AM I RIGHT?????

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AM I RIGHT?????

Mathematics
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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?

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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?!
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?
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
if I was doing that problem I would distribute the 1/3 all around \[A = [(\frac{1}{3}a-\frac{1}{3}b)+\frac{1}{3}c]\]
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 \[3A-a+b=c\]
if I plug this guy back into the original equation I should have A = A or something\[A = \frac{1}{3}[(a-b)+3A-a+b]\]
\[A = \frac{1}{3}[(b-b)+3A-a+a]\] \[A = \frac{1}{3}[3A] \rightarrow A = \frac{3A}{3} \rightarrow A=A\]
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 ^^
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
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

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