1. **State the problem:** Solve the system of equations for $x$, $y$, and $z$ given by: $$\begin{cases} y + 2 = 5 \\ 2x - y = 45 \\ 3y - 4z = 12 \end{cases}$$ 2. **Solve the first equation for $y$:** $$y + 2 = 5 \implies y = 5 - 2 = 3$$ 3. **Substitute $y=3$ into the second equation to find $x$:** $$2x - 3 = 45 \implies 2x = 48 \implies x = 24$$ 4. **Substitute $y=3$ into the third equation to find $z$:** $$3(3) - 4z = 12 \implies 9 - 4z = 12 \implies -4z = 3 \implies z = -\frac{3}{4}$$ 5. **Final solution:** $$x = 24, \quad y = 3, \quad z = -\frac{3}{4}$$