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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

(c+2)+3c

We add all the numbers together, and all the variables

3c+(c+2)

We get rid of parentheses

3c+c+2

We add all the numbers together, and all the variables

4c+2

Back to the equation:

-(4c+2)



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-6c+9=0

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

-6c=-9

c=-9/-6

c=1+1/2

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