Question

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. What was his jogging rate?

Answer 1

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.

Step-by-step explanation:

Assume his jogging rate is x mph.

`(29.25)/(x)-(29.25)/(x+13)=3`
`\begin{gathered} (9.75)/(x)-(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}`