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

$6$ mph.
distance = (rate)$\times$(time) $\Rightarrow$ time = $\displaystyle \frac{\text{distance}}{\text{rate}}$ Let x be the rate of the home to campus leg. The rate of the campus to home leg is $x-9.$ We are given: total time = $\displaystyle \frac{5}{\text{rate of H to C}} + \displaystyle \frac{5}{\text{rate of C to H.}}=\frac{7}{6}$ That is, $\displaystyle \frac{5}{x}+\frac{5}{x-9} =\frac{7}{6}\qquad$ ... $/\times 6x(x-9)$ $30(x-9)+30x=7x(x-9) \qquad$ ... distribute... $30x-270+30x=7x^{2}-63x \qquad$ ... add $270-60x$ $0=7x^{2}-123x+270$ ... Trying with synthetic division, we find: $$\begin{array}{rrrrrr} 15 & | & 7 & -123 & 270 & \\ & & & 105 & -270 & \\ & & -- & -- & -- & \\ & & 7 & -18 & 0 & \end{array}$$ $0=(x-15)(7x-18)\qquad$ ... $x=15$ mph, or $\displaystyle \frac{18}{7}$ mph. The rate of the return leg is $15-x:$ $15-9=6,$ or $\displaystyle \frac{18}{7}-9$, which is negative , so we discard this possibility. So, the rate of the return leg is $6$ mph.