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|dw:1433060237198:dw|

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

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

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

Yes sorry I forgot to put that there

|dw:1433061203222:dw|

That's how I write my A's

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

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

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

Yes

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

Yes and we have to solve for C

ok so what do we do first?

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

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

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

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

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

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

what are your answer choices for this problem?

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

I thought it was the second choice

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

c = 3A – (a – b)
c= 3A-a+b

But why?

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

medal? ^^

and fan :)

yay ^^

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

you made it simple lol thanks for helping me again!

^^ it's a very important rule

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

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