document.write( "Question 512669: On a 50 mile bike ride Irena averaged 4 mph faster for the first 36 miles than she did for the last 14 miles. The entire trop took 3 hours. Find her speed for the 36 miles \n" ); document.write( "
Algebra.Com's Answer #342809 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! On a 50 mile bike ride Irena averaged 4 mph faster for the first 36 miles than she did for the last 14 miles. \n" ); document.write( " The entire trip took 3 hours. Find her speed for the 36 miles \n" ); document.write( ": \n" ); document.write( "let s = the speed for the 1st 36 mi \n" ); document.write( ": \n" ); document.write( "Write a time equation; time = dist/speed \n" ); document.write( ": \n" ); document.write( "36mi time + 14 mi time = 3 hrs \n" ); document.write( " \n" ); document.write( "Multiply by s(s-4) \n" ); document.write( "s(s-4)* \n" ); document.write( "cancel the denominators \n" ); document.write( "36(s-4) + 14s = 3s(s-4) \n" ); document.write( "36s - 144 + 14s = 3x^2 - 12s \n" ); document.write( "50s - 144 = 3x^2 - 12s \n" ); document.write( "arrange as a quadratic equation on the right \n" ); document.write( "0 = 3s^2 - 12s - 50s + 144 \n" ); document.write( "3x^2 - 62s + 144 = 0 \n" ); document.write( "you can use the quadratic formula, but this will factor to: \n" ); document.write( "(3s-8)(s-18) = 0 \n" ); document.write( "The reasonable solution \n" ); document.write( "s = 18 mph for the 1st 36 mi \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this by finding the time at each speed \n" ); document.write( "36/18 = 2 hrs \n" ); document.write( "14/14 = 1 hr \n" ); document.write( " \n" ); document.write( " |