A jogger ran 8 miles and then walked 6 miles. The jogger's running speed was 5 miles per hour faster than her walking speed. The total time for jogging and walking was 2 hours. What was the jogger's walking speed and jogging speed?

Let 'r' denote running and 'w' denote walking.
Let D be distance travelled. So,
#(D_r)=8#mi
#(D_w)=6#mi

Let "S" be the speed. So,
And since speed of running is 5mph faster than walking. So,
#(S_r)=S_w + 5mph#

Let T be the time.So,
Total time taken is 2 hrs.
Since the time includes #T_r# and #T_w#. We could say,
#(T_r)+(T_w)=2#
# or, (T_r)=2-(T_w)#

We know,
#S = (DeltaD)/(Deltat)#
#(S_r)=8/(2-T_w)# and
#(S_w)=6/(T_w)#

As mentioned
#(S_r)=(S_w)+5mph#
# or, 8/(2-T_w)=5+6/(T_w)#
For sake of ease let's suppose #(T_w)=x#
So,
#8/(2-x)=5+6/x#
# or, 8/(2-x)=(5x+6)/x#
Cross multiplying,
# or, 8x=(5x+6)(2-x)#
# or, 8x=4x-(5x^2)+12#
# or, 8x-4x+(5x^2)-12=0#
# or, (5x^2)+4x-12=0#
Factorization by splitting the middle term,
# or, (5x^2)+(10-6)x-12=0#
# or, (5x^2)+10x-6x-12=0#
# or, 5x (x+2)-6 (x+2)=0#
Use reverse of distributive property.
# or, (5x-6)(x+2)=0#

Since time is positive we must choose such binomial from above that yields positive value.
Choose
#(5x-6)=0#
#x=6/5#
#x=(T_w)=6/5#hrs

So,
#S_w=(deltaD_w)/(deltaT_w)#
#=6/(6/5)#
#=30/6#
#=5# mph

Again #S_r=5mph+S_w#
#=5mph+5mph#
#=10mph#

#:.S_r=10#mph
#:.S_w=5#mph