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What's wrong here?

  1. Sep 8, 2004 #1

    Alkatran

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    I took a test in math to determine where I was etc etc...
    anyways, the following question came up:

    How many answers are there to the following system of equations?
    x^2 + 5y = 30
    x^2 + (y-3)^2 = 9

    a) 0
    b) 1
    c) 2 (my answer)
    d) 3 (correct answer)


    The first thing I noticed was that x was the same thing in both cases, the answer would not be changed because of it. As such, x could be almost any value (infinite answers). So obviously, they wanted how many solutions for Y.

    My work (in my head) was:
    x^2 + 5y = 30
    x^2 + (y-3)^2 = 9

    *subtract equations from each other
    5y - (y-3)^2 = 21

    *break (y-3)^2
    5y - (y^2 - 6y + 9) = 21
    5y - y^2 + 6y - 9 = 21
    11y - y^2 = 30

    *place ^2 in positive and use ax^2 + bx + c format
    y^2 - 11y + 30 = 0

    *check for 2 or 1 answers by checking if the sqr() portion of quadratic formula is non zero
    sqr(121 - 4*30*1) = sqr(1) = 1

    since 1 <> 0, and the answer will be +- 1, there are two answers.

    Where is my mistake? :frown:
     
    Last edited: Sep 8, 2004
  2. jcsd
  3. Sep 8, 2004 #2
    [tex]\begin{align*}\\y=\frac{30-x^2}{5}\\
    x^2+[\frac{30-x^2-15}{5}]^2=9\\
    x^2(x^2-5)=0\\
    x=0\ or \x=\pm\sqrt{5}\end{align*}[/tex]

    When you substitute y=5 which you have found into the first equation, you find 2 solution for x and for y=6, you find x=0.
     
  4. Sep 8, 2004 #3

    JasonRox

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    I was thinking...
    - 2nd Degree Equation
    - 1st Degree Equation

    ...3 solutions.

    I can be wrong with the 1st degree equation.
     
  5. Sep 8, 2004 #4

    Alkatran

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    Oh, I get it. I see where I made the mistake... tricky little son of a...
     
  6. Sep 8, 2004 #5

    Tide

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    One is a parabola and the other is a circle. A quick sketch of the graphs will tell you how many (real) solutions there are! :-)
     
  7. Sep 8, 2004 #6

    Alkatran

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    I see, x is limited by y's limitation... it's a weird concept at first.
     
  8. Sep 8, 2004 #7

    JasonRox

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    I can the same answers as the guy above.

    I did the following:

    Take [tex]x^2 + (y-3)^2 = 9[/tex], and you get [tex]x^2 = y^2 + 6y[/tex].

    Take the x^2 and put it in the other equation, and you get...

    [tex]y^2 - 6y + 5y = 30[/tex] to... [tex](y - 6)(y - 5) = 0[/tex]

    Now that you know y... plug in the numbers for x, and there you go.
     
  9. Sep 8, 2004 #8

    JasonRox

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    I was wrong about the 1st degree. :(

    I was doubting it completely because the answer was 3, and I just took the easy way out.

    Look at the bright side, you warmed up your brain today.
     
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