# find solution

#### Albert

##### Well-known member
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

##### Well-known member
Hello, Albert!

I have a very looong solution.

Find solutions of $x,y,z$

.$$\begin{array}{cccc}\frac {1}{x}+\frac {1}{y+z} &=& \frac {1}{2} & [1[ \\ \frac {1}{y}+\frac {1}{z+x}&=&\frac {1}{3} & [2] \\ \frac {1}{z}+\frac {1}{x+y}&=&\frac {1}{4} & (3) \end{array}$$

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

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

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

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

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

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

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

Staff member

#### Albert

##### Well-known member
Nicely done!
very good solution there