document.write( "Question 1128097: Doug travels 3 times as fast as Mike. Traveling in opposite directions, they are 260 miles apart after 6.5 hours. Find their rates of travel. \n" ); document.write( "
Algebra.Com's Answer #744566 by josmiceli(19441)\"\" \"About 
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Add their speeds
\n" ); document.write( "Let \"+s+\" = Mike's speed in mi/hr
\n" ); document.write( "\"+260+=+%28+s+%2B+3s+%29%2A6.5+\"
\n" ); document.write( "\"+260+=+4s%2A6.5+\"
\n" ); document.write( "\"+26s+=+260+\"
\n" ); document.write( "\"+s+=+10+\"
\n" ); document.write( "and
\n" ); document.write( "\"+3s+=+30+\"
\n" ); document.write( "10 mi/hr and 30 mi/hr
\n" ); document.write( "---------------------------
\n" ); document.write( "check with separate rates:
\n" ); document.write( "Mike's equation:
\n" ); document.write( "(1) \"+d+=+s%2A6.5+\"
\n" ); document.write( "Doug's equation:
\n" ); document.write( "(2) \"+260+-+d+=+3s%2A6.5+\"
\n" ); document.write( "----------------------------
\n" ); document.write( "Plug (1) into (2)
\n" ); document.write( "(2) \"+260+-+6.5s+=+19.5s+\"
\n" ); document.write( "(2) \"+26s+=+260+\"
\n" ); document.write( "(2) \"+s+=+10+\"
\n" ); document.write( "OK\r
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