Chapter 8 - Systems of Linear Equations and Problem Solving - 8.4 Systems of Equations in Three Variables - 8.4 Exercise Set - Page 538: 26

$$z=-3,\:y=0.6666667,\:x=0.6$$

Solving the system of equations using substitution, we find: $$ 3\cdot \frac{7-6\left(-\frac{5z-1}{24}\right)-z}{10}-\frac{12}{5}\left(-\frac{5z-1}{24}\right)+\frac{2}{5}z=-1 \\ z=-3 \\ y=-\frac{5\left(-3\right)-1}{24}=\frac{2}{3} \\ x=\frac{7-6\cdot \frac{2}{3}-\left(-3\right)}{10}=.6$$