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Math Help - Solving Quadratic Systems of Equations

  1. #1
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    Solving Quadratic Systems of Equations

    I have this system:

    x^2+2y^2-4x+6y-5=0
    -x+y-4=0

    And I have to find the intersecting points of the two equations by using substitution and only substitution.

    I solved the bottom equation for y getting y=x+4. From there, this is how I progressed:

    x^2+2(x+4)^2-4x+6(x+4)-5=0
    x^2+2(x^2+8x+16)-4x+6x+24-5=0
    x^2+2x^2+16x+32-4x+6x+24-5=0

    Combining like terms:

    3x^2+18x+51=0

    Pulling out a 3 gives me:

    3(x^2+6x+17)=0

    From here, I'm not sure what to do. Am I on the right track or am I doing this completely wrong? If the latter, please explain to me where I made an error so it doesn't happen again.

    Thank you.
    Last edited by mathgeek777; May 10th 2008 at 03:09 PM. Reason: Changed the two typos and replaced the - with a zero (no clue how that got in there...)
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  2. #2
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    Hello,

    Quote Originally Posted by mathgeek777 View Post
    I have this system:

    x^2+2y^2-4x+6y-5=0
    -x+y-4=0

    And I have to find the intersecting points of the two equations by using substitution and only substitution.

    I solved the bottom equation for y getting y=x-4. From there, this is how I progressed:

    x^2+2(x+4)^2-4x+6(x{\color{red}+}4)-5=-
    x^2+2(x^2+8x+16)-4x+6x+24-5=0
    x^2+2x^2+16x+32-4x+6x+24-5=0

    Combining like terms:

    3x^2+18x+51=0

    Pulling out a 3 gives me:

    3(x^2+6x+17)=0

    From here, I'm not sure what to do. Am I on the right track or am I doing this completely wrong? If the latter, please explain to me where I made an error so it doesn't happen again.

    Thank you.
    Little typo in red, not changing what follows.

    Well, I don't see any mistake that far.

    I've substituted y, and I also get complex solutions :-)
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  3. #3
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    Since I couldn't find any obvious real solution, I put it down as no real solutions, checked the back of the book, and it is indeed no real solutions.

    Thanks for the help.
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