SOLUTION: Chris went on a bike ride of 24 miles. He realized that if he had gone 6 mph faster, he would have arrived 9 hours sooner. How fast did he actually ride?

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Question 1029727: Chris went on a bike ride of 24 miles. He realized that if he had gone 6 mph faster, he would have arrived 9 hours sooner. How fast did he actually ride?
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
rt = d
d = 24
rt = 24

(r+6)(t-9) = 24
simplify to get rt - 9r + 6t - 54 = 24

rt = 24
solve for r to get r = 24/t

replace r with 24/t in the equation of rt - 9r + 6t - 54 = 24 to get:
24t/t - 216/t + 6t - 54 = 24

multiply both sides of this equation by t to get:
24t - 216 + 6t^2 - 54t = 24t

subtract 24t from both sides of this equation and combine like terms and reorder the terms in descending order of degree to get:
6t^2 - 54t - 216 = 0

factor out the gcf of 6 to get:
t^2 - 9t - 36 = 0

factor this quadratic equation to get:
(t-12)(t+3)

solve for t to get:
t = 12 or t = -3
t = -3 can't be good because time is not negative, so you get t = 12.

when t = 12, rt = 24 becomes r*12 = 24 and solve for r to get r = 2.

you have r = 2
t = 12

rt = d becomes 2*12 = 24, confirming this is good.

r+6 = 8
t-9 = 3

(r+6)(t-9) = d becomes 8*3 = 24, confirming this is also good.

your solution is that he was riding at 2 miles per hour.


Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

Chris went on a bike ride of 24 miles. He realized that if he had gone 6 mph faster, he would have arrived 9 hours sooner. How fast did he actually ride?
Let his actual speed be S

Actual time taken: 

If he'd traveled 6 mph faster, his speed would've been: S + 6

Time that he would’ve taken, travelling at the faster speed: 

We then get the following TIME equation: 

24(S + 6) = 24S + 9S(S + 6) ------- Multiplying by LCD, S(S + 6)







 ----- Factoring out GCF, 9



(S - 2)(S + 8) = 0

S, or actual speed =  mph 

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