Your question is incomplete, I hope its asking for alfonso's rate
So lets assume alfonso's rate be = x miles per hour
It says when he cycles in the direction of wind, 2 miles per hours is added to his rate which we assumed as x. So 2 is added to x, so speed in direction of wind will be (x+2)
Distance travelled in direction of wind is 38 miles
We have formula for distance speed time as
d = r × t
or [tex] \frac{d}{r} = t [/tex]------------------------------------------(1)
so simply plug distance 38 as distance in d place, (x+2) as speed in r place, so we get time t in direction of wind as
[tex] t = \frac{38}{(x+2)} [/tex]----------------------------------------(2)
Similarly we will work when he goes against the wind. Then 2 miles per hour is subracted from his rate which we asumed as x.
So 2 is subtracted form x. So his speed against the wind will be (x-2).
Distance travelled against the wind is 30 miles
so again plug 30 as distance in d place and (x-2) as speed in r place in formula given by (1)
so we get time t against the wind as
[tex] t = \frac{30}{(x-2)} [/tex]----------------------------------------(3)
As its given in question that time taken is same for with the wind and against wind so we will equate both time expression in equation (2) and equation (3)
[tex] \frac{38}{(x+2)} = \frac{30}{(x-2)} [/tex]
Now we have to solve for x
So first cross multiply as shown
38 × (x-2) = 30 × (x+2)
38x - 76 = 30x + 60
38x - 76 -30x = 30x + 60 -30x
8x -76 = 60
8x -76 +76 = 60 +76
8x = 136
[tex] \frac{8x}{8} = \frac{136}{8} [/tex]
x = 17
So alfonso's rate is 17 miles per hour
Hope this helps !!