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

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



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

We move all terms to the left:

13-(2c+2)-(2(2c+2)+3c)=0

We get rid of parentheses

-2c-(2(2c+2)+3c)-2+13=0



We calculate terms in parentheses: -(2(2c+2)+3c), so:

2(2c+2)+3c

We add all the numbers together, and all the variables

3c+2(2c+2)

We multiply parentheses

3c+4c+4

We add all the numbers together, and all the variables

7c+4

Back to the equation:

-(7c+4)



We add all the numbers together, and all the variables

-2c-(7c+4)+11=0

We get rid of parentheses

-2c-7c-4+11=0

We add all the numbers together, and all the variables

-9c+7=0

We move all terms containing c to the left, all other terms to the right

-9c=-7

c=-7/-9

c=7/9

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