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Math Help - equation of a line

  1. #1
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    equation of a line

    Could you tell me the most effective way to quickly solve this question? Thank you.



    In the xy plane, does the line with equation y=3x+2 contain the point (r,s)?

    1) (3r+2-s)(4r+9-s)=0

    2) (4r-6-s)(3r+2-s)=0


    A) Statement 1) alone is sufficient
    B) Statement 2) alone is sufficient
    C) Statements 1) and 2) together are sufficient
    D) Each statement alone is sufficient
    E) Statement 1) and 2) together are not sufficient
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  2. #2
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    Quote Originally Posted by simone View Post
    Could you tell me the most effective way to quickly solve this question? Thank you.



    In the xy plane, does the line with equation y=3x+2 contain the point (r,s)?

    1) (3r+2-s)(4r+9-s)=0

    2) (4r-6-s)(3r+2-s)=0


    A) Statement 1) alone is sufficient
    B) Statement 2) alone is sufficient
    C) Statements 1) and 2) together are sufficient
    D) Each statement alone is sufficient
    E) Statement 1) and 2) together are not sufficient
    A) Statement 1 introduces one equation in two variables. It cannot be solved to provide a specific set of values r and s.

    B) Same comment.

    C) They are probably sufficient. You would have to solve the two equations simultaneously (and I'm too lazy right now to do that.) But it should be possible, barring something bizarre happening.

    D) See A) and B)

    E) This depends on if you can get a point out of the simultaneous equations.

    I'd say the answer is overwhelmingly likely to be C). Try it and find out.

    -Dan
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  3. #3
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    If simone has written the equations correctly, then C is the correct answer,
    If \left( {r,s} \right)\mbox{ on } y = 3x + 2\;  \Rightarrow \; s = 3r + 2\quad or\quad 3r + 2 - s = 0.
    Note that \left( {3r + 2 - s} \right) is factor of both equations 1) & 2).
    Either \left( {3r + 2 - s} \right) = 0 or \left( {4r + 9 - s} \right) = 0
    And either \left( {3r + 2 - s} \right) = 0 or \left( {4r - 6 - s} \right) = 0
    BUT it is impossible that together we can have \left( {4r - 6 - s} \right) = 0 and \left( {4r + 9 - s} \right) = 0.
    Hence \left( {3r + 2 - s} \right) = 0.
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  4. #4
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    thank you Plato, very elegant and effective instead of comparing the 2 equations as a whole.
    Do you mind taking a look at the other question I posted? thank you.
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