40-(3c+)=2(c+6)+c

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Solution for 40-(3c+)=2(c+6)+c equation: 



40-(3c+)=2(c+6)+c

We move all terms to the left:

40-(3c+)-(2(c+6)+c)=0

We add all the numbers together, and all the variables

-(+3c)-(2(c+6)+c)+40=0

We get rid of parentheses

-3c-(2(c+6)+c)+40=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+40=0

We add all the numbers together, and all the variables

-6c+28=0

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

-6c=-28

c=-28/-6

c=4+2/3

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