Solving $x,y,z$: Equations (1), (2), (3)

[Albert1]

find solutions of $x,y,z$

$\dfrac {1}{x}+\dfrac {1}{y+z}=\dfrac {1}{2}-----(1)$

$\dfrac {1}{y}+\dfrac {1}{z+x}=\dfrac {1}{3}-----(2)$

$\dfrac {1}{z}+\dfrac {1}{x+y}=\dfrac {1}{4}-----(3)$

 

[soroban]

Hello, Albert!

I have a very looong solution.

Find solutions of $x,y,z$

.\begin{array}{cccc}\frac {1}{x}+\frac {1}{y+z} &amp;=&amp; \frac {1}{2} &amp; [1[ \\<br /> <br /> \frac {1}{y}+\frac {1}{z+x}&amp;=&amp;\frac {1}{3} &amp; [2] \\<br /> <br /> \frac {1}{z}+\frac {1}{x+y}&amp;=&amp;\frac {1}{4} &amp; (3) \end{array}

Note that: .x,y,z\, \ne\,0.

\begin{array}{cccccccccc}[1]\!:\;2y + 2x + 2x \:=\:x(y+z) &amp;\Rightarrow&amp; x + y + z \:=\: \frac{xy+xz}{2} &amp; [4] \\<br /> [2]\!:\; 3x + 3x + 3y \:=\:y(x+z) &amp;\Rightarrow &amp; x+y+z \:=\:\frac{xy+yz}{3} &amp; [5] \\<br /> [3]\!:\; 4x+3y+4z \:=\:z(x+y) &amp; \Rightarrow &amp; x+y+z \:=\:\frac{xz+yz}{4} &amp; [6] \end{array}

\text{From }[4],[5],[6]\!:\;\underbrace{\frac{xy+xz}{2}}_{[7]} \:=\:\underbrace{\frac{xy + yz}{3}}_{[8]} \:=\:\underbrace{\frac{xz+yz}{4}}_{[9]}

[7]=[8]\!:\;3xy + 3xz \:=\:2xy + 2yz \;\;\;\Rightarrow\;\;\; x \:=\:\frac{2yz}{y+3x}\;\;[10]

[8] = [9]\!:\;4xy + 4yz \:=\:3xz + 3yz \;\;\;\Rightarrow\;\;\;x \:=\:\frac{yz}{3x-4y}\;\;[11]

[10]=[11]\!:\;\frac{2yz}{y+3x} \:=\:\frac{yz}{3z-4y} \quad\Rightarrow\quad z \:=\:3y\;\;[12]

Substitute into [10]: .x \:=\:\frac{2y(3y)}{y+3(3y)} \quad\Rightarrow\quad x \:=\:\tfrac{3}{5}y\;\;[13]

Substitute [12] and [13] into [4]:
.. 2y + 2(3y) + 2\left(\tfrac{3}{5}y\right) \;=\; \left(\tfrac{3}{5}y\right)y + \left(\tfrac{3}{5}y\right)(3y)
. . \tfrac{46}{5}y \:=\:\tfrac{12}{5}y^2 \quad\Rightarrow\quad 6y^2\:=\:23y \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{23}{6}}

Substitute onto [13}: .x \:=\:\tfrac{3}{5}\left(\tfrac{23}{6}\right) \quad\Rightarrow\quad \boxed{x \:=\:\frac{23}{10}}

Substitute into [12]: .z \:=\:3\left(\tfrac{23}{6}\right) \quad\Rightarrow\quad \boxed{z \:=\:\frac{23}{2}}

 

[Opalg]

http://www.mymathforum.com/viewtopic.php?t=36640&p=151646

 

[Albert1]

http://www.mymathforum.com/viewtopic.php?t=36640&p=151646

Nicely done!
very good solution there :)