= Solution
A <stability boundary in a mineral activity diagram> must also be stable against the third mineral. Use the <common-potential test for a three-mineral activity diagram>. For the two corundum-producing directions,
$$
\frac{\Delta_rG_a}{2RT\ln10}=y-(3.44-x),\qquad
\frac{\Delta_rG_b}{2RT\ln10}=y-(-7.59-3x).
$$
Negative values favour <corundum>. For the albite-to-nepheline conversion, $\Delta_rG/(RT\ln10)=2x+11.03$, so low silica activity favours <nepheline> and high silica activity favours <albite>. These comparisons give:
* \b[Nepheline–corundum: stable for $x<-5.515$], metastable on the opposite extension.
* \b[Albite–corundum: stable for $x>-5.515$], metastable on the opposite extension.
* \b[Nepheline–albite: stable for $y>8.955$] along $x=-5.515$, metastable below the intersection.
All three coexist at the intersection. Pairwise mass-action equations alone do not distinguish the stable segments from their <metastable> extensions.
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