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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document.write( "What I have written so far:
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document.write( "48 mi--against 4 mph slower
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document.write( "60 mi--with and 4 mph faster \n" );
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Algebra.Com's Answer #587027 by lwsshak3(11628)![]() ![]() ![]() 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. \n" ); document.write( "*** \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( " \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 \n" ); document.write( " |