What is the sum of real numbers satisfying the equation $\left \lfloor \frac{6x+5}{8}  \right \rfloor = \frac{15x-7}{5}$?

$ 
\textbf{(A)}\ 2
\qquad\textbf{(B)}\ \frac{81}{90}
\qquad\textbf{(C)}\ \frac{7}{15}
\qquad\textbf{(D)}\ \frac{4}{5}
\qquad\textbf{(E)}\ \frac{19}{15}
$
1. Start with the given equation:
   \[
   \left \lfloor \frac{6x+5}{8} \right \rfloor = \frac{15x-7}{5}
   \]
   Let \( k = \frac{15x-7}{5} \). Since \( k \) is an integer, we can write:
   \[
   15x - 7 = 5k \implies x = \frac{5k + 7}{15}
   \]

2. Substitute \( x = \frac{5k + 7}{15} \) into the left-hand side of the original equation:
   \[
   \frac{6x + 5}{8} = \frac{6 \left( \frac{5k + 7}{15} \right) + 5}{8} = \frac{2(5k + 7) + 5 \cdot 15}{8 \cdot 15} = \frac{10k + 14 + 75}{120} = \frac{10k + 89}{120}
   \]

3. Since \( \left \lfloor \frac{6x+5}{8} \right \rfloor = k \), we have:
   \[
   k \leq \frac{10k + 89}{120} < k + 1
   \]

4. Solve the inequalities:
   \[
   k \leq \frac{10k + 89}{120} \implies 120k \leq 10k + 89 \implies 110k \leq 89 \implies k \leq \frac{89}{110} \approx 0.809
   \]
   \[
   \frac{10k + 89}{120} < k + 1 \implies 10k + 89 < 120k + 120 \implies 89 < 110k + 120 \implies -31 < 110k \implies k > -\frac{31}{110} \approx -0.282
   \]

5. Since \( k \) must be an integer, the only possible values for \( k \) within the range \(-0.282 < k \leq 0.809\) are \( k = 0 \) and \( k = 1 \).

6. For \( k = 0 \):
   \[
   x = \frac{5(0) + 7}{15} = \frac{7}{15}
   \]

7. For \( k = 1 \):
   \[
   x = \frac{5(1) + 7}{15} = \frac{12}{15} = \frac{4}{5}
   \]

8. The sum of the real numbers satisfying the equation is:
   \[
   \frac{7}{15} + \frac{4}{5} = \frac{7}{15} + \frac{12}{15} = \frac{19}{15}
   \]

The final answer is \(\boxed{\frac{19}{15}}\)