AM I RIGHT?????

- anonymous

AM I RIGHT?????

- Stacey Warren - Expert brainly.com

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- katieb

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- anonymous

|dw:1433060237198:dw|

- anonymous

c = 3A – (a – b)
c = 3A – a – b<<<
c = 3a + 3b + A
c = 3Aab

- UsukiDoll

What is this problem anyway?! are you solving for c?

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## More answers

- UsukiDoll

and you forgot to distribute the negative sign
-(a-b) = -a+b

- anonymous

Yes sorry I forgot to put that there

- UsukiDoll

|dw:1433061203222:dw|

- anonymous

That's how I write my A's

- UsukiDoll

\[A = \frac{1}{3}[(a-b)+c]\] so what is the question? solve for c?!

- UsukiDoll

we don't even know what the original question is besides am I right? That doesn't give any information. What is the question and what is the goal of this?

- Michele_Laino

step #1
in order to find c, we have to multiply both sides by 3, what do you get?

- UsukiDoll

first of all, op is your question how do I solve in terms of c? is that why we have c = your answer

- anonymous

Yes

- UsukiDoll

ok then. so we have this equation
\[A = \frac{1}{3}[(a-b)+c]\]

- anonymous

Yes and we have to solve for C

- UsukiDoll

ok so what do we do first?

- anonymous

When I first did it I got c= -a+b+ 1/3A

- anonymous

But it wasnt in the answer choices so I thought of the one that looked similar to it

- UsukiDoll

if I was doing that problem I would distribute the 1/3 all around
\[A = [(\frac{1}{3}a-\frac{1}{3}b)+\frac{1}{3}c]\]

- UsukiDoll

then I would shift the 1/3a and 1/3b to the left

- UsukiDoll

\[A -\frac{1}{3}a+\frac{1}{3}b=\frac{1}{3}c\]

- UsukiDoll

and then multiply 3 on both sides
\[3A-a+b=c\]

- UsukiDoll

if I plug this guy back into the original equation I should have A = A or something\[A = \frac{1}{3}[(a-b)+3A-a+b]\]

- UsukiDoll

\[A = \frac{1}{3}[(b-b)+3A-a+a]\]
\[A = \frac{1}{3}[3A] \rightarrow A = \frac{3A}{3} \rightarrow A=A\]

- UsukiDoll

a - a and -b+b cancel out I am left with 3A and I can just divide 3

- UsukiDoll

what are your answer choices for this problem?

- anonymous

c = 3A – (a – b)
c = 3A – a – b
c = 3a + 3b + A
c = 3Aab

- anonymous

I thought it was the second choice

- UsukiDoll

\[3A-a+b=c\] it's the first choice if you distribute the negative

- UsukiDoll

c = 3A – (a – b)
c= 3A-a+b

- anonymous

But why?

- anonymous

No wait I figured out why and how you did it thank you !

- UsukiDoll

medal? ^^

- anonymous

and fan :)

- UsukiDoll

yay ^^

- UsukiDoll

whenever there is a negative sign on the right side of the equation you have to distribute it
Ex
5x-(2x^2+4)
5x-2x^2-4

- anonymous

Ohh I thought you had to do a whole different complicated equation

- anonymous

you made it simple lol thanks for helping me again!

- UsukiDoll

^^ it's a very important rule

- anonymous

I'm going to start doing problems like these so I wont forget the rule

- UsukiDoll

^_^ you can write a testimonial about me if you want xD

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