Let x be Olivia's rate (in miles per hour).

Then the Inez's rate is (x+2) mph, according to the condition.


Then you can write this "distance" equations


    (25/60 + t)*x = 5,     (1)

     t*(x+2)      = 5      (2)


In these equations, " t " is the time after Inez' start to the catching moment;

the left  side of the equation (1) is the distance ran by Olivia, while 
the right side of the equation (2) is the distance ran by Inez from the starting point 

to the catching up point, and equations state that each of these distances is 5 miles long.


So, this system of two non-linear equations is one possible setup for the problem.



Another possible setup is this equation


     -  = ,     (3)


saying that the difference of two traveled times,  Olivia's time    and  Inez's time  ,  is  25 minutes, 
or    of an hour.

This equation (3) is called the "time" equation.


I prefer this second setup and will show you on how to solve this equation (3).


Multiply both sides by  60x*(x+2)  to get

    300(x+2) - 300x = 25x*(x+2)

    300x + 600 - 300x = 25x^2 + 50x

    25x^2 + 50x - 600 = 0

    x^2 + 2x - 24 = 0

    (x+6)*(x-4) = 0


The roots are -6 and 4, and we accept only the positive value of 4 as the Olivia' speed.


ANSWER.  Olivia' speed is 4 mph;  Inez' speed is 4+2 = 6 mph.


CHECK.   Olivia's time is   hours = 1 hour and 15 minutes = 60 + 15 = 75 minutes.

         Inez's time is   hours = 50 minutes,

         and the difference is  75 minutes MINUS 50 minutes = 25 minutes,

         EXACTLY and PRECISELY as the problem states.