Results 1 to 4 of 4

Math Help - [SOLVED] New Math Problem

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    16

    [SOLVED] New Math Problem

    1. \sqrt{3+2\sqrt{2}}-\sqrt{2}= ?
    2. y+\frac{2}{x+z}=4
    5y+\frac{18}{2x+y+z}=18
    \frac{8}{x+z}-\frac{6}{2x+y+z}=3
    Then the value of y+\sqrt{x^2-2xz+z^2} = ??
    Last edited by Grandad; February 20th 2010 at 05:57 AM. Reason: Fixed LaTeX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    629
    Hello, wizard654zzz!

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

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

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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    629
    Hello again, wizard654zzz!

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


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


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

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

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


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

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


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


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

    Substitute: . 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]: . 1 + z \:=\:2 \quad\Rightarrow\quad \boxed{z \:=\:1}


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

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2009
    Posts
    16

    Thank you

    Mr.Soroban Thank you for your Help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need math problem solved to score a date!
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 14th 2009, 06:10 PM
  2. Replies: 1
    Last Post: April 25th 2009, 02:41 PM
  3. [SOLVED] Little Math Problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 21st 2008, 12:14 PM
  4. [SOLVED] 4th grade math problem
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 8th 2007, 07:22 PM
  5. [SOLVED] math word problem - need help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 30th 2005, 09:36 AM

Search Tags


/mathhelpforum @mathhelpforum