Question 149687
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

Hope this helps------ptaylor