Using distance=rate multiplied by time as the formula,
mountaineer climbed 1,000 feet at a rate of x feet per hour
1000 = xt1 <--eq.1
"He climbed an additional 5,000 feet at a different rate."
5000 = x't2 <--eq.2
"This rate was 10 feet per hour less than twice the first rate."
x' = 2x - 10 <--eq.3
We now have 3 equations/expressions to solve for the number of hours:
From eq. 1 and 2:
Total time = t1 +t2 = t
t = 1000/x + 5000/x'
but x' = 2x - 10
t = 1000/x + 5000/(2x-10)
t = 1000/x + 5000/[2(x-5)]
t = 1000/x + 2500/(x-5) <--ANSWER
