document.write( "Question 1031938: A child is running on a moving sidewalk in an airport. When he runs against the sidewalk's motion, he travels 39ft in 13 seconds. When he runs with the sidewalk's motion, he travels 65ft in the same amount of time. What is the rate of the child on a still sidewalk and what is the rate of a moving sidewalk? \n" ); document.write( "
Algebra.Com's Answer #646613 by addingup(3677)\"\" \"About 
You can put this solution on YOUR website!
I just answered this question, I don't know if you posted it twice or if someone else had the same question:
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\n" ); document.write( "r= rate of child on a still sidewalk
\n" ); document.write( "w= walkway speed
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\n" ); document.write( "formula: distance = rate*time
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\n" ); document.write( "13(r-w) = 39
\n" ); document.write( "13(r+w) = 65
\n" ); document.write( "Divide all by 13 in order to simplify the equations:
\n" ); document.write( "r-w = 3
\n" ); document.write( "r+w = 5 Add both equations to eliminate one variable (-w cancels +w):
\n" ); document.write( "2r = 8
\n" ); document.write( "r = 4 This is the speed of the child. Now substitute to find w:
\n" ); document.write( "13(4-w) = 39
\n" ); document.write( "52-13w = 39
\n" ); document.write( "-13r = -13
\n" ); document.write( "w = 1 This is the speed of the sidewalk.
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\n" ); document.write( "With the sidewalk:
\n" ); document.write( "4+1=5; 5*13= 65 correct
\n" ); document.write( "Against the sidewalk:
\n" ); document.write( "4-1=3; 3*13=39 also correct \n" ); document.write( "
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