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Math Help - Having Trouble

  1. #1
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    Having Trouble

    Hi all,

    I am just new to this site and it looks like such an awesome resource.

    Can anyone help me with this???

    If x + y = 1 and x + y = 19, find the value of x + y


    Last edited by Jameson; April 15th 2009 at 05:31 PM.
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  2. #2
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    Quote Originally Posted by Joel View Post
    Hi all,

    I am just new to this site and it looks like such an awesome resource.

    Can anyone help me with this???

    If x + y = 1 and x + y = 19, find the value of x + y
    \left\{\begin{array}{lclcr}<br />
x&+&y&=&1\\<br />
x^3&+&y^3&=&19<br />
\end{array}\right.

    \Rightarrow x^3+(1-x)^3=19,

    and after expanding and simplifying,

    3x^2-3x-18=0

    \Rightarrow3(x-3)(x+2)=0.

    You should be able to finish.
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  3. #3
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    Hello, Joel!

    Welcome aboard!

    I have a back-door approach . . .


    If . \begin{array}{cccc}x + y &=&  1 & {\color{blue}[1]} \\ x^3 + y^3 &=& 19 & {\color{blue}[2]} \end{array} . find the value of x^2+y^2.

    Cube [1]: . (x+y)^3 \:=\:1^3\quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:1

    \text{We have: }\;\underbrace{(x^3 + y^3)}_{\text{This is 19}} + \;3xy\underbrace{(x+y)}_{\text{This is 1}} \:=\:1 \quad\Rightarrow\quad 19 \;+ \;3xy \:=\:1 \quad\Rightarrow\quad xy \:=\:-6


    Square [1]: . (x+y)^2 \:=\:1^2 \quad\Rightarrow\quad x^2+2xy + y^2 \:=\:1

    \text{We have: }\;x^2 + y^2 + 2(xy) \:=\:1
    . . . . . . . . . . . . . . \uparrow
    . . . . . . . . . . . . ^{\text{This is }\text{-}6}

    Therefore: . x^2+y^2 - 12 \:=\:1 \quad\Rightarrow\quad \boxed{x^2+y^2 \:=\:13}



    This method is guarenteed to impress/surprise/terrify your teacher.
    .
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Joel!

    Welcome aboard!

    I have a back-door approach . . .
    I like your solution. I wish I had thought about the problem a bit more before going down the obvious path.
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