Question 7621
As you say, 
let Elizabeth speed = x (in miles)
let Rick's speed = x+5


Elizabeth rode 40 miles, Rick rode 80 miles


Elizabeth's time = t (in hours)
Rick's time = t+1

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we want to find their speeds, but we also do not know their times, so we will get an equation for each based upon speed = distance/time and then re-arrange for t and then equate, to get rid of t. This will just leave one equation with the unknown, x, which we will solve. 


That is the theory :-)


Right then, first Elizabeth:


speed = distance/time 
x = 40/t --> t = 40/x


Now Rick:


speed = distance/time
x+5 = 80/(t+1)
t+1 = 80/(x+5)
t = 80/(x+5) - 1


so we have {{{40/x = 80/(x+5) - 1}}}. Now it is just algebraic manipulation...


{{{40/x + 1 = 80/(x+5)}}}
{{{40/x + x/x = 80/(x+5)}}}
{{{(x+40)/x = 80/(x+5)}}}. Now cross multiply


{{{(x+40)(x+5) = 80x}}}
{{{x^2 + 45x + 200 = 80x}}}
{{{x^2 - 35x + 200 = 0}}}


put this into the quadratic formula to give {{{x = (35 +- 5sqrt(17))/2}}}.


This means that x = 27.8 mph or x=7.192mph.


For a person on a cycle, i would suggest that 7.192mph is the answer required. Then Rick's speed would be 12.192mph.


Please check my working though, just in case i have made an error somewhere.


jon.