# SOLUTION: A bicycle trip of 120 miles would have taken 3 hours less if the average speed had been increased by 2 miles per hour. Find the average speed of the bicycle.

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 Question 164477: A bicycle trip of 120 miles would have taken 3 hours less if the average speed had been increased by 2 miles per hour. Find the average speed of the bicycle.Found 2 solutions by checkley77, themathprof:Answer by checkley77(12569)   (Show Source): You can put this solution on YOUR website!120/X=[120/(X+2)]+3 X[120/(X+2)+3=120 120X/(X+2)+3X=120 [(120X+3X(X+2)]/(X+2)=120 (120X+3X^2+6X)/(X+2)=120 126X+3X^2=120(X+2) 3X^2+126X=120X+240 3X^2+126X-120X-240=0 3X^2+6X-240=0 3(X^2+2X-80)=0 3(X+10)(X-8)=0 X-8=0 X=8 MPH IS THE SLOWER TRIP. 8+2=10 MPH FOR THE FASTER TRIP. 120/8=15 HOURS. 120/10=12 HOURS. 240/X=27 27X=240 X=240/27 X=8.89 MPH IS THE AVERAGE SPEED. Answer by themathprof(21)   (Show Source): You can put this solution on YOUR website!A bicycle trip of 120 miles would have taken 3 hours less if the average speed had been increased by 2 miles per hour. Find the average speed of the bicycle. x= avg speed of bicycle The distance is the same (120) before and after the increase The Rate changes from x to x+2 Note that in each case the time is Distance/Rate Rate x Time = Distance Before increase x....120/x.....120 After increase x+2...120/x+2..120 New Time is 3 less 120/(x+2) = 120/x -3 Multiply each term by LCM=x(x+2) x(x+2)(120)/(x+2) = x(x+2)(120)/x -3(x)(x+2) Cancel denominators 120x=120(x+2)-3x(x+2) 120x=120x+240-3x^2-6x 0=-3x^2-6x+240 divide by -3 0=x^2+2x-80 Factor right side 0=(x-8)(x+10) x-8=0 .... x=8 avg speed of bicycle x+10=0 ... x=-10 (reject, speed can't be negative!) CHECK Rate x Time = Distance Before increase 8....120/8=15.....120 After increase 8+2=10...120/8+2=12..120 Time decreases 3 from 15 to 12 See more word problems worked out step by step in my math911 free tutorial software. Download, install, activate in minutes. www.math911.com No Charge! Tell your classmates, friends, (and teacher) Prof Martin Weissman mathprof@hotmail.com