document.write( "Question 230819: An express and local train leave Grayslake at 3pm and head for Chicago 50 miles away. The express travels twice as fast as the local and arrives 1 hour ahead of the local. Find the speed of each train \n" ); document.write( "

Algebra.Com's Answer #170991 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"s\"$ = the speed of the local in mi/hr
\n" ); document.write( "Then $\"2s\"$ = speed of the express
\n" ); document.write( "Let $\"t\"$ = time in hrs it takes for local to arrive in Chicago
\n" ); document.write( "Then $\"t+-+1\"$ time in hrs for express to arrive in Chicago
\n" ); document.write( "They both travel the same distance, $\"d+=+50\"$ mi
\n" ); document.write( "Now I can write an equation for each train
\n" ); document.write( "For local:
\n" ); document.write( "(1) $\"50+=+st\"$
\n" ); document.write( "For express:
\n" ); document.write( "(2) $\"50+=+2s%2A%28t+-+1%29\"$
\n" ); document.write( "---------------------
\n" ); document.write( "From (2)
\n" ); document.write( " $\"50+=+2st+-+2s\"$
\n" ); document.write( "Substitute from (1)
\n" ); document.write( " $\"50+=+2%2A50+-+2s\"$
\n" ); document.write( " $\"2s+=+100+-+50\"$
\n" ); document.write( " $\"2s+=+50\"$
\n" ); document.write( " $\"s+=+25\"$
\n" ); document.write( "and
\n" ); document.write( " $\"2s+=+50\"$
\n" ); document.write( "The speed of the local is 25 mi/hr and
\n" ); document.write( "the speed of the express is 50 mi/hr
\n" ); document.write( "check answer:
\n" ); document.write( "(1) $\"50+=+st\"$
\n" ); document.write( " $\"50+=+25t\"$
\n" ); document.write( " $\"t+=+2\"$
\n" ); document.write( "(2) $\"50+=+2s%2A%28t+-+1%29\"$
\n" ); document.write( " $\"50+=+2%2A25%2A%282-1%29\"$
\n" ); document.write( " $\"50+=+50%2A1\"$
\n" ); document.write( " $\"50+=+50\"$
\n" ); document.write( "OK
\n" ); document.write( "
\n" ); document.write( " \n" ); document.write( "

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