56-(2c+3)=4(c+5)+c

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Solution for 56-(2c+3)=4(c+5)+c equation: 



56-(2c+3)=4(c+5)+c

We move all terms to the left:

56-(2c+3)-(4(c+5)+c)=0

We get rid of parentheses

-2c-(4(c+5)+c)-3+56=0



We calculate terms in parentheses: -(4(c+5)+c), so:

4(c+5)+c

We add all the numbers together, and all the variables

c+4(c+5)

We multiply parentheses

c+4c+20

We add all the numbers together, and all the variables

5c+20

Back to the equation:

-(5c+20)



We add all the numbers together, and all the variables

-2c-(5c+20)+53=0

We get rid of parentheses

-2c-5c-20+53=0

We add all the numbers together, and all the variables

-7c+33=0

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

-7c=-33

c=-33/-7

c=4+5/7

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