## Geometry: Common Core (15th Edition)

a. $\frac{a}{13}$ = $\frac{7}{b}$ b. $\frac{a-7}{7}$ = $\frac{13-b}{b}$ c. $\frac{7}{a}$ = $\frac{b}{13}$
We are given the proportion $\frac{a}{7}$ = $\frac{13}{b}$ Cross multiply ab = 7 x 13 ab=91 Divide both sides of ab=91 by 13b $\frac{ab}{13b}$ = $\frac{91}{13b}$ Reducing leaves you with $\frac{a}{13}$ = $\frac{7}{b}$ b. Start with the initial proportion of $\frac{a}{7}$ = $\frac{13}{b}$ Subtract the denominator from the numberator. $\frac{a-7}{7}$ = $\frac{13-b}{b}$ c. Reverse the original proportion $\frac{a}{7}$ = $\frac{13}{b}$ is the same as $\frac{7}{a}$ = $\frac{b}{13}$