(3c+12)+2c=3c+(12+2c)

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Solution for (3c+12)+2c=3c+(12+2c) equation: 



(3c+12)+2c=3c+(12+2c)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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



We calculate terms in parentheses: -(3c+(2c+12)), so:

3c+(2c+12)

We get rid of parentheses

3c+2c+12

We add all the numbers together, and all the variables

5c+12

Back to the equation:

-(5c+12)



We add all the numbers together, and all the variables

5c-(5c+12)+12=0

We get rid of parentheses

5c-5c-12+12=0

We add all the numbers together, and all the variables

=0

c=0/1

c=0

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