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13-1(2c+2)=2(c+2)+3c
We move all terms to the left:
13-1(2c+2)-(2(c+2)+3c)=0
We calculate terms in parentheses: -(2(c+2)+3c), so:We get rid of parentheses
2(c+2)+3c
We add all the numbers together, and all the variables
3c+2(c+2)
We multiply parentheses
3c+2c+4
We add all the numbers together, and all the variables
5c+4
Back to the equation:
-(5c+4)
-1(2c+2)-5c-4+13=0
We add all the numbers together, and all the variables
-5c-1(2c+2)+9=0
We move all terms containing c to the left, all other terms to the right
-5c-1(2c+2)=-9
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