Show that $xz +x'y + zy = xz + x'y$ (Boolean Algebra)

Question

My best guess is that somehow $$xz + x'y = zy$$ I know that $$A + A'B = A + B$$ but can that be applied here since there is an extra variable? Is there a relevant rule I'm missing?

Question 25 from https://web.mit.edu/6.111/www/s2007/PSETS/pset1.pdf

Answer 1

Since $x + x' = 1$, one gets \begin{align} xz +x'y + zy &= xz +x'y + (x + x')zy \\ &= xz + x'y + xzy + x'zy \\ &= xz(1 + y) + x'y(1+z) \\ &= xz + x'y \end{align}