document.write( "Question 349512: Bill leaves his house for Makayla's house, riding his bicycle at 8 miles per hour. At the same time, Mikayla leaves her house, heading for bill's house walking at 3 miles per hour. Considering they live 8.25 miles apart please write a linear system that shows at what time they would meet. \n" ); document.write( "
Algebra.Com's Answer #249973 by ankor@dixie-net.com(22740)\"\" \"About 
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Bill leaves his house for Makayla's house, riding his bicycle at 8 miles per
\n" ); document.write( " hour. At the same time, Mikayla leaves her house, heading for bill's house
\n" ); document.write( "walking at 3 miles per hour.
\n" ); document.write( "Considering they live 8.25 miles apart please write a linear system that shows at what time they would meet.
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\n" ); document.write( "Let t = time B bikes and time M walks
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\n" ); document.write( "When they meet their total distance will = 8.25 miles
\n" ); document.write( "Write a distance equation; dist = speed * time
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\n" ); document.write( "B's dist + M's dist = 8.25 mi
\n" ); document.write( "8t + 3t = 8.25
\n" ); document.write( "11t = 8.25
\n" ); document.write( "t = \"8.25%2F11\"
\n" ); document.write( "t = .75 hrs or .75(60) = 45 minutes for them to meet
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\n" ); document.write( "Check by finding the actual dist each traveled
\n" ); document.write( "8(.75) = 6 mi
\n" ); document.write( "3(.75) = 2.25
\n" ); document.write( "-------------
\n" ); document.write( "total: 8.25 mi; confirms our solution of t = .75 \n" ); document.write( "
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