document.write( "Question 581312: Dave and sara live 50 miles apart and travel towards each other on their bikes. Sara leaves on time and dave leaves 2 hours later but tries to make up for it by riding 7 mph faster than sara.Find the speed of both sara and dave on their bikes, if sara is on the bike for 4 hours. \n" ); document.write( "
Algebra.Com's Answer #371555 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+s+\" = Sarah's speed in mi/hr
\n" ); document.write( "\"+s+%2B+7+\" = Dave's speed in mi/hr
\n" ); document.write( "given:
\n" ); document.write( "Sarah gets a head start of \"+d+=+2s+\"
\n" ); document.write( "They are \"+50+-+2s+\" miles apart when Dave leaves
\n" ); document.write( "----------
\n" ); document.write( "Now I can think of Sarah as standing still and
\n" ); document.write( "Dave moving towards her at the sum of their speeds
\n" ); document.write( "If Sarah is on her bike for \"+4+\" hrs, then
\n" ); document.write( "when Dave starts, they both travel for \"+2+\" more hrs
\n" ); document.write( "\"+50+-+2s+=+%28+s+%2B+s+%2B+7+%29%2A2+\"
\n" ); document.write( "\"+50+-+2s+=+%28+2s+%2B+7+%29%2A2+\"
\n" ); document.write( "\"+50+-+2s+=+4s+%2B+14+\"
\n" ); document.write( "\"+6s+=+36+\"
\n" ); document.write( "\"+s+=+6+\"
\n" ); document.write( "\"+s+%2B+7+=+13+\"
\n" ); document.write( "Sarah's speed is 6 mi/hr
\n" ); document.write( "Dave's speed is 13 mi/hr
\n" ); document.write( "check:
\n" ); document.write( "Their separate distances are:
\n" ); document.write( "Dave:
\n" ); document.write( "\"+%28+s+%2B+7%29%2A2+=+13%2A2%0D%0A%7B%7B%7B+13%2A2+=+26+\" mi
\n" ); document.write( "Sarah:
\n" ); document.write( "\"+2s+%2B+s%2A2+=+4s+\"
\n" ); document.write( "\"+4s+=+4%2A6+\"
\n" ); document.write( "\"+4%2A6+=+24+\" mi
\n" ); document.write( "and
\n" ); document.write( "\"+24+%2B+26+=+50+\" mi
\n" ); document.write( "OK
\n" ); document.write( "
\n" );