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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

2(c+2)+3+c

determiningTheFunctionDomain
2(c+2)+c+3

We add all the numbers together, and all the variables

c+2(c+2)+3

We multiply parentheses

c+2c+4+3

We add all the numbers together, and all the variables

3c+7

Back to the equation:

-(3c+7)



We add all the numbers together, and all the variables

-2c-(3c+7)+11=0

We get rid of parentheses

-2c-3c-7+11=0

We add all the numbers together, and all the variables

-5c+4=0

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

-5c=-4

c=-4/-5

c=4/5

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