## Algebra 2 (1st Edition)

$$z=-2,\:y=3,\:x=1$$
To solve, we first must isolate one of the variables to solve for that variable. Then, we use that value to solve for the other two variables: $$\frac{10-\left(-\frac{-5z-22}{4}\right)+2z}{3}+4\left(-\frac{-5z-22}{4}\right)+3z=7 \\ z=-2$$ Thus: $$y=-\frac{-5\left(-2\right)-22}{4}=3$$ $$x=\frac{10-3+2\left(-2\right)}{3}=1$$