document.write( "Question 574461: during the first part of the trip a canoeist travels 40 miles at a certain speed the canoeist travels 20 miles on the seconf part of the trip at a speed 5 mph slower the total time for the trip is 3 hours what was the speed on each part of the trip? \n" ); document.write( "

Algebra.Com's Answer #369105 by mananth(16946) $\"\"$ $\"About$

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Forward 40 miles
\n" ); document.write( "Returning speed to DC 20 miles
\n" ); document.write( "
\n" ); document.write( "Forward speed x mph
\n" ); document.write( "return speed x-5 mph
\n" ); document.write( "Total time =3hours
\n" ); document.write( "Time forward 40 / x
\n" ); document.write( "time return 20/(x-5)
\n" ); document.write( "
\n" ); document.write( "add up the time =3 hours
\n" ); document.write( "
\n" ); document.write( "40/x + 20/(x-5)=3
\n" ); document.write( "LCD = (x)*(x-5)
\n" ); document.write( "multiply the equation by the LCD
\n" ); document.write( "we get
\n" ); document.write( "40*(x-5 )+20x = 3
\n" ); document.write( "40x-200 +20x=3x^2-15x
\n" ); document.write( "75x-200 =3x^2
\n" ); document.write( "3x^2-75x+200= 0
\n" ); document.write( "3X^2-75 x+200 = 0
\n" ); document.write( "Find the roots of the equation by quadratic formula
\n" ); document.write( "
\n" ); document.write( "a= 3 b= -75 c= 200
\n" ); document.write( "
\n" ); document.write( "b^2-4ac= 5625 - 2400
\n" ); document.write( "b^2-4ac= 3225
\n" ); document.write( " $\"sqrt%28%093225%09%29\"$ = 56.79 2.99
\n" ); document.write( " $\"x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29\"$
\n" ); document.write( " $\"x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29\"$
\n" ); document.write( "x1=( 75 + 56.79 )/ 6
\n" ); document.write( "x1= 21.96
\n" ); document.write( "x2=( 75 -56.79 ) / 6
\n" ); document.write( "x2= 3.04
\n" ); document.write( "Forward speed = 22 mph
\n" ); document.write( "return speed = 17 mph
\n" ); document.write( " \n" ); document.write( "

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