= Solution
Let $\ell_N(x)=3.44-x$ and $\ell_A(x)=-7.59-3x$. The <common-potential test for a three-mineral activity diagram> gives relative potentials, in one positive energy unit,
$$
\Phi_C=0,\qquad\Phi_N=\ell_N-y,\qquad\Phi_A=\ell_A-y.
$$
The stable phase minimizes these potentials. Consequently the \b[stable <nepheline>–<corundum> branch has $x\le-5.515$], the \b[stable <albite>–<corundum> branch has $x\ge-5.515$], and the \b[stable <nepheline>–<albite> branch has $x=-5.515$, $y\ge8.955$]. Draw them solid.
The other halves of each line are \b[metastable]: a third mineral has lower potential there. In particular, below the triple point the vertical equality describes two phases both less stable than <corundum>. Extend those halves as dashed lines. This thermodynamic comparison prevents the incorrect use of all three complete pairwise lines as stable field boundaries.
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