Solve for c.
A=1/3[(a-b)+c]

a. c = 3A-(a-b)
b. c = 3A-a-b
c. c = 3a+3b=A
d. c = 3Aab

Anyone who can tell me the answer and explain how they got it will get a medal. Thanks!

Question

Solve for c.
A=1/3[(a-b)+c]

a. c = 3A-(a-b)
b. c = 3A-a-b
c. c = 3a+3b=A
d. c = 3Aab

Anyone who can tell me the answer and explain how they got it will get a medal. Thanks!

anonymous · Answer

Dude...come on...basic simplification....

anonymous · Answer

@tanmaykhandait  You don't have to be rude. If you don't want to answer the question then just ignore it.

anonymous · Answer

You are looking to get the value for c. So we must get c to one side of the equation. But first lets get rid of the fraction (1/3) and make it a value of 1. How would we do that?

anonymous · Answer

Well I get that you would divide 1/3 by both sides and get 3A = (a-b) + c but after that I don't understand what to do.

anonymous · Answer

Great.  now we have to get a-b to the left side of the equation. You don't need the parentheses so if we have 3A=a+b+c what do we do to get a and b to the left side?

anonymous · Answer

Sorry should be 3A = a-b+c...

anonymous · Answer

So it would be c = 3A-a+b?

anonymous · Answer

Wait it would be c = 3A - (a+b) right?

anonymous · Answer

Yes but that can also be c=3A-(a-b)   do you see how that works as we distribute the minus sign over (a-b)...Good job

anonymous · Answer

Okay, thanks for your help!