2022 AMC 10A Problem 3

Let $x, y,$ and $z$ be the three numbers. The conditions from the problem give us the following relations: $$\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)}. \end{aligned}$$ Rearranging $(3),$ we get $y = z + 40.$ Plugging this new equation and $(2)$ into $(1),$ we get $$6z + z + 40 + z = 96$$ $$8z + 40 = 96$$ $$8z = 56 \Rightarrow z = 7.$$ From this, we get that $$x = 6 \cdot z = 6 \cdot 7 = 42$$ and $$y = z + 40 = 7 + 40 = 47.$$ Therefore, $y - x = 47 - 42 = 5.$ Thus, E is the correct answer.