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cutiecomittee123

  • one year ago

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

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  1. anonymous
    • one year ago
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    All coordinates in quadrant 1 are positive

  2. cutiecomittee123
    • one year ago
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    good to know

  3. cutiecomittee123
    • one year ago
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    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.

  4. anonymous
    • one year ago
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    \(x^2-y^2=(x-y)(x+y)=25\) now use the second equation x+y=25 hope this is enough, :)

  5. cutiecomittee123
    • one year ago
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    thanks for that but what do I do with that information?

  6. anonymous
    • one year ago
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    next step would be: x-y=1 x+y=25

  7. anonymous
    • one year ago
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    then 2x=26 x=13 y=12

  8. cutiecomittee123
    • one year ago
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    so you basically took the square root of the top equation

  9. anonymous
    • one year ago
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    i used the fact that a^2-b^2=(a-b)(a+b)

  10. cutiecomittee123
    • one year ago
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    how did 25 go to 1?

  11. anonymous
    • one year ago
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    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

  12. anonymous
    • one year ago
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    got it?

  13. cutiecomittee123
    • one year ago
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    okay

  14. perl
    • one year ago
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    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.

  15. cutiecomittee123
    • one year ago
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    Yeah I figured it out though:) (13,12)

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