document.write( "Question 910599: Mark works out for 50 minutes by biking and jogging. He bikes at an average rate of 1200 feet per minute for x minutes. He jogs at an average rate of 900 feet per minute. He wants to travel a combined 10 miles. How many minutes did Mark spend biking?
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Algebra.Com's Answer #552574 by ankor@dixie-net.com(22740)\"\" \"About 
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Mark works out for 50 minutes by biking and jogging.
\n" ); document.write( " He bikes at an average rate of 1200 feet per minute for x minutes.
\n" ); document.write( " He jogs at an average rate of 900 feet per minute.
\n" ); document.write( " He wants to travel a combined 10 miles.
\n" ); document.write( " How many minutes did Mark spend biking?
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\n" ); document.write( "Let j = time jogging (in minutes)
\n" ); document.write( "let b = time biking
\n" ); document.write( "then
\n" ); document.write( "j + b = 50
\n" ); document.write( "j = (50-b), use this form for substitution
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\n" ); document.write( "Change 10 mi to ft: 10 * 5280 = 52,800 ft
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\n" ); document.write( "1200b + 900j = 52800
\n" ); document.write( "Simplify, divide by 100
\n" ); document.write( "12b + 9j = 528
\n" ); document.write( "substitute (50-b) for j
\n" ); document.write( "12b + 9(50-b) = 528
\n" ); document.write( "12b + 450 - 9b = 528
\n" ); document.write( "12b - 9b = 528 - 450
\n" ); document.write( "3b = 78
\n" ); document.write( "b = 78/3
\n" ); document.write( "b = 26 min biking
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\n" ); document.write( "Check this, we know that he spent 50 - 26 = 24 min jogging
\n" ); document.write( "26*1200 = 31200 ft biking
\n" ); document.write( "24*900 = 21600 ft jogging
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\n" ); document.write( "total dist: 52800 which is 10 mi
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