Equations with the floor function I am trying to solve the following problem:

For what real numbers $x$ is: $\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=0$?

I have no idea how to do this, please help me
 A: As for integer $a, \lfloor{y+a}\rfloor=\lfloor y\rfloor+a$
$$6+3=\lfloor{2x}\rfloor-3\lfloor x\rfloor$$
Now let $x=I+f$
If $f<.5,$
$$9=2I-3I\iff I=?$$
Else $f\ge.5,$  $$9=2I+1-3I\iff I=?$$
A: Hint:
We know:

$x,y\in [z,z+1)$ then $\lfloor x\rfloor=\lfloor y\rfloor$
$\lfloor a+b\rfloor=\lfloor a\rfloor+b$ for integer $b$

With those 2:
$\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=\lfloor2x\rfloor-3-3(\lfloor x\rfloor+2)=\lfloor2x\rfloor-3\lfloor x\rfloor-9=0\\\implies\lfloor2x\rfloor-3\lfloor x\rfloor=9$
Now divide it into 2 cases, when $\lfloor x\rfloor=\lfloor x+0.5\rfloor$ and when $\lfloor x\rfloor+1=\lfloor x+0.5\rfloor$
Can you continue from here?

Explanation about the cases:
I got to the equation $\lfloor2x\rfloor-3\lfloor x\rfloor=9$, let's look only on $\lfloor2x\rfloor$, if $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$ then $\lfloor2x\rfloor=2\lfloor x\rfloor+1$ elsewhere $\lfloor2x\rfloor=2\lfloor x\rfloor$
Solution to case one:
If  $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$:
$$\lfloor2x\rfloor-3\lfloor x\rfloor=9\\2\lfloor x\rfloor+1-3\lfloor x\rfloor=9\\\lfloor x\rfloor=-8\\x=-8+c,c\in[0,1)\\\text{we know that $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$ so:}\\\lfloor -8+c+0.5\rfloor=\lfloor-8+c\rfloor+1=\lfloor-8+c+1\rfloor\\c\in[0.5,1)$$
