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Math Help - Line intersection

  1. #1
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    Line intersection

    This is the question:

    4x + 3y = 24, x^2 + y^2 = 25

    Find where the line intersects the circle.

    I tried solving for y, the substituting it into the circle equation, but I don't get the correct answer. It ends up becomes a messy quadratic equation.

    Can someone help me out? Thanks.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    This is the question:

    4x + 3y = 24, x^2 + y^2 = 25

    Find where the line intersects the circle.

    I tried solving for y, the substituting it into the circle equation, but I don't get the correct answer. It ends up becomes a messy quadratic equation.

    Can someone help me out? Thanks.
    Please show all the work you have done.
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  3. #3
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    3y = -4x + 24
    y = (-4/3)x + 8

    Sub this into the circle eqn:

    x^2 + \frac{-4}{3}x^2 + \frac{64}{3}x + 64 = 25

    -\frac{1}{3}x^2 + \frac{64}{3}x = -39

    x^2 - 64x = 117

    x^2 - 64x - 117 = 0

    I don't know where to go from here. The quadratic formula seems to make a mess.
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  4. #4
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    Quote Originally Posted by Glitch View Post
    3y = -4x + 24
    y = (-4/3)x + 8

    Sub this into the circle eqn:

    x^2 + \frac{-4}{3}x^2 + \frac{64}{3}x + 64 = 25 Mr F says: This is wrong. And who knows why since you have not shown all your work (as I requested).

    [snip]
    Actually I do know why, but I want to see every step that lead you to this. (And when you start typing those steps you will probably find the mistake yourself, which is another reason why I asked you to show all your work).

    BY the way: http://www.wolframalpha.com/input/?i...B+y%5E2+%3D+25
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  5. #5
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    Sorry, I this is the bit I'm missing:

    x^2 + y^2 = 25 (circle eqn)

    x^2 + ((-4/3)^2 + 8)^2 = 25 (substituting into circle eqn)

    Am I using the wrong expansion method?
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  6. #6
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    Quote Originally Posted by Glitch View Post
    Sorry, I this is the bit I'm missing:

    x^2 + y^2 = 25 (circle eqn)

    x^2 + ((-4/3)^2 + 8)^2 = 25 (substituting into circle eqn)

    Am I using the wrong expansion method?
    So you went straight from x^2 + ((-4/3)^2 + 8)^2 = 25 to x^2 + \frac{-4}{3} x^2 + \frac{64}{3} + 64 = 25, did you?

    For the last time, please show ALL your work:
    Quote Originally Posted by Mr F, who said (bold face added)
    [snip]but I want to see every step that lead you to this.
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  7. #7
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    Yes I did, using this equation:

    (a + b)^2 = a^2 + 2ab + b^2
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  8. #8
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    Quote Originally Posted by Glitch View Post
    Yes I did, using this equation:

    (a + b)^2 = a^2 + 2ab + b^2
    OK, for your question what is a and what is b?

    Now, please show me exactly how you calculated a^2 + 2ab (b^2 = 64 was correct). Show every detail. Pretend you are explaining it to a 14 year old student.
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  9. #9
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    I squared -\frac{4}{3}x to get \frac{16}{9}x^2, which is a^2.

    For 2ab, I did 2*(\frac{-4}{3}* 8) = \frac{64}{3}x

    Hmm, looks like I squared it wrong before. I think I also expanded it using the wrong equation, since (-4/3 + 8)^2 is equivallent to (8 - 4/3)^2, which expands using the (a - b)^2 equation.

    I think I've got the correct answers now.
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  10. #10
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    Quote Originally Posted by Glitch View Post
    I squared -\frac{4}{3}x to get \frac{16}{9}x^2, which is a^2.

    For 2ab, I did 2*(\frac{-4}{3}* 8) = \frac{64}{3}x Mr F says: This should be negative.

    Hmm, looks like I squared it wrong before. I think I also expanded it using the wrong equation, since (-4/3 + 8)^2 is equivallent to (8 - 4/3)^2, which expands using the (a - b)^2 equation.

    I think I've got the correct answers now.
    ..
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