Algebra Word Problem (distance)

Question

Bryce went on a bike ride of 72 miles. He realized that if he had gone 18 mph faster, he would have arrived 9 hours sooner. How fast did he actually ride?

 

Can someone help me set this up? Please! :)

I understand the general idea of most of the distance problems, but I always make mistakes when it comes to the + hours sooner

Answer 1

 

Let x = his rate for the trip (initially)

Let t = his time in hours for the trip (initially)

Let "x + 18" = his rate had he traveled 18 miles per hour faster

Let "t - 9" = his time had he arrived 9 hours sooner

The distance of the trip = 72

 

Since rate times time = distance,

 

For the initial trip.

 

    x times t = 72

    xt = 72

 

If he were going 18 miles per hour faster,

 

    (x + 18)(t - 9) = 72

 

    Multiplying and simplifying, we have

 

    xt -9x + 18t - 162 = 72

    xt -9x + 18t = 234

 

    Note in the earliar equation that "xt" = 72 ...

    Solving for s in that equation, we have "x = 72/t" ...

    Substituting those values into the above equation, we have

 

    72 - 9(72/t) + 18t = 234

    72 - 648/t + 18t = 234

    -648/t + 18t = 234

    -648/t + 18t²/t = 234

    (18t² -648)/t = 234

 

     Cross multiply

 

     18t²  -648 = 234t

     18t² -234t -648 = 0

 

     Divide alll terms by 18

 

     t² -13t -36 = 0

     

     Use quadratic formula to solve the equation

 

     t = 15.35 hours