You need to use general intervals. Remember that:
[tex]
[3 x] = N \in \mathbb{Z}, \; N \le 3 x < N + 1 \Leftrightarrow \frac{N}{3} \le x < \frac{N + 1}{3}[/tex]
So, consider the following 3 families of intervals:
First family:
[tex]
N = 3 k, k \le x < k + \frac{1}{3}, \; \mathbb{Z}[/tex]
Then you have:
[tex]
[x] = k, [3 x] = N = 3 k[/tex]
and your equation reads:
[tex]
2 k - 3 (3 k) + 7 = 0[/tex]
Second family:
[tex]
N = 3 k + 1, k + \frac{1}{3} \le x < k + \frac{2}{3}[/tex]
Then, you have:
[tex]
[x] = k, [3 x] = N = 3 k + 1[/tex]
and your equation reads:
[tex]
2 k - 3 ( 3 k + 1) + 7 = 0[/tex]
Third family:
[tex]
N = 3 k + 2, k + \frac{2}{3} \le x < k + 1[/tex]
Then you have:
[tex]
[x] = k, [3 x] = N = 3 k + 2[/tex]
and your equation reads:
[tex]
2 k - 3 (3 k + 2) + 7 = 0[/tex]
Thus, you get three cases that exhaust the whole set of real numbers and for each case you get an equation. The solution to those equations has to be an integer!