document.write( "Question 895049: Natasha drove from Bedingfield to Portsmouth at an average speed of 100 km/h to attend a job interview. On the way back she decided to slow down to emjoy the scenery, so she drove at just 75 km/h. Her trip involved a total of 3.5 hours of driving time. What is the distance between Bedingfield and Portsmouth? \n" ); document.write( "
Algebra.Com's Answer #542564 by josmiceli(19441)\"\" \"About 
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Let \"+t%5B1%5D+\" = time in hrs going
\n" ); document.write( "Let \"+t%5B2%5D+\" = time in hrs coming back
\n" ); document.write( "Driving from Bedingfield to Portsmouth:
\n" ); document.write( "(1) \"+d=+100t%5B1%5D+\"
\n" ); document.write( "Driving back:
\n" ); document.write( "(2) \"+d+=+75t%5B2%5D+\"
\n" ); document.write( "and, also:
\n" ); document.write( "(3) \"+t%5B1%5D+%2B+t%5B2%5D+=+3.5+\"
\n" ); document.write( "--------------------------
\n" ); document.write( "By substitution:
\n" ); document.write( "(1) \"+75t%5B2%5D+=+100t%5B1%5D+\"
\n" ); document.write( "(1) \"+t%5B1%5D+=+%283%2F4%29%2At%5B2%5D+\"
\n" ); document.write( "Substitute (1) into (3)
\n" ); document.write( "(3) \"+%283%2F4%29%2At%5B2%5D+%2B+t%5B2%5D+=+3.5+\"
\n" ); document.write( "(3) \"+3t%5B2%5D+%2B+4t%5B2%5D+=+14+\"
\n" ); document.write( "(3) \"+7t%5B2%5D+=+14+\"
\n" ); document.write( "(3) \"+t%5B2%5D+=+2+\"
\n" ); document.write( "and
\n" ); document.write( "(3) \"+t%5B1%5D+=+3.5+-+t%5B2%5D+\"
\n" ); document.write( "(3) \"+t%5B1%5D+=+3.5+-+2+\"
\n" ); document.write( "(3) \"+t%5B1%5D+=+1.5+\"
\n" ); document.write( "------------------
\n" ); document.write( "(1) \"+d=+100t%5B1%5D+\"
\n" ); document.write( "(1) \"+d+=+100%2A1.5+\"
\n" ); document.write( "(1) \"+d+=+150+\"
\n" ); document.write( "The distance between Bedingfield and Portsmouth
\n" ); document.write( "is 150 km
\n" ); document.write( "--------------
\n" ); document.write( "check:
\n" ); document.write( "(2) \"+d+=+75t%5B2%5D+\"
\n" ); document.write( "(2) \"+d+=+75%2A2+\"
\n" ); document.write( "(2) \"+d+=+150+\"
\n" ); document.write( "OK
\n" ); document.write( "
\n" );