document.write( "Question 148319: You can walk 10 miles in the same time that it takes to travel 15 miles by bicycle. If the bike's rate is 3 miles per hour faster than your wlaking rate, fing the average rate for each. \n" ); document.write( "
Algebra.Com's Answer #108733 by ankor@dixie-net.com(22740)\"\" \"About 
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You can walk 10 miles in the same time that it takes to travel 15 miles by bicycle. If the bike's rate is 3 miles per hour faster than your walking rate, find the average rate for each.
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\n" ); document.write( "Let x = walking rate
\n" ); document.write( "then
\n" ); document.write( "(x+3) = biking rate
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\n" ); document.write( "Write a time equation: Time = \"dist%2Frate\"
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\n" ); document.write( "Biking time = walking time
\n" ); document.write( "\"15%2F%28%28x%2B3%29%29\" = \"10%2Fx\"
\n" ); document.write( "Cross multiply
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\n" ); document.write( "15x = 10(x+3)
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\n" ); document.write( "15x = 10x + 30
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\n" ); document.write( "15x - 10x = 30
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\n" ); document.write( "5x = 30
\n" ); document.write( "x = \"30%2F5\"
\n" ); document.write( "x = 6 mph, that's really fast walking!
\n" ); document.write( "then
\n" ); document.write( "6 + 3 = 9 mph is the bike rate
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\n" ); document.write( "Check solution by finding the time for each:
\n" ); document.write( "15/9 = 1.67 hrs
\n" ); document.write( "10/6 = 1/67 hrs; confirm our solutions
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