document.write( "Question 960542: Trina can ride 48 mi on her bike against the wind. With the wind at her back, she rides 4 mph faster and can ride 60 mi in the same amount of time. Find her speed riding against the wind, and her speed riding with the wind.\r
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\n" ); document.write( "\n" ); document.write( "What I have written so far:
\n" ); document.write( "48 mi--against 4 mph slower
\n" ); document.write( "60 mi--with and 4 mph faster
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Algebra.Com's Answer #587027 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
Trina can ride 48 mi on her bike against the wind. With the wind at her back, she rides 4 mph faster and can ride 60 mi in the same amount of time. Find her speed riding against the wind, and her speed riding with the wind.
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\n" ); document.write( "let x=Trina's speed against the wind
\n" ); document.write( "x+4=Trina's speed with the wind
\n" ); document.write( "travel time=distance/speed
\n" ); document.write( "..
\n" ); document.write( "\"48%2Fx=60%2F%28x%2B4%29\"
\n" ); document.write( "60x=48x+192
\n" ); document.write( "12x=192
\n" ); document.write( "x=16
\n" ); document.write( "16+4=20
\n" ); document.write( "Trina's speed against the wind=16 mph
\n" ); document.write( "Trina's speed with the wind=20 mph
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