Question 111616
distance(d)=rate(r) times time(t) or d=rt; r=d/t and t=d/r

Let d=distance the student jogs
Then 8-d=distance the student walks

Time spent walking=(8-d)/5

Time spent jogging=d/9

Now we are told that the time spent walking plus the time spent jogging = 1 hour. So:

(8-d)/5+d/9=1  multiply each term by 45 (LCM)
9(8-d)+5d=45  get rid of parens

72-9d+5d=45  subtract 72 from both sides

72-72-9d+5d=45-72  collect like terms

-4d=-27  divide both sides by -4

d=6.75 mi-----------------------distance student jogs

8-6.75=1.25 mi--------------------distance student walks

CK

6.75/9+1.25/5=1

0.75+0.25=1
1=1


Hope this helps-----ptaylor