(c=3)-2c-(1-3c)=2

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Solution for (c=3)-2c-(1-3c)=2 equation: 



(c=3)-2c-(1-3c)=2

We move all terms to the left:

(c-(3)-2c-(1-3c))=0

We add all the numbers together, and all the variables

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



We calculate terms in parentheses: +(c-3-2c-(-3c+1)), so:

c-3-2c-(-3c+1)

determiningTheFunctionDomain
c-2c-(-3c+1)-3

We add all the numbers together, and all the variables

-1c-(-3c+1)-3

We get rid of parentheses

-1c+3c-1-3

We add all the numbers together, and all the variables

2c-4

Back to the equation:

+(2c-4)



We get rid of parentheses

2c-4=0

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

2c=4

c=4/2

c=2

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