Solving Hard System

Solve the system

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

$\dfrac{1}{y}+\dfrac{1}{x+z}=\dfrac{1}{3}$

$\dfrac{1}{z}+\dfrac{1}{x+y}=\dfrac{1}{4}$

Method I (Long and tedious but workable solution):

If we let $x+y+z=m$, we have

$\dfrac{1}{x}+\dfrac{1}{m-x}=\dfrac{1}{2}$

$\dfrac{m}{x(m-x)}=\dfrac{1}{2}$

$2m=mx-x^2$

$x^2=m(x-2)$

$m=\dfrac{x^2}{x-2}---(1)$

and similarly,

$m=\dfrac{y^2}{y-3}---(2)$

$m=\dfrac{z^2}{z-4}---(3)$

To relate $x$ and $y$, we equate (1) and (2):

$\dfrac{x^2}{x-2}=\dfrac{y^2}{y-3}$

$\implies y=\dfrac{x^2\pm x\sqrt{x^2-12(x-2)}}{2(x-2)}$

To relate $x$ and $z$, we equate (1) and (3):

$\dfrac{x^2}{x-2}=\dfrac{z^2}{z-4}$

$\implies z=\dfrac{x^2\pm x\sqrt{x^2-16(x-2)}}{2(x-2)}$

Next, we add these all up so that we can make another equation based on the fact $x+y+z=m=\dfrac{x^2}{x-2}$:

$x+\dfrac{x^2\pm x\sqrt{x^2-12(x-2)}}{2(x-2)}+\dfrac{x^2\pm x\sqrt{x^2-16(x-2)}}{2(x-2)}=\dfrac{x^2}{x-2}
$

Clearing fractions by multiplying both sides by the quantity $2(x-2)$ we get:

$x(2(x-2))+x^2\pm x\sqrt{x^2-12(x-2)}+x^2\pm x\sqrt{x^2-16(x-2)}=x^2(2)$

$2x^2-4x+2x^2\pm x\sqrt{x^2-12(x-2)}\pm x\sqrt{x^2-16(x-2)}=2x^2$

$2x^2-4x+\pm x\sqrt{x^2-12(x-2)}\pm x\sqrt{x^2-16(x-2)}=0$

$x(2x-4+\pm \sqrt{x^2-12(x-2)}\pm \sqrt{x^2-16(x-2)})=0$

Divide through by $x$ since $x\ne 0$ yields

$2x-4+\pm \sqrt{x^2-12(x-2)}\pm \sqrt{x^2-16(x-2)}=0$

$2(x-2)+\pm \sqrt{x^2-12(x-2)}\pm \sqrt{x^2-16(x-2)}=0$

Now, we let $x-2=k$, we get

$2k+\pm \sqrt{(k+2)^2-12k}\pm \sqrt{(k+2)^2-16k)}=0$

$2k+\pm \sqrt{(k+2)^2-12k}=\mp \sqrt{(k+2)^2-16k)}$

Square both sides gives:

$4k^2\pm4k\sqrt{(k+2)^2-12k}+(k+2)^2-12k=(k+2)^2-16k$

$4k^2+4k=\mp4k\sqrt{(k+2)^2-12k}$

$4k(k+1)=\mp4k\sqrt{(k+2)^2-12k}$

$k+1=\mp\sqrt{(k+2)^2-12k}$

Square both sides again to get:

$(k+1)^2=(k+2)^2-12k$

$k^2+2k+1=k^2+4k+4-12k$

$10k=3$

$k=\frac{3}{10}$

Hence,

$x-2=k=\dfrac{3}{10} $

$x=\dfrac{23}{10}$

$y=\dfrac{x^2\pm x\sqrt{x^2-12(x-2)}}{2(x-2)}=\dfrac{23}{6}$ or $\dfrac{69}{5} $

$z=\dfrac{x^2\pm x\sqrt{x^2-16(x-2)}}{2(x-2)}=\dfrac{23}{2}$ or $\dfrac{92}{15}$

Now, we need to check their validity by substituting them back into the original given equation, we see that only the set of

$x=\dfrac{23}{10} \:,\: y=\dfrac{23}{6} \:,\: z=\dfrac{23}{2}$ is true.