distance = speed * time

notice that hour and minutes are differencet unit.

15minutes = .25hr

let a be the speed of the wind

when the airplane travels with the wind its resultant speed is a + 25. (resultant means sum). Now that the airplane returns, it travels agains the wind, the resultant speed of the airplane is a – 25.

now for the time

let t be the time it takes the airplane to travel one city to another in the same direction of the wind. “The trip out takes 15 mintues less time than the retuning flight” tells you the time it takes the airplane to return is 15 minutes (.25hr) MORE than it takes to travel out, thus, the time is t + .25

traveling out:

700 = (a + 25)t

return:

700 = (a – 25) (t + .25)

solve for t

700 = (a + 25)t

t = 700 / (a + 25)

plug t back in the for t in the other equation

700 = (a – 25) (t + .25)

700 = (a – 25) ( 700/(a + 25) + .25)

distrubute

700 = 700a/(a + 25) + .25a – 17,500/(a + 25) – 6.25

multiply both sides by (a +25)

700(a + 25) = 700a + .25a(a + 25) – 17,500 – 6.25(a + 25)

distribute

700a+17,500 =700a+.25a^2 +6.25a – 17,500-6.25a -156.25

combine like terms

700a + 17,500 = .25a^2 + 17,343.75

subtract 700a for both sides and subtract 17,500 for both sides

0 = .25a^2 – 35156.25

35156.25 = .25a^2

a^2 = 140625

a = 375

so the speed of the airplane is 375 mi/hr

d = t1 * (v + 25)

700 = t1 * (v + 25)

Similarly, on the return trip,

700 = t2 * (v – 25)

And we know that

t2 = t1 + 15

Now we have three equations with three variables. Solve by substitution.