Equations Inequalities and VIC's page 120/equation top of pa

[wignewton]

How do you arrive at z(1-y\100) from z-(y\100)z? I come to the formula z-(zy\100). How are you taking to z off the numerator and making it one?

[esledge]

The key is that there's a z in each term of the difference: z*A - z*B.

We aren't "taking z off the numerator," we are factoring it out: z*A - z*B = z*(A-B)

Here are the steps:
[z] - [(y/100)*z]
=[z*1] - [z*(y/100)]
= z * [1 - (y/100)]

Your expression was equivalent to the original also, but you hadn't factored the z out:

[z] - [(y/100)z]
= [z] - [zy/100] {This is where you stopped}
=[z*1] - [z*(y/100)]
= z * [1 - (y/100)]