Question 151934: Two planes leave simultaneously from the same airport, one flying due north and the other due east. The northbound plane is flying 50 mph faster than the east bound plane. After 3 hours, the planes are 2440 miles apart. Find the speed of each plane.
3(x+50 +x)=2440
x+50 +x =813.33
2x +50= 813.33
2x=763.33
x = 381.67 northbound plane: 431.33
eastbound plane: 381.33
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Two planes leave simultaneously from the same airport, one flying due north and the other due east. The northbound plane is flying 50 mph faster than the east bound plane. After 3 hours, the planes are 2440 miles apart. Find the speed of each plane.
:
I think you got off on the wrong track here, see if this makes sense.
:
Let x = speed of the eastbound plane
then
(x+50) = speed of the northbound
:
Distance
3x = dist flown by the eastbound plane
3(x+50) or (3x+150) = distance of the northbound
:
Solving it as a right triangle
(3x)^2 + (3x+150)^2 = 2440^2
;
9x^2 + 9x^2 + 900x + 22500 = 5953600
:
18x^2 + 900x + 22500 - 5953600 = 0
:
18x^2 + 900x - 5931100 = 0
:
Solve this quadratic equation using the quadratic formula:
The positive solution: x = 549.57; speed of the eastbound plane
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