## anonymous one year ago AM I RIGHT?????

1. anonymous

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2. anonymous

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

3. UsukiDoll

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

4. UsukiDoll

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

5. anonymous

Yes sorry I forgot to put that there

6. UsukiDoll

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7. anonymous

That's how I write my A's

8. UsukiDoll

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

9. 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?

10. Michele_Laino

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

11. UsukiDoll

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

12. anonymous

Yes

13. UsukiDoll

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

14. anonymous

Yes and we have to solve for C

15. UsukiDoll

ok so what do we do first?

16. anonymous

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

17. anonymous

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

18. UsukiDoll

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

19. UsukiDoll

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

20. UsukiDoll

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

21. UsukiDoll

and then multiply 3 on both sides $3A-a+b=c$

22. UsukiDoll

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

23. UsukiDoll

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

24. UsukiDoll

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

25. UsukiDoll

26. anonymous

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

27. anonymous

I thought it was the second choice

28. UsukiDoll

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

29. UsukiDoll

c = 3A – (a – b) c= 3A-a+b

30. anonymous

But why?

31. anonymous

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

32. UsukiDoll

medal? ^^

33. anonymous

and fan :)

34. UsukiDoll

yay ^^

35. 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

36. anonymous

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

37. anonymous

you made it simple lol thanks for helping me again!

38. UsukiDoll

^^ it's a very important rule

39. anonymous

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

40. UsukiDoll

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