To train for the running portion of the race,she runs 7 miles each day over the same course. The first 3 miles of the course is on level ground, while the last 4 miles is downhill. She runs 2 miles per hour slower on level ground than she runs downhill?

So we know that the total run will take 1 hour. 3 miles will be at #x-2# mph and 4 miles will be at #x# mph.

Therefore, the amount of time she spends doing the first 3 miles is #3 / (x-2)# hours and she spends #4 / x# hours running the rest. Since that adds up to 1 hour,

#3/(x-2) + 4/x = 1#
#(3x + 4(x-2))/(x(x-2)) = 1 #
#3x + 4x-8 = x(x-2)#
#0 = x^2 - 9x +8#
By quadratic formula,
#x = (9 pm sqrt(81 - 4 cdot 8))/(2) = (9 pm 7)/2 #
Since we want the faster speed, #x = 16 / 2 = 8#.

Therefore, she runs on flat ground at 6 miles per hour and downhill at 8 miles per hour.

We can check this: it would take her 30 minutes to run 3 miles at 6 miles per hour and it would take her 30 minutes to run 4 miles at 8 miles per hour, so it would take her an hour to run all 7 miles, as we wanted.

Units of measurement in green
Values in red

Let the 'velocity' on level ground be #x#
Let the time run for level ground be #t_L#
Let the time run for downhill be #t_d#

For level ground #(color(red)(x)color(white)(.)color(green)(m/cancel(h)^1))xx(color(red)(t_L)color(white)(.)color(green)(cancel(h)^1))=color(red)(3)color(white)(.)color(green)( m)color(white)(.) Equation(1) #

For down hill # (color(red)(2x)color(white)(.) color(green)(m/cancel(h)^1)) xx (color(red)(t_d)color(white)(.)color(green)(cancel(h)^1)) = color(red)(4)color(white)(.)color(green)( m)color(white)(.) .......Equation(2) #

Total time running #->1 hour -> color(red)(t_L+t_d=1)color(white)(.)color(green)( h)" "..Equation(3)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of "x)#

Consider #Equation(3)#

Subtract #t_d# from both sides.

#color(red)(t_L=1-t_d)" "..................Equation(3_a)#

Using #Equation(3_a)# substitute for #t_L# in #Equation(1)#
This changes the problem into simultaneous equations.

#x(1-t_d)=3" ".........................Equation(1_a)#
#2xt_d=4" "..................................Equation(2)#

#-xt_d+x=3" ".........................Equation(1_a) #
#color(white)(.)2xt_dcolor(white)("ddd")=4" "...........................Equation(2)#

Multiply #Eqn(1_a)# by 2

#-2xt_d+2x=6" ".........................Equation(1_a) #
#ul(color(white)(..)2xt_dcolor(white)("dddd")=4" "...........................Equation(2))#
#color(white)("ddddddd")2x=10 larrEqn(1_a)+Eqn(2)#

Divide both sides by 2

#color(blue)(x=5)#

#color(blue)("=========================================")#

#color(blue)("Determine the time running on the level")#

Let velocity (spead or rate) be represented by #v#

Substitute for #x=5# in #Equation(1)# to determine #t_L#

For level ground #(color(red)(x)color(white)(.)color(green)(m/cancel(h)^1))xx(color(red)(t_L)color(white)(.)color(green)(cancel(h)^1))=color(red)(3)color(white)(.)color(green)( m)color(white)(.) Equation(1) #

#5t_L=3 => t_L=3/5 " hours"#

Thus velocity (rate) #->"velocity"xx"time"="distance"#

#color(white)("dddddddddddddddddddd")vcolor(white)("dd.")xxcolor(white)("dd")t_L=3#

#v=3-:t_L#

#v=3-:3/5#

#v=3xx5/3 = 5 mph#