40-(3c+4)=(c+5)+c

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Solution for 40-(3c+4)=(c+5)+c equation: 



40-(3c+4)=(c+5)+c

We move all terms to the left:

40-(3c+4)-((c+5)+c)=0

We get rid of parentheses

-3c-((c+5)+c)-4+40=0



We calculate terms in parentheses: -((c+5)+c), so:

(c+5)+c

We add all the numbers together, and all the variables

c+(c+5)

We get rid of parentheses

c+c+5

We add all the numbers together, and all the variables

2c+5

Back to the equation:

-(2c+5)



We add all the numbers together, and all the variables

-3c-(2c+5)+36=0

We get rid of parentheses

-3c-2c-5+36=0

We add all the numbers together, and all the variables

-5c+31=0

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

-5c=-31

c=-31/-5

c=6+1/5

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