SOLUTION: I cannot figure out where I am going wrong on this problem. I distributed solving the parentheses first, but no matter what I do, I keep getting this problem wrong. 3x+2[1-3(x+2

Question 452437: I cannot figure out where I am going wrong on this problem. I distributed solving the parentheses first, but no matter what I do, I keep getting this problem wrong.
3x+2[1-3(x+2)]=2x
This is my work..
3x+2[1-3(x+2)]=2x
3x+2[1-3x-6]=2x
3x+2-6x-12=2x
I get x= -10. Could you tell me where I am going wrong? I thought you needed to do inside the parentheses first, then inside the brackets.
Thank you in advance.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I cannot figure out where I am going wrong on this problem. I distributed solving the parentheses first, but no matter what I do, I keep getting this problem wrong.
3x+2[1-3(x+2)]=2x
This is my work..
3x+2[1-3(x+2)]=2x
3x+2[1-3x-6]=2x
3x+2-6x-12=2x
---------------
-3x -10 = 2x
I get x= -10.
---------------
3x+2[1-3(x+2)]=2x
2[1-3(x+2)] = -x
2 - 3(x+2) = -x
- 3(x+2) = -x-2
3(x+2) = x+2
3x+6 = x+2
2x+4 = 0
x = -2

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