document.write( "Question 181114: two military squads are 26 miles apart. they plan to rendezvous at some intermediate point. if one squad hikes 2/3 mph faster than the other, and they meet in 3 hrs, what is the rate of the faster squad? \n" ); document.write( "
Algebra.Com's Answer #135786 by ankor@dixie-net.com(22740)\"\" \"About 
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two military squads are 26 miles apart. they plan to rendezvous at some
\n" ); document.write( " intermediate point. if one squad hikes 2/3 mph faster than the other,
\n" ); document.write( " and they meet in 3 hrs, what is the rate of the faster squad?
\n" ); document.write( ":
\n" ); document.write( "Let s = the speed of the faster squad
\n" ); document.write( "then
\n" ); document.write( "(s-\"2%2F3\") = speed of the slow-pokes
\n" ); document.write( ":
\n" ); document.write( "The total dist of both squads = 26 mi, they both travel for 3 hrs:
\n" ); document.write( "write a distance equation dist = time * speed
\n" ); document.write( ":
\n" ); document.write( "3s + 3(s-\"2%2F3\") = 26
\n" ); document.write( "Multiply what's inside the brackets
\n" ); document.write( "3s + 3s - 2 = 26
\n" ); document.write( "6s = 26 + 2
\n" ); document.write( "6s = 28
\n" ); document.write( "s = \"28%2F6\"
\n" ); document.write( "s = 14/3 or 4\"2%2F3\" mph is the speed of the faster squad
\n" ); document.write( ":
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\n" ); document.write( "check solution by finding the dist traveled by each (slower squad = 4 mph)
\n" ); document.write( "3 * 4 = 12
\n" ); document.write( "3 * \"14%2F3\" = 14
\n" ); document.write( "------------
\n" ); document.write( "Total 26 mi
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