Question 253210
This is a rate x time = distance problem. Here is the table based on the question:
wind . . . . . rate . . . . . . . . . time . . . . . . . . . . .distance
with . . . . .  r + 180 . . . . . . .  t . . . . . . . . . . . . . .  7
against . . . r - 180 . . . . . . . . t . . . . . . . . . . . . . . 5
The time is distance / rate or
7/(r+180) and 5/(r-180).
Since the times were the same set these fractions equal to each other as:
7/(r+180)  = 5/(r-180).
Cross multiply to get
7r - 1260 = 5r + 900
solve for r to get
2r = 2160
r = 1080.
The wind speed is 1080 mph.