Solve the system of equations. Express numbers as integers or simplified fractions. \[ \begin{aligned} 6 x+9 y & =-4 \\ 12 x+5 z & =0 \\ 6 y+5 z & =-8\end{aligned} \]

To solve the system of equations: 1. **First, simplify the first equation:** \[ 6x + 9y = -4. \] Divide every term by 3: \[ 2x + 3y = -\frac{4}{3}. \, (1) \] 2. **Next, simplify the second equation:** \[ 12x + 5z = 0. \] This can be rewritten as: \[ z = -\frac{12}{5}x. \, (2) \] 3. **Substituting (2) into the third equation:** \[ 6y + 5z = -8. \] Substituting \(z\): \[ 6y + 5\left(-\frac{12}{5}x\right) = -8. \] Simplifying: \[ 6y - 12x = -8. \] Dividing the entire equation by 6: \[ y - 2x = -\frac{4}{3}. \, (3) \] 4. **Now we have two simplified equations:** - From (1), \(2x + 3y = -\frac{4}{3}\). - From (3), \(y - 2x = -\frac{4}{3}\). 5. **Now let's solve for \(y\) using (3):** Rearranging (3): \[ y = 2x - \frac{4}{3}. \] 6. **Substituting \(y\) back into equation (1):** \[ 2x + 3\left(2x - \frac{4}{3}\right) = -\frac{4}{3}. \] This simplifies to: \[ 2x + 6x - 4 = -\frac{4}{3}. \] Combine like terms: \[ 8x - 4 = -\frac{4}{3}. \] Add 4 to both sides: \[ 8x = 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3}. \] Dividing by 8: \[ x = \frac{8}{3 \cdot 8} = \frac{1}{3}. \] 7. **Now substitute \(x\) back to find \(y\):** Substitute \(x = \frac{1}{3}\) into \(y = 2x - \frac{4}{3}\): \[ y = 2\left(\frac{1}{3}\right) - \frac{4}{3} = \frac{2}{3} - \frac{4}{3} = -\frac{2}{3}. \] 8. **Substituting \(x\) into (2) to find \(z\):** From (2): \[ z = -\frac{12}{5}\left(\frac{1}{3}\right) = -\frac{12}{15} = -\frac{4}{5}. \] Finally, the solution to the system of equations is: \[ x = \frac{1}{3}, \quad y = -\frac{2}{3}, \quad z = -\frac{4}{5}. \]