:how do i solve this: if james goes 15 miles faster per hour, he will go 260 miles in 20 minutes less than normal. what is the original rate?

15 mi/h= 15mi/60 min= 1/4 (mi/min)

x is  original speed (mi/min)
t  is original time
x*t =260

(x+1/4)=5x/4  is new speed
t-20 - new time
(x+1/4)(t-20)=260

xt = (x+1/4)(t-20)
xt=xt + t/4 - 20x - 5
t/4=20x+5
t=80x +20 

We can substitute t=80x +20 into equation x*t =260
x*(80x+20)=260
80x²+20x-260=0 
4x²+x-13=0

D=b² - 4ac = 1+4*4*13= 209
x=(-1+/-√209)/2*4
x=1.68 mi/min

1.68mi/min * 60 min/1h≈100 mi/h

Check
260/100=2.6h time with old speed
260/(115)=2.26 h time with new speed
2.6-2.26 = 0.34 h difference between old and new time 
0.34h*60min/1h≈20 min difference between old and new time in minutes