You can set up a table, Taylor, using the relationship distance = rate x time
For the bike ride, the distance = 6 mi., the rate can be expressed as the variable x, and the time can be expressed as the variable t. Plugging this into the above equation gives us:
6 = xt
For the walk, the distance = 1 mi., the rate can be expressed as x - 6, and the time can be expressed as 1 - t. This will yield the following equation:
1 = (x - 6)(1 - t)
When we have a system of equations, it can be helpful to use either the Substitution or Elimination Method to solve the system. In this example, we can use the Substitution Method by solving the first equation for x.
x = 6/t
Now, let's substitute this into the second equation:
1 = (6/t - 6)(1 - t)
This is now an equation with only one unknown, which is t. We can use FOIL to expand this binomial expression.
1 = 6/t - 6 - 6 + 6t
1 = 6/t - 12 + 6t
If we multiply both sides of the equation by t, we get:
t = 6 - 12t + 6t²
6t² - 13t + 6 = 0
(2t - 3)(3t - 2) = 0
t = 3/2 and t = 2/3 are solutions to this quadratic equation. However, the only solution that is plausible is t = 2/3. Why? Because if we attempt to use t = 3/2, then the time Roberto spend walking would be negative (1 - 3/2 = -1/2), and time cannot be negative.
Now that we have the time t, we can determine the rate (or speed) for the bike ride and the walk. Since we previously said that x = 6/t, then:
x = 6/(2/3) = 6 * 3/2 = 18/2 = 9 mph
The above tells us that Roberto rode the bike at 9 mph. The rate at which he walks is 6 mph less than his speed riding the bike. So his walking speed is 9 - 6 = 3 mph.
