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

z=$\frac{29}{6}$
Add $\frac{5}{2}$ to both sides of the equation to isolate the variable. $\frac{7}{3}$=-$\frac{5}{2}$+z $\frac{7}{3}$+$\frac{5}{2}$=z $\frac{7(2)}{3(2)}$+$\frac{5(3)}{2(3)}$=z $\frac{14}{6}$+$\frac{15}{6}$=z $\frac{29}{6}$=z Check the answer. $\frac{7}{3}$=-$\frac{5}{2}$+($\frac{29}{6}$) $\frac{7}{3}$=-$\frac{15}{6}$+$\frac{29}{6}$ $\frac{7}{3}$=$\frac{14}{6}$ $\frac{7}{3}$=$\frac{7}{3}$