Chapter 1 - Expressions, Equations, and Inequalities - 1-4 Solving Equations - Lesson Check - Page 30: 5

$k=\frac{1}{3}z+2$

First, we add 2z to both sides to isolate k with its coefficient. We have $6k-2z=12$ is $6k=2z+12$. Now, we divide both sides by 6 to fully isolate k. We have $\frac{6k}{6}=\frac{2z+12}{6}$ is $k=\frac{1}{3}z+2$