 Solution

At $138{}^{\circ }\mathrm{C}$, one line is tangent to all three Gibbs free energy curves at approximately $x_{\mathrm{Sn}}=0.03$ for (Bi), $0.61$ for liquid, and $0.93$ for $(\beta \mathrm{Sn})$. These three phases can coexist at the eutectic point. Between the two solid compositions, the equilibrium phase fractions depend on how much liquid has already transformed; exactly at the invariant temperature, the overall composition alone does not fix all three fractions. At the eutectic overall composition, cooling just below the temperature eliminates the liquid and leaves the two solids.