it took a faster runner 10 seconds longer to run a distance of 1500 ft. Then it took a slower runner to run 1000 ft. If the rate of faster runner was 5 feet per second more than the slower runner, what was the rate of each runner
I'll even set it up for you.
$\displaystyle 1000 = r * t $
Solve for t
$\displaystyle t = \frac{1000}{r} $
Plug this value in for t in equation 1
$\displaystyle 1500 = (r + 5)(\frac{1000}{r} + 10) $
Foil it
$\displaystyle 1500 = 1000 + 10r + \frac{5000}{r} + 50 $
Combine like terms
$\displaystyle 450 = 10r + \frac{5000}{r} $
Multiply through by an r
$\displaystyle 0 = 10r^2 - 450r + 5000 $
Divide by 10
$\displaystyle r^2 - 45r + 500 = 0 $
Solve the quadratic. You will get two rates. One will be the rate for runner 1. The other will be the rater for runner 2.
25 and 20??? is that how easy it is???? but why does my formula doesnt work the way it should be???
distace rate time
faster 1500 ft x+5 1500/x+5
slower 1000 ft X 1000 / x
equation: distance of faster is equal to distance travveled by slower:
1500/ x +5 = (1000 / x) + 10
plus 10 to make the slower runner distance 1500 feet also... why my equation doesnt work?? but its correct right?
Given $\displaystyle \frac{1500}{x + 5} = \frac{1000}{x} + 10 $
Multiply by the common denominator $\displaystyle (x)(x+5) $
$\displaystyle 1500x = 1000(x+5) + 10(x)(x+5) $
Multiply the expressions out
$\displaystyle 1500x = 1000x + 5000 + 10(x^2 + 5x) $
$\displaystyle 1500x = 1000x + 5000 + 10x^2 + 50x $
Combine like terms
$\displaystyle 450x = 10x^2 + 5000 $
It should look familiar from here.