The following information is provided in the question:
![\begin{gathered} Total\text{ }Distance=178\text{ }miles \\ Velocity\text{ }of\text{ }Run=7mph \\ Velocity\text{ }of\text{ }Bicycling=30mph \\ Time\text{ }Taken=9\text{ }hours \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rkwy5553l9uubkgi1icqdtsg3syeflfofi.png)
Recall the formula relating distance, speed, and time:
![s=(d)/(t)](https://img.qammunity.org/2023/formulas/physics/high-school/hs3capot9u20ahgs6jaudlrhvo6wv8t7f0.png)
Let the distance for the run be x. This means that the distance for the bicycle race will be 178 - x.
Were we to use the provided information to calculate the time for each part of the race, we would have:
![\begin{gathered} time\text{ }of\text{ }run=(x)/(7) \\ time\text{ }of\text{ }bicycle\text{ }race=(178-x)/(30) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/51nka87ax86maskxurb4rbzia4ftit3msn.png)
Since the total time is 9, we have:
![(x)/(7)+(178-x)/(30)=9](https://img.qammunity.org/2023/formulas/mathematics/college/cu705wqan4ynra0875hc7sois4z7luz8nh.png)
Solving, we have:
![\begin{gathered} \mathrm{Multiply\:by\:LCM=}210 \\ (x)/(7)\cdot \:210+(178-x)/(30)\cdot \:210=9\cdot \:210 \\ 30x+7\left(-x+178\right)=1890 \\ Simplify \\ 30x-7x+1246=1890 \\ 30x-7x+1246=1890 \\ 23x=644 \\ \mathrm{Divide\:both\:sides\:by\:}23 \\ x=28 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w33cuze9m9j1u89nnralywok45naiw0hgu.png)
Therefore, the distance for the bicycle race will be:
![\Rightarrow178-28=150](https://img.qammunity.org/2023/formulas/mathematics/college/29bturo1dw8j2xiakpr56zadnneh4ww74z.png)
The distance of the run is 28 miles.
The distance of the bicycle race is 150 miles.