A hiker walked for two days. On the second day the hiker walked four hours longer and his speed was 2 miles per hour faster than that of the first day. If during the two days he walked a total of 104 miles and spent a total of 20 hours, what was his average speed (miles/hour) on the first day?

Explanation: Let \(t_1\) and \(s_1\) be the time and speed on the first day. Let \(t_2\) and \(s_2\) be the time and speed on the second day. Given: \(t_1 + t_2 = 20\) hours, and \(t_2 = t_1 + 4\). \(t_1 + (t_1 + 4) = 20 \Rightarrow 2t_1 = 16 \Rightarrow t_1 = 8\) hours. \(t_2 = 8 + 4 = 12\) hours. Given: \(s_2 = s_1 + 2\). Total distance: \(s_1 t_1 + s_2 t_2 = 104\). \(s_1 (8) + (s_1 + 2) (12) = 104\). \(8s_1 + 12s_1 + 24 = 104\). \(20s_1 = 80\). \(s_1 = 4\) miles/hour.