SOLUTION: When Acme Airlines changed to planes that could fly 100 km/h faster than its old ones, the time of its 2800 km Dallas-Seattle flight was reduced by 30 minutes. Find the speed of th

Question 69074This question is from textbook Algebra and Trigonometry: Structure and Method
: When Acme Airlines changed to planes that could fly 100 km/h faster than its old ones, the time of its 2800 km Dallas-Seattle flight was reduced by 30 minutes. Find the speed of the new planes. This question is from textbook Algebra and Trigonometry: Structure and Method

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x= speed of old ones
x+100= speed of new ones
distance (d)=rate(r) times time (t) or d=rt and t=d/r
2800/x= time it took old planes to make the trip
2800/(x+100)= time it takes the new planes to make the trip
Now we are told that the new planes make the trip 30 min (0.5 hr) faster than the old planes. So our equation to solve is:

2800/x=(2800/(x+100))+0.5 multiply both sides by x(x+100)
2800(x+100)=2800x+0.5x(x+100) clear parens
2800x+280000=2800x+0.5x^2+50x subtract 2800x and also 280000 from both sides
2800x-2800x+280000-280000=2800x-2800x-280000+0.5x^2+50x collect like terms
0=0.5x^2+50x-280000 or multiply both sides by 2
x^2+100x-560000=0 Quadratic in standard form
This quadratic can be factored:
(x+800)(x-700)=0
discount negative value for speed
So
x=700 km/hr ------------------------------speed of old planes
x+100=800 km/hr speed of new planes
Ck
2800/x=4 hrs
2800/(x+100)=3.5 hrs-----30 min less ---checks

Hope this helps----ptaylor

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