Given:
An express train travels 200miles between two cities.
Let, s be the train speed in level terrain.
(s-10) be the speed in the mountains.
The distance on level terrain is 200-108=92 miles
The equation becomes,
[tex]\begin{gathered} Time=\frac{Dis\tan ce}{\text{sped}} \\ \frac{108}{s-10}+\frac{92}{s}=5 \end{gathered}[/tex]Solve the equation,
[tex]\begin{gathered} \frac{108}{s-10}+\frac{92}{s}=5 \\ \frac{100s+92(s-10)}{s(s-10)}=5 \\ 108s+92s-920=5s^2-50s \\ 5s^2-250s+920=0 \\ s^2-50s+184=0 \\ s^2-46s-4s+184=0 \\ s(s-46)-4(s-46)=0 \\ (s-46)(s-4)=0 \\ \Rightarrow s=46,4 \end{gathered}[/tex]But he appropriate value of s should be 46.
Since for s=4, s-10=-6 that is not possible.
So, the speed of the train on level terrain is 46 mph
