What's the value of $x$ if $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and...

Question

Given $x, y, z\in \mathbb{R}$, such that $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and $ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}$ and $\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}$. Find the value of $x$.

Answer 1

Let $xy=c$, $xz=b$ and $yz=a$.

Thus, $$\frac{x+y+z}{c+b}=\frac{1}{2},$$ $$\frac{x+y+z}{c+a}=\frac{1}{3}$$ and $$\frac{x+y+z}{a+b}=\frac{1}{4},$$ which gives $$\frac{c+a}{c+b}=\frac{3}{2}$$ and $$\frac{a+b}{c+b}=2.$$ From here we obtain $$b=\frac{3}{5}a$$ and $$c=\frac{1}{5}a$$ or $$xz=\frac{3}{5}yz$$ and $$xy=\frac{1}{5}yz,$$ which gives $$y=\frac{5}{3}x$$ and $$z=5x.$$ Id est, $$\frac{1}{x}+\frac{1}{\frac{5}{3}x+5x}=\frac{1}{2},$$ which gives $$x=2.3.$$

Answer 2

I suggest to go for $X=\frac 1x$, $Y=\frac 1y$ and $Z=\frac 1z$ because it has the advantage to simplify the denominators.

The equations become

$\dfrac{XY+YZ+ZX}{Y+Z}=\dfrac 12\quad;\quad\dfrac{XY+YZ+ZX}{X+Z}=\dfrac 13\quad;\quad\dfrac{XY+YZ+ZX}{X+Y}=\dfrac 14$

And it gives a simple system to solve

$\begin{cases} Y+Z=2k\\ X+Z=3k\\ X+Y=4k\\ XY+YZ+ZX=k\end{cases}$

For instance $(2)-(1)$ gives $X-Y=k$ and reporting in $(3)$ gives $2Y=3k$.

This is all similar easy calculation and we end up with $X=\frac 52 k,Y=\frac 32k,Z=\frac 12k$ and $\dfrac{23k^2}4=k$

Finally $x=\frac 1X=\frac 25\times\frac{23}{4}=\frac {23}{10}$.