During the first part of a trip Trudys Honda traveled 120 mi

During the first part of a trip, Trudy\'s Honda traveled 120 mi at a certain speed. Trudy then drove another 100 mi at a speed that was 10 mph slower. If the total time of Trudy\'s trip was 4 hr, what was her speed on each part of the trip?
R=D/T
R= 120/t
R=100/t-10
120/x=100/x-10
x(x-10)/1/120/x x(x-10)/1/100/x-10

Solution

Distance(D) equals Rate(R) times Time(T) or D=RT; R=D/T and T=D/R
I HAVE FOUND THAT IF YOU LAY OUT YOUR PROBLEM BY DESCRIBING WHAT EACH UNKNOWN REPRESENTS IT IS GENERALLY LESS CONFUSING. SEE BELOW:
Let R=Trudy\'s speed during the 120 mi stretch
Then R-10=Trudy\'s speed during the 100 mi stretch
Trudy\'s time during the 120 mi stretch=120/R
Trudy\'s time during the 100 mi stretch = 100/(R-10)
Now we are told that the above times add up to 4 hours, so our equation to solve is:
120/R + 100/(R-10)=4 multiply each term by R(R-10)
120(R-10)+100R=4R(R-10) get rid of parens
120R-1200+100R=4R^2-40R subtract 4R^2 from and add 40R to each side
120R-1200+100R+40R-4R^2=4R^2-4R^2-40R+40R collect like terms
-4R^2+260R-1200=0 divide each term by -4
R^2-65R+300=0------------------------quadratic in standard form and it can be factored:
(R-60)(R-5)=0
R=60 mph-----------------------------rate during 120 mi stretch
R-10=60-10=50 mph-----------------rate during 100 mi stretch
and R=5 mph
R-10=-5 mph-----------no good!!!!! can\'t have negative speed in this problem
CK
120/60=2 hours
100/50=2 hours
2+2=4
4=4