In the example that is mentioned in the problem, solving for $y$ in the equation $x - y = 0$ yields, $y = x$, so you can turn the equation $x+y+z=1$ into an equation that has only $x$ and $z$: $x+x+z = 1$, so $\displaystyle x = \frac{1-z}{2}$. Since we already knew that $x=y$, we have the line is determined by the equation $\displaystyle x = y = \frac{z-1}{2}$. Note we have $$\text{something in terms of only x} = \text{something in terms of only y} = \text{something in terms of only z}.$$