document.write( "Question 908494: Dan left downtown greensboro, and three hours later, Sara left going 51 mph faster to catch up. after another two hours, Sara caught up. Find dan's average speed \n" ); document.write( "
Algebra.Com's Answer #551072 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
Dan left downtown greensboro, and three hours later, Sara left going 51 mph faster to catch up. after another two hours, Sara caught up.
\n" ); document.write( " Find dan's average speed
\n" ); document.write( ":
\n" ); document.write( "let s = Dan's speed
\n" ); document.write( "then
\n" ); document.write( "(s+51) = Sara's speed
\n" ); document.write( ":
\n" ); document.write( "From the information given, we know that S's travel time is 2 hrs, D's is 5 hrs
\n" ); document.write( ":
\n" ); document.write( "When S catches D, they will have traveled the same distance
\n" ); document.write( "Write a distance equation; dist = time * speed
\n" ); document.write( "5s = 2(s+51)
\n" ); document.write( "5s = 2s + 102
\n" ); document.write( "5s - 2s = 102
\n" ); document.write( "3s = 102
\n" ); document.write( "s = 102/3
\n" ); document.write( "s = 34 mph is D's speed
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Confirm this, find the distance each traveled, should be equal
\n" ); document.write( "5 * 34 = 170 mi
\n" ); document.write( "2 * 85 = 170 mi \n" ); document.write( "
\n" );