2(19-c)=2(6c+2)+3c

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Solution for 2(19-c)=2(6c+2)+3c equation: 



2(19-c)=2(6c+2)+3c

We move all terms to the left:

2(19-c)-(2(6c+2)+3c)=0

We add all the numbers together, and all the variables

2(-1c+19)-(2(6c+2)+3c)=0

We multiply parentheses

-2c-(2(6c+2)+3c)+38=0



We calculate terms in parentheses: -(2(6c+2)+3c), so:

2(6c+2)+3c

We add all the numbers together, and all the variables

3c+2(6c+2)

We multiply parentheses

3c+12c+4

We add all the numbers together, and all the variables

15c+4

Back to the equation:

-(15c+4)



We get rid of parentheses

-2c-15c-4+38=0

We add all the numbers together, and all the variables

-17c+34=0

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

-17c=-34

c=-34/-17

c=+2

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