# SOLUTION: Solve the problem. A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10m

Algebra ->  Algebra  -> Test -> SOLUTION: Solve the problem. A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10m      Log On

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 Click here to see ALL problems on test Question 309607: Solve the problem. A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10miles per hour faster than his rate walking, wat was each rate?Answer by mananth(12270)   (Show Source): You can put this solution on YOUR website!A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10miles per hour faster than his rate walking, wat was each rate? let his rate of walking be x mph His rate of riding will be x + 10 mph . time taken for walking = 8/x hours time taken for riding = 12/x+10 . Time walking + time riding = total time 8/x + 12/ x+10 = 5 hours LCM = x(x+10) 8(x+10) + 12x / x (x+10)= 5 8x +80 +12x = 5x^2 +50x 5x^2+30x-80=0 x^+6x-16 = 0 x^2+8x-2x-16=0 x(x+8)-2(x+8)=0 (x+8)(x-2)=0 x=2 which is positive He walks at 2 mph rides at 12 mph