The binary lever rule is a component balance across a tie line

At fixed temperature and pressure, a nonreacting binary system separates into two equilibrium phases, $\alpha$ and $\beta$. Let $x$ denote the mole fraction of component 1. The endpoints of the relevant tie line are \[ x_\alpha=0.20,\qquad x_\beta=0.70, \] and a closed sample has overall composition $z=0.40$ and total amount $n=10.0\ \mathrm{mol}$. Let $f_\alpha=n_\alpha/n$ and $f_\beta=n_\beta/n$. 1. Starting only from the total-mole balance and the component-1 balance, derive formulas for $f_\alpha$ and $f_\beta$ in terms of $z,x_\alpha,x_\beta$. Do not quote the lever rule without deriving it. 2. Evaluate both phase fractions and both phase amounts for the stated sample. 3. Independently check the answer by inventorying component 1 in each phase and comparing the sum with the overall inventory. 4. Explain geometrically why the numerator for the $\beta$ fraction is the distance from $x_\alpha$ to $z$, not the distance from $z$ to $x_\beta$. State what happens at $z=x_\alpha$ and $z=x_\beta$, and why the two-phase formulas are not to be extrapolated to $z$ outside $[x_\alpha,x_\beta]$. Keep phase fractions distinct from the compositions inside each phase. This is a material-balance exercise on a supplied tie line, not a calculation of a Clapeyron coexistence slope or a van-der-Waals spinodal.

Write $n=n_\alpha+n_\beta$ and $nz=n_\alpha x_\alpha+n_\beta x_\beta$, then divide both equations by $n$.