document.write( "Question 1063671: To train for the bicycle portion of the race, Jerri rides 24 miles out a straight road, then turns around and rides 24 miles back. The trip is out against the wind, whereas the trip back is with the wind. If she rides 10mph faster with the wind then she does against the wind, and the complete trip out and back takes 2 hours, how fast does she ride when she rides against the wind? \n" ); document.write( "

Algebra.Com's Answer #678804 by jorel1380(3719) $\"\"$ $\"About$

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Let s be Jerri's speed in still air, and w be the speed of the wind. Then:
\n" ); document.write( "s+c-10=s-c
\n" ); document.write( "2c=10
\n" ); document.write( "c=5
\n" ); document.write( "Then, to find out her speed in still air:
\n" ); document.write( "24/s+5 + 24/s-5=2
\n" ); document.write( "24(s-5)+24(s+5)=2s²-50
\n" ); document.write( "2s²-48s-50=0
\n" ); document.write( "s²-24s-25=0
\n" ); document.write( "(s-25)(s+1)=0
\n" ); document.write( "s=25 or -1
\n" ); document.write( "Throwing out the negative result from the answer, we get Jerri's speed in still air as 25 mph; therefore, her speed against the wind is 25-5, or 20 mph. ☺☺☺☺ \n" ); document.write( "

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