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13-(2c+2)=2(c+2+3c)
We move all terms to the left:
13-(2c+2)-(2(c+2+3c))=0
We add all the numbers together, and all the variables
-(2c+2)-(2(4c+2))+13=0
We get rid of parentheses
-2c-(2(4c+2))-2+13=0
We calculate terms in parentheses: -(2(4c+2)), so:We add all the numbers together, and all the variables
2(4c+2)
We multiply parentheses
8c+4
Back to the equation:
-(8c+4)
-2c-(8c+4)+11=0
We get rid of parentheses
-2c-8c-4+11=0
We add all the numbers together, and all the variables
-10c+7=0
We move all terms containing c to the left, all other terms to the right
-10c=-7
c=-7/-10
c=7/10
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