document.write( "Question 317855: A student walks and jogs to college each day. she averages 5km/h walking and 9km/h jogging. The distancefrom home to college is 8km and she makes the trip in 1 hour. How far does the student jog?\r
\n" ); document.write( "\n" ); document.write( "I can't seem to get it set up right to get the correct calculation. I used 5km(X) +9 km(y) = 8km. \r
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Algebra.Com's Answer #227613 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
let her walk x km.
\n" ); document.write( "she will jog 8-x km
\n" ); document.write( "..
\n" ); document.write( "walking speed = 5km/hr
\n" ); document.write( "jogging speed = 9km/hr
\n" ); document.write( "..
\n" ); document.write( "distance to college 8km.\r
\n" ); document.write( "\n" ); document.write( "walking time + jogging time = total time
\n" ); document.write( "x/5+ 8-x /9 =1
\n" ); document.write( "9x+40-5x /45=1
\n" ); document.write( "4x+40=45
\n" ); document.write( "4x=5
\n" ); document.write( "x= 5/4 km
\n" ); document.write( "she walks 5/4 km
\n" ); document.write( "1.25 km
\n" ); document.write( "so she jogs 8-5/4
\n" ); document.write( "32-5 /4
\n" ); document.write( "27/4
\n" ); document.write( "6.75 km
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