One approach is to look for a pattern

1 day: cost = $z$ dollars

2 days: cost = $z + \frac{z}{6}$ dollars

3 days: cost = $z + \frac{z}{6}+ \frac{z}{6}$ dollars

4 days: cost = $z + \frac{z}{6}+\frac{z}{6}+\frac{z}{6}$ dollars

Notice that the number of times we add z/6 is 1 less than the number of vacation days
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y days: cost = $z + (y-1)(\frac{z}{6})$ dollars

Check answer choices....$z + (y-1)(\frac{z}{6})$ is not an option. So, we must simplify this expression.

Take: $z + (y-1)(\frac{z}{6})$

Rewrite $z$ as $\frac{6z}{6}$ to get: $\frac{6z}{6} + \frac{(y-1)z}{6}$

Simplify: $\frac{6z}{6} + \frac{yz-z}{6}$

Simplify: $\frac{6z+(yz-z)}{6}$

Simplify: $\frac{5z+yz}{6}$

Answer: D

Cheers,
Brent