-40-(3c+4)=2(c+3)+c

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Solution for -40-(3c+4)=2(c+3)+c equation: 



-40-(3c+4)=2(c+3)+c

We move all terms to the left:

-40-(3c+4)-(2(c+3)+c)=0

We get rid of parentheses

-3c-(2(c+3)+c)-4-40=0



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

2(c+3)+c

We add all the numbers together, and all the variables

c+2(c+3)

We multiply parentheses

c+2c+6

We add all the numbers together, and all the variables

3c+6

Back to the equation:

-(3c+6)



We add all the numbers together, and all the variables

-3c-(3c+6)-44=0

We get rid of parentheses

-3c-3c-6-44=0

We add all the numbers together, and all the variables

-6c-50=0

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

-6c=50

c=50/-6

c=-8+1/3

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