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

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The discussion focuses on solving the equations involving variables x, y, and z defined by three reciprocal relationships. The equations lead to a series of transformations and substitutions, ultimately resulting in expressions for x, y, and z in terms of each other. After a series of calculations, the values are determined as y = 23/6, x = 23/10, and z = 23/2. The solution showcases the methodical approach to finding the values while ensuring that none of the variables are zero. The final results are presented clearly, confirming the successful resolution of the equations.
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find solutions of $x,y,z$

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

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

$\dfrac {1}{z}+\dfrac {1}{x+y}=\dfrac {1}{4}-----(3)$
 
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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}}
 
http://www.mymathforum.com/viewtopic.php?t=36640&p=151646
 
Opalg said:
http://www.mymathforum.com/viewtopic.php?t=36640&p=151646
Nicely done!
very good solution there :)
 
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