40-(3c+4)=20(c+6)+c

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Solution for 40-(3c+4)=20(c+6)+c equation: 



40-(3c+4)=20(c+6)+c

We move all terms to the left:

40-(3c+4)-(20(c+6)+c)=0

We get rid of parentheses

-3c-(20(c+6)+c)-4+40=0



We calculate terms in parentheses: -(20(c+6)+c), so:

20(c+6)+c

We add all the numbers together, and all the variables

c+20(c+6)

We multiply parentheses

c+20c+120

We add all the numbers together, and all the variables

21c+120

Back to the equation:

-(21c+120)



We add all the numbers together, and all the variables

-3c-(21c+120)+36=0

We get rid of parentheses

-3c-21c-120+36=0

We add all the numbers together, and all the variables

-24c-84=0

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

-24c=84

c=84/-24

c=-3+1/2

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