Problem source

The numbers $x,y$ and $z$ are non-zero, and satisfy
\[
2a-3y=\frac{\left(z-x\right)^{2}}{y}\quad\mbox{ and }\quad2a-3z=\frac{\left(x-y\right)^{2}}{z},
\]
for some number $a$. If $y\neq z$, prove that 
\[
x+y+z=a,
\]
and that 
\[
2a-3x=\frac{\left(y-z\right)^{2}}{x}.
\]
Determine whether this last equation holds \textit{only} if $y\neq z$.

Solution source

\begin{align*}
&& \begin{cases} 2a-3y=\frac{\left(z-x\right)^{2}}{y} \\
2a-3z=\frac{\left(x-y\right)^{2}}{z}
 \end{cases} \\
\Rightarrow && \begin{cases} 2ay-3y^2=\left(z-x\right)^{2} \\
2az-3z^2=\left(x-y\right)^{2}
 \end{cases} \\
\Rightarrow && 2a(y-z)-3(y+z)(y-z) &= (z-x+x-y)(z-x-x+y) \\
\Rightarrow && (y-z)(2a-3y-3z) &= (z-y)(z-2x+y) \\
\Rightarrow && 2a-3y-3z &= 2x-y-z \tag{$y \neq z$} \\
\Rightarrow && a &= x+y+z \\
\end{align*}

This is is our first result.

\begin{align*}
&& 2a-3y-3z &= 2x-y-z \\
\Rightarrow && 2a-3y-3x &= 3z-y-x \\
\Rightarrow && (y-x)2a-3(y-x)(y+x) &= (y-x)(2z-x-y) \\
\Rightarrow && 2a(y-x)-3(y^2-x^2) &= (z-y)^2-(x-z)^2 \\
\Rightarrow && 2ax - 3x^2 &= (y-z)^2 \\
\Rightarrow && 2a - 3x &= \frac{(y-z)^2}{x} 
\end{align*}

Suppose $x = \frac23 a, y = z = \frac16 a$ then all equations are satisfied, but $y = z$.