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:
![\begin{gathered} D_1=S_1\cdot t_1 \\ D_1=33\cdot t_1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/67gorijk511gxim7sqbp3h5qot9ixes9g5.png)
for interval 2:
![\begin{gathered} D_2=S_2\cdot t_2 \\ D_2_{}=5\cdot t_2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mgbd4vnb0ybko2w87mevafpoq9y0ejzn9n.png)
for the whole trip: -Eq 1. Distance
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/m76foo2ipj9xoh6sro1sf9p1929ryslgzn.png)
and also: -Eq 2. Time
![\begin{gathered} t=t_1+t_2 \\ 6=t_1+t_2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kbqby36kg0qg49swi4ef3dsvmbutk0ujlh.png)
Now we have a system of 2 equations with 2 unknowns.
Let's solve it!
![\begin{gathered} 156=33t_1+5t_2 \\ t_1=(156-53t_2)/(33) \\ (156-5t_2)/(33)+t_2=6 \\ t_2=(3)/(2) \\ t_1=(156-5\cdot(3)/(2))/(33) \\ t_1=(9)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zds80ccxxci84kgi1yvk9kq4ecpojotujo.png)
We can see that he spent 4.5 hours riding the bike and 1.5 h walking