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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

2(c+3)+6

We multiply parentheses

2c+6+6

We add all the numbers together, and all the variables

2c+12

Back to the equation:

-(2c+12)



We add all the numbers together, and all the variables

-1c-(2c+12)+16=0

We get rid of parentheses

-1c-2c-12+16=0

We add all the numbers together, and all the variables

-3c+4=0

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

-3c=-4

c=-4/-3

c=1+1/3

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