# Thread: [SOLVED] New Math Problem

1. ## [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 $y+\sqrt{x^2-2xz+z^2}$ = ??

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

3. Hello again, wizard654zzz!

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

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

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

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

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

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

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

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

4. ## Thank you

Mr.Soroban Thank you for your Help