Solving equations including floor function. I got a little trouble solving equations that involve floor function in an efficient way.
For example :
$$
\left\lfloor\frac{x+3}{2}\right\rfloor = \frac{4x+5}{3}
$$
In the one above, I get that you basically let $$ \frac{4x+5}{3} = k $$ and then inserting $k$ in the left side, take $k = 8l, 8l+1$, and so on and test it.
If there's a better solution to the one above plese tell me. 
My main problem is when it comes down to functions that have multiple floors such as : 
$$
\left\lfloor \frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}{6}\right\rfloor = \frac{3x-5}{2}
$$
Using the same method for each of them and then intersecting the solutions should give me the right answer but is there a faster way to solve equations like this ? 
 A: As the left hand side is integer, so should be $\dfrac{3x-5}2\iff2|3(x-1)\iff2|(x-1)\implies x$ is odd (assuming $x$ to be an integer)
Again as lcm$(3,6)$ we need to test for $x\equiv0,1,2,3,4,5\pmod6$
But as $x$ is odd, $x\equiv1,3,5\pmod6$
If $x=6b+1$
$$\left\lfloor\frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}{6}\right\rfloor=\left\lfloor\frac{6b+1+1}{3}\right\rfloor + \left\lfloor\frac{2(6b+1)+5}{6}\right\rfloor=2b+(2b+1)=4b+1$$
and $$\dfrac{3x-5}2=\dfrac{3(6b+1)-5}2=9b-1$$
and so on
If $x$ is not necessarily an integer, $\dfrac{3x-5}2+I\iff x=\dfrac{5+2I}3$
Check for $I\equiv0,1,2\pmod3$
A: Since $\dfrac{4x+5}3 = \left\lfloor \dfrac{x+3}2 \right \rfloor$ is an integer, we need $3$ to divide $4x+5$, i.e.,
$$\dfrac{4x+5}3 = m \in \mathbb{Z} \implies x = \dfrac{3m-5}4 \text{ where }m \in \mathbb{Z}$$
Hence,
$$\left\lfloor \dfrac{x+3}2 \right\rfloor = m \implies \dfrac{x+3}2 = m + e \implies x+3 = 2m+2e \implies x = 2m-3+2e$$
where $e \in[0,1)$.
Hence, we need
$$\dfrac{3m-5}4 = 2m-3+2e \implies 3m-5 =8m-12+8e \implies 8e = -5m+7$$ This gives us $e = \dfrac{7-5m}8$. Since $e \in [0,1)$, we have $7-5m \in [0,8) \implies m = 0,1$. Hence, $$x=-\dfrac54,-\dfrac12$$
A: **Below I have provided a solution
$\mathbf{\left(\frac{4x+5}{3}=t\right)}$
$,\,t\in\mathbb z$
$\rightarrow$
$$x=\left(\frac{3t-5}{4}\right)$$
$$\lfloor\left(\frac{\frac{3t-5}{4}+3}{2}\right)\rfloor=t$$
$\rightarrow$
$$\lfloor\left(\frac{3t+7}{8}\right)\rfloor=t$$
$$t\le\left(\frac{3t+7}{8}\right)\lt t+1$$
$$t\le\left(\frac{7}{5}\right) ,t\gt\left(\frac{-1}{5}\right)$$
$\rightarrow$
$$t=0 , 1$$
$$x=\left(\frac{-5}{4}\right) or  x=\left(\frac{-1}{2}\right)$$
thank you for your attention
