## Intermediate Algebra for College Students (7th Edition)

Distribute $\frac{1}{3}$ on the left side and $\frac{1}{5}$ on the right side, then combine like terms to obtain: $\frac{1}{3}(6z) + \frac{1}{3}(12) = \frac{1}{5}(20z) +\frac{1}{5}(30) - 8 \\2z + 4 = 4z+6-8 \\2z + 4 = 4z - 2$ Subtract $2z$ and add $2$ on both sides of the equation to obtain: $4+2=4z-2z \\6 = 2z \\\frac{6}{2} = \frac{2z}{2} \\3 = z$ The given equation has one solution so it is a conditional equation.