This diagram represents the problem
We know that distance = speed*time; D=S*t
Total distance: 156 miles
time: 6 h
Speed1: 33 miles/h
Speed2: 5 miles/h
for interval 1:
[tex]\begin{gathered} D_1=S_1\cdot t_1 \\ D_1=33\cdot t_1 \end{gathered}[/tex]for interval 2:
[tex]\begin{gathered} D_2=S_2\cdot t_2 \\ D_2_{}=5\cdot t_2 \end{gathered}[/tex]for the whole trip: -Eq 1. Distance
[tex]\begin{gathered} D=D_1+D_2 \\ D=33\cdot t_1+5\cdot t_2 \\ 156=33\cdot t_1+5\cdot t_2 \end{gathered}[/tex]and also: -Eq 2. Time
[tex]\begin{gathered} t=t_1+t_2 \\ 6=t_1+t_2 \end{gathered}[/tex]Now we have a system of 2 equations with 2 unknowns.
Let's solve it!
[tex]\begin{gathered} 156=33t_1+5t_2 \\ t_1=\frac{156-53t_2}{33} \\ \frac{156-5t_2}{33}+t_2=6 \\ t_2=\frac{3}{2} \\ t_1=\frac{156-5\cdot\frac{3}{2}}{33} \\ t_1=\frac{9}{2} \end{gathered}[/tex]We can see that he spent 4.5 hours riding the bike and 1.5 h walking
