This is a classic "rate times time equals distance" word problem. If I write that in letters: rt=d, where rate=r, time=t, and distance=d.
In this word problem, the rate is made up of two things: the speed of the helicopter (lets call that r) and the speed of the wind (which is 6mph). When the helicopter is traveling with the wind, it goes (r+6)mph and when it is traveling against the wind,
the helicopter goes (r-6)mph.
OK, we know the helicopter goes 270miles with the wind, so let's set up this "rt=d" equation filling in the numbers we know:
(r+6)t=270
OK, we also know the helicopter goes 180miles against the wind, so that "rt=d" equation looks like this:
(r-6)t=180
In both of those equations, we know that t=t since in this word problem the times are the same, but we are not given the time. No worries, let's rewrite each of those equations equal to t:
t=270/(r+6)
t=180/(r-6)
(I just divided both sides by the (r +/- 6))
Since t=t, then:
270 =
180
(r+6) (r-6)
Cross-multiply to get rid of the fractions:
270(r-6)=180(r+6)
Distribute:
270r-1620=180r+1080
Gather your "r" terms on one side and the numbers on the other side:
90r=2700
Divide both sides by 90:
r=30.
Therefore, the speed of the helicopter is 30mph.