Eutectic Phase Diagram and Lever Rule A solution having composition $p$ (left side of graph) is cooled to just above the eutectic temperature (point $s$ is at $0.18$ $x_{\ce{Si}}$ and the corresponding tie line intersects the liquidus curve at $0.31$ $x_{\ce{Si}} $); calculate the composition of the solid that separates and that of the liquid that remains.


The liquid phase is obviously 31 % silicon and 69 % gold as given in the question. The solutions manual also gives the weight percentage of the solid phase. Using the lever rule,

$$ m_\text{solid}(0.18)=m_\text{liquid}(0.31-0.18)$$
$$ \frac{m_\text{solid}}{m_\text{liquid}}=\frac{0.13}{0.18} $$
$$ \frac{m_\text{solid}}{m_\text{solid}+m_\text{liquid}} = \frac{0.13}{0.13+0.18}=0.42$$

However, the solution manual states that the wt% of solid phase is 58 %, and that of the liquid phase is 42 %. Is the solutions manual wrong, or am I losing my mind?
You are correct. Another way to calculate it is: $$X_{\mathrm{L}} = \frac{s-b}{e-b} = \frac{0.18-0}{0.31-0}=0.58$$
You can also inspect the diagram visually and see that the $bs$ line is longer than the $se$ line. The lever rule dictates that the line length is proportional to the percentage of the phase opposite from the line. Thus, it is not possible that there is more solid than liquid at this point.

There is definitely a mistake in your textbook's solutions.