3(c+4)=2(c+6)+c

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Solution for 3(c+4)=2(c+6)+c equation: 



3(c+4)=2(c+6)+c

We move all terms to the left:

3(c+4)-(2(c+6)+c)=0

We multiply parentheses

3c-(2(c+6)+c)+12=0



We calculate terms in parentheses: -(2(c+6)+c), so:

2(c+6)+c

We add all the numbers together, and all the variables

c+2(c+6)

We multiply parentheses

c+2c+12

We add all the numbers together, and all the variables

3c+12

Back to the equation:

-(3c+12)



We get rid of parentheses

3c-3c-12+12=0

We add all the numbers together, and all the variables

=0

c=0/1

c=0

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