= Solution
The <stability boundary in a mineral activity diagram> requires the two coexisting minerals to be stable against the third. For the forward dissolution directions of (I) and (II),
$$
\frac{\Delta_rG_I}{2RT\ln10}=y-(3.44-x),\qquad
\frac{\Delta_rG_{II}}{2RT\ln10}=y-(-7.59-3x).
$$
A negative value favours <corundum> over the corresponding <silicate mineral>. For (III),
$$
\frac{\Delta_rG_{III}}{RT\ln10}=2x+11.03,
$$
so low silica activity favours <nepheline> over <albite>, and high silica activity favours <albite>. Combining these comparisons gives the stable portions:
* \b[(I): $x<-5.515$], the nepheline–corundum boundary to the left of the intersection.
* \b[(II): $x>-5.515$], the albite–corundum boundary to its right.
* \b[(III): $x=-5.515$, $y>8.955$], the nepheline–albite boundary above the intersection.
The opposite extension of each line is <metastable> because the third mineral has lower thermodynamic potential there. All three minerals can coexist at the intersection in this fixed-temperature, fixed-pressure activity model.
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