Let's use the variable x to represent the speed of the scooter and y to represent the speed of the bicycle.
For a same time t, the scooter travels 22 mi and the bicycle travels 8 mi, so we can write the following equation:
![\begin{gathered} distance=speed\cdot time\\ \\ 22=x\cdot t\\ \\ t=(22)/(x)\\ \\ 8=y\cdot t\\ \\ t=(8)/(y)\\ \\ (22)/(x)=(8)/(y) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d2uz833i3lgaptpcrrtaerfw0zw5rcjsas.png)
Then, if the rate of the scooter is 6 mph more than twice the rate of the bicycle, we have the following equation:
![x=2y+6\\](https://img.qammunity.org/2023/formulas/mathematics/college/wxz5kiwiey1s3pybr3ueemc2315n4ex0a4.png)
Using this value of x in the first equation, let's solve it for y:
![\begin{gathered} (22)/(2y+6)=(8)/(y)\\ \\ 22y=8(2y+6)\\ \\ 22y=16y+48\\ \\ 6y=48\\ \\ y=8\text{ mph} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/z5atahpghprqk7n1q2uprzns9y9qx4sxo4.png)
Now, calculating the value of x, we have:
![\begin{gathered} x=2y+6\\ \\ x=16+6\\ \\ x=22\text{ mph} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h4ekebmjezlo5i5d8mod0m3mj15wbc9x05.png)
Therefore the scooter's rate is 22 mph and the bicycle's rate is 8 mph.