42-(3c+4)=(c+4)+c

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Solution for 42-(3c+4)=(c+4)+c equation: 



42-(3c+4)=(c+4)+c

We move all terms to the left:

42-(3c+4)-((c+4)+c)=0

We get rid of parentheses

-3c-((c+4)+c)-4+42=0



We calculate terms in parentheses: -((c+4)+c), so:

(c+4)+c

We add all the numbers together, and all the variables

c+(c+4)

We get rid of parentheses

c+c+4

We add all the numbers together, and all the variables

2c+4

Back to the equation:

-(2c+4)



We add all the numbers together, and all the variables

-3c-(2c+4)+38=0

We get rid of parentheses

-3c-2c-4+38=0

We add all the numbers together, and all the variables

-5c+34=0

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

-5c=-34

c=-34/-5

c=6+4/5

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