first off, let's split the triplet into two equations, then from there on we'll do substitution.
[tex]\cfrac{y}{x-z}=\cfrac{x}{y}=\cfrac{x+y}{z}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{\cfrac{y}{x-z}=\cfrac{x}{y}\implies }y^2=\underline{x^2-xz} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 2nd equation}}{\cfrac{x}{y}=\cfrac{x+y}{z}\implies }xz=xy+y^2\implies \stackrel{\textit{substituting for }y^2}{xz=xy+(\underline{x^2-xz})}[/tex]
[tex]2xz=xy+x^2\implies 2xz=x(y+x)\implies \cfrac{2xz}{x}=y+x \\\\\\ 2z=y+x\implies 2=\cfrac{y+x}{z}\implies 2=\cfrac{x+y}{z} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{}{ \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases}}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=2 \end{cases}\implies \begin{cases} \cfrac{y}{x-z}=2\\[2em] \cfrac{x}{y}=2 \end{cases}[/tex]
that of course, is only true if x + y, or our numerator doesn't turn into 0, if it does then our fraction becomes 0 and our equation goes south. Keeping in mind that x,y and z are numeric values that correlate like so.
