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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.
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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