Question 166507
The goal of this problem is to find the distance "d" of {{{d=rt}}}



Let t=time it took to get to school 



So this means that it took {{{27-t}}} minutes to get home



Now convert 10 mi/hr to mi/min to get {{{10/60=1/6}}} mi/min. So going to school, he's running {{{1/6}}} mi/min (or in decimal form 0.1667 mi/min)



Now convert 8 mi/hr to mi/min to get {{{8/60=2/15}}} mi/min. So coming home, he's running {{{2/15}}} mi/min (or in decimal form 0.1333 mi/min)

<table border="1">
<th>Direction</th><th>Distance</th><th>Rate/Speed</th><th>Time</th>
<tr><td>Going to School</td><td>d miles</td><td>1/6 mi/min</td><td>t min</td></tr>
<tr><td>Coming Home</td><td>d miles</td><td>2/15 mi/min</td><td>27-t min</td></tr>
</table>



Going to School:


{{{d=rt}}} Start with the distance rate time formula



{{{d=(1/6)t}}} Plug in {{{r=1/6}}}


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Coming Home:


{{{d=rt}}} Start with the distance rate time formula



{{{d=(2/15)(27-t)}}} Plug in {{{r=8}}} and replace "t" with {{{27-t}}}



{{{(1/6)t=(2/15)(27-t)}}} Now plug in {{{d=(1/6)t}}} (the "going to school" equation)



{{{5t=4(27-t)}}} Multiply both sides by the LCD 30 to clear out the fractions.



-----------------------------------


{{{5t=108-4t}}} Distribute.



{{{5t+4t=108}}} Add {{{4t}}} to both sides.



{{{9t=108}}} Combine like terms on the left side.



{{{t=(108)/(9)}}} Divide both sides by {{{9}}} to isolate {{{t}}}.



{{{t=12}}} Reduce.



So he took 12 minutes to get to school.



{{{d=(1/6)t}}} Go back to the first equation



{{{d=(1/6)(12)}}} Plug in {{{t=12}}}



{{{d=12/6}}} Multiply



{{{d=2}}} Reduce



So the distance from his home to school is 2 miles. Now since he went to school and came home he traveled that distance twice. So this means that he traveled a total distance of 4 miles