Algebra PLEASE help!
posted by Faith on .
Pete's Plymouth travels 200 miles at a certain speed. If the car had gone 10 mph faster, the trip would have taken 1 hour less. Find Pete's speed.
I am very confused about how to solve this problem! Please explain well. Thank you so much!!! :)

let his original speed be x mph
then his original time is 200/x hours
let his second speed be x+10 mph
then his second time is 200/(x+10) hours
so 200/x  200/(x+10) = 1
multiply both sides by x(x+10)
200x + 2000  200x = x^2 + 10x
x^2 + 10x  2000 = 0
(x40)(x+50) = 0
x = 40 or x= 50 (negative speed makes no sense)
so Pete's speed is 40 mph.
check: at 40 mph, it would take him 200/40 = 5 hours
at 50 mph it would have taken him 200/50 = 4 hours.
OK then! 
Lat V be the slower speed and T be the time it takes to go 200 miles at that speed. You can write these two equations:
T = 200/V
T = 200/(V + 10) + 1
Now get rid of the T and create and solve an equation for v alone.
200 [1/V  1/(V+10) = 1
200(V + 10  V) = V(V+10)
V^2 + 10V 2000 = 0
(V40)(V+50) = 0
V = 40 since only positive roots are allowed.