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 ? 

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, $$x=-\dfrac54,-\dfrac12$$
