This is a plea for readers solutions!
Solve $[20 \, x−3] = [15 \, x + 27]$, where $[x]$ is the integer part of $x$.
I have my solution which is spread over five intervals and can solve any $[a \, x+b] = [c \, x + d]$ if $a,b,c,d$ are known integers.
What I wish to do is set down a general solution for integer $a, b, c, d, ....$ this is proving 'awkward'. So, I would welcome readers methods to solve the equation in order that it may furnish me with ideas for a general solution. The number of intervals in which the solutions lie varies depending on the values of $a, b, c, d$.
$$\lfloor 20x−3 \rfloor = \lfloor 15x + 27 \rfloor$$
Let's start by wondering if there is a solution where $20x-3$ and $15x + 27$ are both integers. If that were the case, then we would get
\begin{align} 20x−3 &= 15x + 27 \\ 5x &= 30 \\ x & = 6 \end{align}
This is the union of these sets \begin{align} \end{align}
And we see that $20x-3 = 15x + 27 = 117$, so both sides are both equal to the integer $117$.
So lets move our origin to the point $x=6$.
Let $x = t+6$. Then we get
\begin{align} \lfloor 20x−3 \rfloor &= \lfloor 15x + 27 \rfloor \\ \lfloor 20t+120−3 \rfloor &= \lfloor 15t + 90 + 27 \rfloor \\ \lfloor 20t+117 \rfloor &= \lfloor 15t + 117 \rfloor \\ \lfloor 20t \rfloor + 117 &= \lfloor 15t \rfloor + 117 \\ \lfloor 20t \rfloor &= \lfloor 15t \rfloor \end{align}
Let $n$ be an integer. Then
$\lfloor 15t \rfloor = n \implies n \le 15t < n + 1 \implies \frac{n}{15} \le t \lt \frac{n+1}{15} $ $\lfloor 20t \rfloor = n \implies n \le 20t < n + 1 \implies \frac{n}{20} \le t \lt \frac{n+1}{20}$
For non negative $n$, the intersection $$\bigg[\frac{n}{20}, \frac{n+1}{20}\bigg) \cap \bigg[\frac{n}{15}, \frac{n+1}{15}\bigg) = \bigg[\frac{n}{15}, \frac{n+1}{20}\bigg)$$ must be non empty.
This will work until
\begin{align} \frac{n}{15} &\ge \frac{n+1}{20} \\ 20n &\ge 15n + 15 \\ n &\ge 3 \end{align}
For negative $n$, the interserction $$\bigg[\frac{n}{15}, \frac{n+1}{15}\bigg) \cap \bigg[\frac{n}{20}, \frac{n+1}{15}\bigg) = \bigg[\frac{n}{20}, \frac{n+1}{15}\bigg)$$ must be non empty.
This will work until
\begin{align} \frac{n}{20} &\ge \frac{n+1}{15} \\ 15n &\ge 20n + 20 \\ n &\le -4 \end{align}
So the solution set is $$\bigcup_{n=-3}^{-1}\left[6+\frac{n}{20}, 6+\frac{n+1}{15} \right) \cup \bigcup_{n=0}^{2}\left[6+\frac{n}{15}, 6+\frac{n+1}{20} \right) $$
This is the union of these six intervals
\begin{array}{l} \left[6-\frac{3}{20}, 6-\frac{2}{15} \right) \\ \left[6-\frac{2}{20}, 6-\frac{1}{15} \right)\\ \left[6-\frac{1}{20}, 6 \right)\\ \left[6+\frac{0}{15}, 6+\frac{1}{20} \right) \\ \left[6+\frac{1}{15}, 6+\frac{2}{20} \right) \\ \left[6+\frac{2}{15}, 6+\frac{3}{20} \right) \end{array}