Given:
Wyatt jogged to his friend's house 29.25 miles away and then got a ride back home.
It took him 3 hours longer to jog there than ride back.
His jogging rate was 13 mph slower than the rate when he was riding.
Required:
To find the jogging rate.
Explanation:
Assume his jogging rate is x mph.
[tex]\frac{29.25}{x}-\frac{29.25}{x+13}=3[/tex][tex]\begin{gathered} \frac{9.75}{x}-\frac{9.75}{x+13}=1 \\ \\ 9.75(x+13)-9.75x=x(x+13) \\ \\ 9.75x+126.75-9.75x=x^2+13x \\ \\ x^2+13x-126.75=0 \end{gathered}[/tex]