Chapter 8 - Systems of Linear Equations and Problem Solving - 8.4 Systems of Equations in Three Variables - 8.4 Exercise Set - Page 538: 32

$$c=-0.25,\:b=-0.5,\:a=0.25$$

Solving the system of equations using substitution, we find: $$ 2c-3\cdot \frac{-16c-25}{42}=1 \\ c=-.25 \\ b=\frac{-16\left(-\frac{1}{4}\right)-25}{42}=-\frac{1}{2} \\ a=\frac{2+3\left(-\frac{1}{2}\right)}{2}=\frac{1}{4} $$