document.write( "Question 111616: A student walks and jogs to college each day. She averages 5km/h walking and 9km/h jogging. The distance from home to college is 8kn, and she makes the trip in 1 hr. How far does the student jog? \n" ); document.write( "
Algebra.Com's Answer #81479 by ptaylor(2198)\"\" \"About 
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distance(d)=rate(r) times time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let d=distance the student jogs
\n" ); document.write( "Then 8-d=distance the student walks\r
\n" ); document.write( "\n" ); document.write( "Time spent walking=(8-d)/5\r
\n" ); document.write( "\n" ); document.write( "Time spent jogging=d/9\r
\n" ); document.write( "\n" ); document.write( "Now we are told that the time spent walking plus the time spent jogging = 1 hour. So:\r
\n" ); document.write( "\n" ); document.write( "(8-d)/5+d/9=1 multiply each term by 45 (LCM)
\n" ); document.write( "9(8-d)+5d=45 get rid of parens\r
\n" ); document.write( "\n" ); document.write( "72-9d+5d=45 subtract 72 from both sides\r
\n" ); document.write( "\n" ); document.write( "72-72-9d+5d=45-72 collect like terms\r
\n" ); document.write( "\n" ); document.write( "-4d=-27 divide both sides by -4\r
\n" ); document.write( "\n" ); document.write( "d=6.75 mi-----------------------distance student jogs\r
\n" ); document.write( "\n" ); document.write( "8-6.75=1.25 mi--------------------distance student walks\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "6.75/9+1.25/5=1\r
\n" ); document.write( "\n" ); document.write( "0.75+0.25=1
\n" ); document.write( "1=1\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor\r
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