Question 133899

Distance(d) equals Rate(r) times Time(t) or d=rt;  r=d/t and t=d/r

Let r= her rate going to NY
Then r-10=her rate returning in the bad wx
Now we are told that her time coming back from NY is 2 hr longer than her time going, well:
Her time going to NY=400/r
Her time coming back=(400/(r-10)
Now if we subtract her time going from her time coming back we should have 2 hr.
So our equation to solve is:
 400/(r-10) - 400/r=2  multiply each term by r(r-10)

400r-400r+4000=2r(r-10) simplify and divide each term by 2

2000=r^2-10r  subtract 2000 from each side

r^2-10r-2000=2000-2000  collect like terms

r^2-10r-2000=0  quadratic in standard form and it can be factored

(r-50)(r+40)=0
Neglect the negative value for r.  Rates in this problem are positive
 r=50 mph---------------------------------- her rate going to NY

r-10=50-10=40 mph-----------------------her rate coming back from NY in the wx

CK

400/40 - 400/50=2
10-8=2
2=2

Hope this helps---ptaylor