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Math Help - stuck on this one - first post

  1. #1
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    Smile stuck on this one - first post

    just joined so don't let me down

    Can't sort this one:

    y = 3x
    2y(squared) - xy = 15

    Tried substituting y = 3x and x = y/3. I only get half of the right answer using x = y/3 as y = -3 and x= -1. Can't seem to get the full right answer.

    Any ideas

    Becca.
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  2. #2
    Senior Member OReilly's Avatar
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    Quote Originally Posted by becca999 View Post
    just joined so don't let me down

    Can't sort this one:

    y = 3x
    2y(squared) - xy = 15

    Tried substituting y = 3x and x = y/3. I only get half of the right answer using x = y/3 as y = -3 and x= -1. Can't seem to get the full right answer.

    Any ideas

    Becca.
    \begin{array}{l}<br />
 y = 3x \\ <br />
 2y^2  - xy = 15 \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 2(3x)^2  - x(3x) = 15 \\ <br />
 2(9x^2 ) - 3x^2  = 15 \\ <br />
 18x^2  - 3x^2  = 15 \\ <br />
 15x^2  = 15 \\ <br />
 x^2  = 1 \\ <br />
 x = 1 \\ <br />
  \\ <br />
 y = 3 \\ <br />
 \end{array}<br />
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by OReilly View Post
    \begin{array}{l}<br />
 y = 3x \\ <br />
 2y^2  - xy = 15 \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 2(3x)^2  - x(3x) = 15 \\ <br />
 2(9x^2 ) - 3x^2  = 15 \\ <br />
 18x^2  - 3x^2  = 15 \\ <br />
 15x^2  = 15 \\ <br />
 x^2  = 1 \\ <br />
 x = 1 \\ <br />
  \\ <br />
 y = 3 \\ <br />
 \end{array}<br />
    Of course, x^2 = 1 also has the solution x = -1, but substitution shows that this x value does not provide a new intersection point.

    -Dan
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  4. #4
    Senior Member OReilly's Avatar
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    Quote Originally Posted by topsquark View Post
    Of course, x^2 = 1 also has the solution x = -1, but substitution shows that this x value does not provide a new intersection point.

    -Dan
    Both solutions (1,3) and (-1,-3) are correct and they providing intersection points.

    Graph is showing that.
    Attached Thumbnails Attached Thumbnails stuck on this one - first post-image1.gif  
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  5. #5
    Super Member
    earboth's Avatar
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    Quote Originally Posted by topsquark View Post
    Of course, x^2 = 1 also has the solution x = -1, but substitution shows that this x value does not provide a new intersection point.
    -Dan
    Hi,

    actually you calculate the intersection between a hyperbola and a straight line. And of course there are two intersection points.
    I've attached a diagram to show you where these points are:
    Attached Thumbnails Attached Thumbnails stuck on this one - first post-hyp_gerade1.gif  
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  6. #6
    Forum Admin topsquark's Avatar
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    Pffl on me. The second equation is 2y^2 - xy = 15. When I did the problem in my head for x = -1 I was using "+xy" Ah well. You win some, you lose some!

    -Dan
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