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Thread: [SOLVED] New Math Problem

  1. #1
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    [SOLVED] New Math Problem

    1. $\displaystyle \sqrt{3+2\sqrt{2}}-\sqrt{2}=$ ?
    2. $\displaystyle y+\frac{2}{x+z}=4$
    $\displaystyle 5y+\frac{18}{2x+y+z}=18$
    $\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3$
    Then the value of $\displaystyle y+\sqrt{x^2-2xz+z^2}$ = ??
    Last edited by Grandad; Feb 20th 2010 at 05:57 AM. Reason: Fixed LaTeX
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  2. #2
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    Hello, wizard654zzz!

    (1) Simplify: .$\displaystyle \sqrt{3+2\sqrt{2}}\:-\:\sqrt{2}$

    I recognized that: .$\displaystyle 3 + 2\sqrt{2} \;=\;\left(1+\sqrt{2}\right)^2$

    So we have: .$\displaystyle \sqrt{(1+\sqrt{2})^2} \,-\,\sqrt{2} \;\;=\;\;(1 + \sqrt{2}) - \sqrt{2} \;\;=\;\;1$

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  3. #3
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    Hello again, wizard654zzz!

    I've solved the second one,
    but I hope someone finds a shorter solution.


    $\displaystyle 2)\;\text{ Given: }\;\begin{Bmatrix}y+\dfrac{2}{x+z} &=& 4 & {\color{blue}[1]} \\ \\[-3mm]
    5y+\dfrac{18}{2x+y+z} &=& 18 & {\color{blue}[2]} \\ \\[-3mm]
    \dfrac{8}{x+z}-\dfrac{6}{2x+y+z}&=& 3 & {\color{blue}[3]} \end{Bmatrix}$


    Find: .$\displaystyle y+\sqrt{x^2-2xz+z^2}\;\;{\color{blue}=\;\;y + \sqrt{(x-z)^2} \;\;=\;\;y + x - z}$

    $\displaystyle \begin{array}{ccccc}
    3\times {\color{blue}[3]}: & \dfrac{24}{x+z} - \dfrac{18}{2x+y+z} &=& 9 \\ \\[-3mm]
    \text{Add {\color{blue}[2]}:} & 5y + \dfrac{18}{2x+y+z} &=& 18 \end{array}$

    We have: .$\displaystyle 5y + \frac{24}{x+z} \;=\;27\;\;[4]$


    $\displaystyle \begin{array}{ccccc}
    12\times {\color{blue}[1]}: & 12y + \dfrac{24}{x+z} &=& 48 \\ \\[-3mm]
    (\text{-}1)\times {\color{blue}[4]}: & -5y - \dfrac{24}{x+z} &=& -27 \end{array}$

    We have: .$\displaystyle 7y \:=\:21 \quad\Rightarrow\quad \boxed{ y \:=\:3}$


    Substitute into [1]: .$\displaystyle 3 + \frac{2}{x+2} \:=\:4 \quad\Rightarrow\quad \frac{2}{x+z} \:=\:1 \quad\Rightarrow\quad x + z \:=\:2\;\;{\color{blue}[4]}$


    $\displaystyle {\color{blue}[2]}:\;\;5y + \frac{18}{2x+y+z} \;=\;18 \quad\Rightarrow\quad 5y + \frac{18}{x+y+(x+z)} \:=\:18$

    Substitute: .$\displaystyle 5(3) + \frac{18}{x+3 + 2} \:=\:18 \quad\Rightarrow\quad \frac{18}{x+5} \:=\:3 \quad\Rightarrow\quad\boxed{ x \:=\:1}$


    Substitute into [4]: .$\displaystyle 1 + z \:=\:2 \quad\Rightarrow\quad \boxed{z \:=\:1}$


    Therefore: .$\displaystyle y + x - z \;=\;3 + 1 - 1 \;=\;\bf{\color{red}3}$

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  4. #4
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    Thank you

    Mr.Soroban Thank you for your Help
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