document.write( "Question 317833: A cyclist travels 54 miles in 6 hours going against the wind and 52 miles in 4 hours with the wind. What is the rate of the cyclist in still air and what is the rate of the wind? \n" ); document.write( "
Algebra.Com's Answer #227615 by mananth(16949) You can put this solution on YOUR website! let speed of cycle be x in still air \n" ); document.write( "speed of wind =y \n" ); document.write( ".. \n" ); document.write( "6(x-y)=54 \n" ); document.write( "4(x+y)52 \n" ); document.write( ".. \n" ); document.write( "6x-6y=54 \n" ); document.write( "x-y=9 \n" ); document.write( "4x+4y=52 \n" ); document.write( "x+y=13 \n" ); document.write( ".. \n" ); document.write( "add the two equations \n" ); document.write( "x-y+x+y=22 \n" ); document.write( "2x=22 \n" ); document.write( "x=11 mph. speed of cycle \n" ); document.write( "x-y=9 \n" ); document.write( "11-y=9 \n" ); document.write( "y=2 \n" ); document.write( "speed of wind = 2 mph \n" ); document.write( " \n" ); document.write( " |