Speed (s) is defined as follows:
[tex]s=\frac{d}{t}[/tex]
where d is distance and t is time.
Let's call x to the distance traveled by one of the cyclists. Combining the distance traveled by both cyclists, they rode 81 miles. Then, if one cyclist rode x miles, then the other one traveled (81 - x) miles, that is,
[tex]\begin{gathered} d_1=81-x \\ d_2=x \\ d_1+d_2=81-x+x=81 \end{gathered}[/tex]
Where d1 and d2 are the distances traveled.
Substituting into the speed formula:
[tex]\begin{gathered} s_1=\frac{d_1}{t}=\frac{81-x}{t} \\ s_2=\frac{d_2}{t}=\frac{x}{t} \end{gathered}[/tex]
They traveled 3 hours, then t = 3 in both equations.
We also know that one cyclist travels 5 miles/h slower than the other one. Therefore:
[tex]\begin{gathered} s_1=s_2-5 \\ \frac{81-x}{t}=\frac{x}{t}-5 \\ \frac{81-x}{3}=\frac{x}{3}-5 \end{gathered}[/tex]
Solving for x:
[tex]\begin{gathered} 3\cdot\frac{81-x}{3}=3\cdot(\frac{x}{3}-5) \\ 81-x=x-15 \\ 81+15=x+x \\ 96=2x \\ \frac{96}{2}=x \\ 48=x \end{gathered}[/tex]
Substituting this result into the speed equations:
[tex]\begin{gathered} s_1=\frac{81-x}{t} \\ s_1=\frac{81-48}{3} \\ s_1=\frac{33}{3} \\ s_1=11\frac{\text{ miles}}{hour} \end{gathered}[/tex][tex]\begin{gathered} s_2=\frac{x}{t} \\ s_2=\frac{48}{3} \\ s_2=16\frac{miles}{hour} \end{gathered}[/tex]
