# Distributive property rule

Algebra

### Distributive property rule

My professor showed me a rule about the distributive property that I’ve read for years but have always overlooked.

We’ve all seen the classic a(b+c)

Apparently the distributive property only respects addition and subtraction. Which means that a(bc) is not distributive.

So if you have this problem:

0.01+2(0.01+0.03)=0.01

And you suddenly realized that you wanted to completely annihilate the zeroes by distributing 100 to both sides of the equation;

100[0.01+2(0.01+0.03)]=0.01(100)

What the hell would you do?

Keep in mind a(b+c) is the distributive property in a mathematical nutshell, and that this is a grade school question, so let’s look at the left side of the equation I.e the part I fucked up on I’m college algebra thanks to never encountering the scenario prior.

a=100
b=0.01

So what is c?

I wrote the answer in my notes but thanks to overestimating my damn memory I didn’t write the notes concisely enough.

Is “c” the product of 2(0.01+0.03)?

Or is “c” (0.01+0.03)?

Or is “c” a choice between “2” and “(0.01+0.03)”?

In other words, how do you properly distribute the 100?

Please help because this principle is so fucking basic that it could kill me in the future.
Guest

### Re: Distributive property rule

a(b+c) = ab + ac

100[0.01 + 2(0.01+0.03)] = 100*0.01 + 100*(2(0.01+0.03))

The + separating blue and green is always the last operation you would if perform if evaluating the part in [brackets] according to order of operations.

Your goal of getting rid of all the decimals can be achieved by reapplying the distributive law to the last part of the expression.

100 (2 (0.01 + 0.03)) = 100 * 2 * 0.01 + 100 * 2 * 0.03 = 2 + 6

Is “c” the product of 2(0.01+0.03)?

This is the best answer, representing the literal application of the distributive law, however ...

Or is “c” (0.01+0.03)?

Or is “c” a choice between “2” and “(0.01+0.03)”?

both of these will give you the same result at the end and are useful shortcuts for multiple applications of the distributive, associative, and commutative laws in a single step.

phw

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