'm not sure if $[x]$ is called whole part of $x$ in English, it's not my native language.
 A: So as $k:=[\frac{x+1}3] \in \mathbb Z$, we have $\frac{x-1}2 =k \in\mathbb Z$ also. So $x = 2k+1$ is an odd integer. We have to solve $[\frac{2k+2}3] = k$, we now consider three cases. 


*

*$k = 3\ell$ for some integer $\ell$: Then 
$$
  \left[\frac{2k+2}3\right] = \left[\frac{6\ell+2}3\right] = \left[2\ell + \frac 23\right]
$$
Now $2\ell$ is an integer, therefore the integer part of this is $\left[2\ell + \frac 23 \right] = 2\ell$. This equals $k = 3\ell$ for $\ell = 0$, which yields $k=0$, hence $x=1$.

*$k = 3\ell + 1$ for some $\ell \in \mathbb Z$, here  $$\left[\frac{2k+2}3\right] = \left[\frac{6\ell+4}3\right] = \left[2\ell + 1 + \frac 13\right] = 2\ell + 1,$$ which equals $k = 3\ell + 1$ for $\ell = 0$, giving $k = 1$, hence $x = 3$.

*$k = 3\ell + 2$ for some $\ell \in \mathbb Z$, here $$\left[\frac{2k+2}3\right] = \left[\frac{6\ell+6}3\right] = \left[2\ell + 2\right] = 2\ell + 2,$$ which equals $k = 3\ell + 2$ for $\ell = 0$, giving $k = 2$, hence $x = 5$.
The only solutios are there fore $1$, $3$ and $5$, hence it is (e) (intersected with $2\mathbb Z+1$).
A: First of all, notice that $\frac{x-1}{2}$ has to be a whole number (an integer).
Therefore, $x-1$ has to be even, and $x$ has to be an odd integer.
We can use the options to eliminate any number which includes an even number. I guess we are talking only about integers, $\mathbb{Z}$, since any non-integer solutions are definitely out.
For example, the region $(2,7)$ in part (a) includes 4 and 6, so (a) is not correct.
This on its own is not enough to come to a solution, but it should allow us to quickly eliminate wrong options.
In fact, we can go through like this and eliminate all the options that you've given - none of them are right, as it stands. 
Perhaps there is a typo in (e)? Since two of the three regions are representing an odd number, but $[5,7)$ incorrectly includes 6. I guess that (e) is supposed to be the correct answer, but maybe it should say $[5,6)$.
