What are the coordinates to the solution that lays in quadrant 1? x^2-y^2=25 x+y=25

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What are the coordinates to the solution that lays in quadrant 1? x^2-y^2=25 x+y=25

Mathematics
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All coordinates in quadrant 1 are positive
good to know
I tried squaring the second equation and then adding both equations together, which essentially canceled out the y^2's so then I solved for x but came up with an odd number.

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Other answers:

\(x^2-y^2=(x-y)(x+y)=25\) now use the second equation x+y=25 hope this is enough, :)
thanks for that but what do I do with that information?
next step would be: x-y=1 x+y=25
then 2x=26 x=13 y=12
so you basically took the square root of the top equation
i used the fact that a^2-b^2=(a-b)(a+b)
how did 25 go to 1?
if in (x-y)(x+y)=25 you substitute the value of (x+y) from second equation, you get (x-y)25=25 or x-y=1
got it?
okay
Nice solution. Another way to do it is solve for x in the second equation and substitute it in the first equation. That gives you a quadratic and more work involved to get the answer.
Yeah I figured it out though:) (13,12)

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