Question

On a particular day, the wind added 2 miles per hour to alfonso's rate when he was cycling with the wind and subtracted 2 miles per hour from his rate on his return trip. alfonso found that in the same amount of time he could cycle 38 miles with the wind, he could go only 30 miles against the wind.

Answer 1

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
`(d)/(r) = t`------------------------------------------(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

`t = (38)/((x+2))`----------------------------------------(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

`t = (30)/((x-2))`----------------------------------------(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)

`(38)/((x+2)) = (30)/((x-2))`

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

`(8x)/(8) = (136)/(8)`

x = 17

So alfonso's rate is 17 miles per hour

Hope this helps !!