The exercise is already scheduled, so we just follow the instruction:
1: Given two points
and
, the equation of the line passing through the two points is given by

In your case, we have

To put this equation into standard form, we have

2: By the exact same logic, the line that represents Kelsa's training progress is given by

Solving for y yields

and finally

3: We have the following system:

Since both equations are solved for y, we deduce that the two right hand sides must be equal:

Multiply both equations by 14 to get rid of the denominators:

Plug this value for x in one of the equations:

4: Since x represents the number of weeks, and x = 7 is the solution of the system, after 7 weeks they will habe the same average minute per mile
5: We have the equations that represent the times of the two girls, depending on time. So, we simply have to plug x=30 in the equations to get the times after 30 weeks: for Alana we have

which evaluates to

And for Kelsa we have

which evaluates to
