In the row of "original drive", we have the first part of the travel, so the time needed was 8 hours. The distance can be found by multiplying the rate and the time:
[tex]\begin{gathered} \text{distance}=\text{rate}\cdot\text{time} \\ \text{distance}=x\cdot8=8x \end{gathered}[/tex]
Then, in the "return drive", the rate is 9 mph higher, so the rate is "x + 9".
Also, the time is 7 hours. So, the distance is:
[tex]\text{distance}=(x+9)\cdot7=7x+63[/tex]
Since the distances are the same, we can write the following equation and solve it for x:
[tex]\begin{gathered} 8x=7x+63 \\ 8x-7x=63 \\ x=63 \\ \\ x+9=63+9=72\text{ mph} \end{gathered}[/tex]
Therefore the average speed on the return drive is 72 mph.
