SOLUTION: I can get about half way on this need a little assistance completing. Problem is..... Two airplanes are leaving O'Hare at the same time, one bound for New York and the other for

Question 111213: I can get about half way on this need a little assistance completing.
Problem is.....
Two airplanes are leaving O'Hare at the same time, one bound for New York and the other for California. The one heading to Califorina as a tail wind and is thus going 25 MPH faster than the one headed to New York. If after one hour they are 575 miles apart, what are their speeds?
What I have so far.....
x^2+y^2=575^2,x^2(x^2+25)^2=575^2,x^2+x^2=50x+625=330,625
2x^2+50x-330,000=0, But i can not get any further and I am not sure if I did what I have so far right. Can you help please??? Thank you SO MUCH!!!
Found 2 solutions by solver91311, checkley71:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I think your actual first step was to assign x to represent the speed of the eastbound aircraft and y to represent the speed of the one going to California. Ok, as far as that goes, but from there on, you went completely off in the wrong direction.
:
The basic formula you need for distance-rate-time questions like this is . let's use these three variables instead of x and y because it is easier to keep things straight in our mind. Let's call the plane going to New York number 1 and the other one number 2.
:
So, to start, we can write two equations:
:
, and

:
But we know that both planes have been flying for 1 hour, so we can now write:
:
, and

:
We also know that aircraft 2 was traveling 25 mph faster than aircraft 1, so we can write:
:
, and

:
Finally, we know that the SUM of the distances the two aircraft traveled was 575 miles, or:
:
, but we can substitute the right side expressions to obtain:
:

:
Now all we have to do is solve:
:

:
So the New York flight is traveling 275 mph, and the California flight is traveling 25 mph faster than that or 300 mph. After one hour, the New York flight will have gone 275 miles, and the California flight, 300 miles: 275 plus 300 is 575, so the answer checks.
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
QUESTION -- ARE THESE PLANES FLYING DUE WEST & DUE EAST?
IF NOT -- WHAT IS THE ANGLE BETWEEN THEIR FLIGHT PATHS?
THIS INFORMATION WOULD HELP DETERMINE HOE TO SOLVE THIS PROBLEM.
IF THEY ARE ON OPOSITE (180 DEGREE) FLIGHT PATHS THEN:
X+(X+25)=575
2X+25=575
2X=575-25
2X=550
X=550/2
X=275 MPH FOR THE SLOWER PLANE.
275+25=300MPH FOR THE FASTER PLANE.
PROOF
275+300=575
575=575
RELATED QUESTIONS
The fable about the tortoise an the hare? what? lol (math)? hello this is a math... (answered by rothauserc,robertb)
The hare moved three times as fast as the tortoise of Aesop's Fables. After 4 minutes,... (answered by Alan3354)
I am needing help with this problem. I am totally confused on how to solve this problem. (answered by checkley77,Alan3354,ptaylor)
I need some assistance on how to factor. This problem is a little more complicated for me (answered by stanbon)
I have a question regarding completing the square. There is no graphing involved, I... (answered by richard1234)
I have worked this problem several times and come up mwith different answers i think i am (answered by checkley77)
Please solve this following problem. A tortoise that can walk three miles in two hours (answered by josmiceli)
Imagine I wrap sheet metal part way around a tube leaving two parallel sides connected by (answered by ikleyn,BM22222)
I have a question about how to set up this problem. We are going back from the start of... (answered by rothauserc)