13-(2c+2)=2(c+3)+3c

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Solution for 13-(2c+2)=2(c+3)+3c equation:



13-(2c+2)=2(c+3)+3c
We move all terms to the left:
13-(2c+2)-(2(c+3)+3c)=0
We get rid of parentheses
-2c-(2(c+3)+3c)-2+13=0
We calculate terms in parentheses: -(2(c+3)+3c), so:
2(c+3)+3c
We add all the numbers together, and all the variables
3c+2(c+3)
We multiply parentheses
3c+2c+6
We add all the numbers together, and all the variables
5c+6
Back to the equation:
-(5c+6)
We add all the numbers together, and all the variables
-2c-(5c+6)+11=0
We get rid of parentheses
-2c-5c-6+11=0
We add all the numbers together, and all the variables
-7c+5=0
We move all terms containing c to the left, all other terms to the right
-7c=-5
c=-5/-7
c=5/7

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