To solve this problem you must analyze the data provided very well.
The distance the cyclist travels across the country is the same as when he travels back home. Let's call d at that distance.
The distance traveled when the cyclist travels on the bicycle is:
[tex]d = 10t_1[/tex] (i)
Where [tex]t_1[/tex] is the time he traveled on the bicycle
Now, the distance traveled by the cyclist back home is:
[tex]d = 3t_2[/tex] (ii)
Where [tex]t_2[/tex] is the time it took the cyclist to walk back home.
The total time that the tour lasted was 6.5 hours. So:
[tex]t_1 + t_2 = 6.5[/tex] (iii)
We know from equation (i) that:
[tex]t_1 = \frac{d}{10}[/tex]
We know from equation (ii) that:
[tex]t_2 = \frac{d}{3}[/tex]
Then, substituting in (iii) we have:
[tex]\frac{d}{10} + \frac{d}{3} = 6.5[/tex].
We cleared d from the equation:
[tex]\frac{3d + 10d}{30} = 6.5[/tex]
[tex]13d = 195\\\\d = \frac{195}{13}\\\\d = 15 miles\\\\ t_1 = \frac{15}{10} = 1.5 h\\\\ t_2 = \frac{15}{3} = 5 h[/tex]
Answer:
The cyclist traveled on his bicycle a distance of 15 miles
