Question 112878
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r

Let r=original rate (speed) of the plane
Then r+20=the increased speed of the plane
Time that elapsed while travelling at the original speed=480/r
Time that elapsed while travelling at the increased speed=(840-480)/(r+20)

Now we are told that the time that elapsed travelling at the original speed plus the time that elapsed travelling at the increased speed equals 5 hours.

So our equation to solve is:

(480/r)+360/(r+20)=5  multiply each term by r(r+20)

480(r+20)+360r=5r(r+20)
480r+9600+360r=5r^2+100r subtract 5r^2 and also 100r from both sides

480r+9600+360r-5r^2-100r=0  collect like terms
-5r^2+740r+9600=0  divide each term by -5

r^2-148r-1920=0------------------quadratic in standard form and it can be factored

(r-160)(r+12)=0

r=160 mph----original rate (speed) of the plane

and
r=-12-----------------------------neglect negative speed

CK

(480/160)+(360/180)=5
3+2=5
5=5


Hope this helps ----ptaylor