motion problem?

Question

lynn drove 225 km to visit a college campus. on the trip home, she averaged 15 km/h more than on the trip to the college. if her total travel time ws 6 3/4 h, what was her average speed on the trip home?

Answer 1

Let v1 be the average speed on the way
let v2 be the average speed on the way home
let t1 be the time to the college
let t2 be the time to get home

v1+15=v2 (given in the problem)
t1+t2=6.75 (given in the problem), or t1=6.75-t2
v1=225/t1 (avg velocity=distance/time)
v2=225/t2 ("), or v2*t2=225, or t2=225/v2

Now solve by substitution:
v2=v1+15 (equation 1)
v2=225/t1+15 (substitute with equation 3)
v2=225/(6.75-t2)+15 (substitute with equation 2, second form)
v2=225/(6.75-(225/v2))+15 (substitute with equation 4, third form)
Now we have only one variable, we just need to simplify (there is probably an easier way to simplify)
v2-15=225/(6.75-(225/v2)) (subtract 15 from both sides)
(v2-15)(6.75-(225/v2))=225 (multiply both sides by 6.75-225/v2
6.75*v2-225-101.25+3375/v2=225 (FOIL)
6.75*v2-551.25=-3375/v2 (re-arrange and simplify)
6.75*v2^2-551.25*v2=-3375 (multiply both sides by v2)
Now this is a quadratic, which you may solve by using any method you prefer. I will spare you the details, as they get ugly, but the possible answers are:
v2=75 or v2=6.67
The second answer can be thrown away since your first velocity would be negative if this were true. Thus the answer is 75kph.

Check:
v2=75
v1=75-15=60
225/60=3.75
225/75=3
3.75+3=6.75 =>Q.E.D.