Question 201779
A basic formula, which we need for this problem, is that distance = (average speed)*(time). d=rt for short.
If we let x = Roger's average speed on the first day, then his average speed on the second day would be (x+15), since we are told he drove 15 miles per hour faster. The time spent traveling on the first day is 2. The time spent traveling on the second day was 30 minutes (or 1/2 hour) less. (We will use 1/2 hour so that all our times are in hours.) 2 - (1/2) = 3/2 so he spent 3/2 hours traveling on the second day.
Using d=rt we can say:<ul><li>the distance on the first day would be 2*x</li><li>On the second day the distance traveled would be (x+15)*(1/2).</li></ul>
But this distance traveled by Roger was the same on both days. This means that
{{{2*x = (x+15)*(3/2)}}}
Now we solve this using Algebra. First we will multiply out the right side using the Distributive Property:
{{{2x = x*(3/2) + 15*(3/2)}}}
Simplifying
{{{2x = (3/2)x + 45/2}}}
Subtracting {{{(3/2)x}}} from both sides we get
{{{(1/2)x = 45/2}}}
Now we divide both sides by {{{1/2}}} (or multiply both sides by the reciprocal of {{{1/2}}} (which is 2)). Either way we get
{{{ x = 45 }}}
If we look back we will see that we decided that "x" was the speed on the first day. So we have found that Roger's average speed on the first day was 45 miles per hour.
But question asks for his distance. (Always remember to check to see if you have answered the question! It is very tempting, once you have "x = ...", to feel like you are finished.) Looking back we can see that his distance on the first day was 2x. (We could use (x+15)*(3/2), instead, but 2x is easier.) So the distance Roger traveled was 2*45 or 90 miles.