document.write( "Question 682001: Two girls, Jen and Amber, went for a walk. Jen begin walking 20 seconds earlier than amber. Jen walked at a speed of 4 feet per second and Amber walked at a speed of 6 feet per second?
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Algebra.Com's Answer #422955 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
The question must be- when do they meet?
\n" ); document.write( "Jen's head start is:
\n" ); document.write( " $\"+4%2A20+=+80+\"$ ft
\n" ); document.write( "Let $\"+d+\"$ = the distance that Amber now has to
\n" ); document.write( "walk after Jen's head start
\n" ); document.write( "Let $\"+t+\"$ = the time for both after Jen's head start
\n" ); document.write( "------------
\n" ); document.write( "Jen's equation:
\n" ); document.write( "(1) $\"+d+-+80+=+4t+\"$
\n" ); document.write( "Amber's equation:
\n" ); document.write( "(2) $\"+d+=+6t+\"$
\n" ); document.write( "-------------
\n" ); document.write( "Substitute (2) into (1)
\n" ); document.write( "(1) $\"+6t+-+80+=+4t+\"$
\n" ); document.write( "(1) $\"+2t+=+80+\"$
\n" ); document.write( "(1) $\"+t+=+40+\"$
\n" ); document.write( "They will meet 40 sec after Amber starts
\n" ); document.write( "Jen walks for $\"+20+%2B+40+=+60+\"$ , or 1 min
\n" ); document.write( "---------------
\n" ); document.write( "check:
\n" ); document.write( "(2) $\"+d+=+6t+\"$
\n" ); document.write( "(2) $\"+d+=+6%2A40+\"$
\n" ); document.write( "(2) $\"+d+=+240+\"$ ft
\n" ); document.write( "and
\n" ); document.write( "(1) $\"+d+-+80+=+4t+\"$
\n" ); document.write( "(1) $\"+d+-+80+=+4%2A40+\"$
\n" ); document.write( "(1) $\"+d+=+80+%2B+160+\"$
\n" ); document.write( "(1) $\"+d+=+240+\"$ ft
\n" ); document.write( "OK
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