3(c+1)+2c=7+3(c-1)

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Solution for 3(c+1)+2c=7+3(c-1) equation: 



3(c+1)+2c=7+3(c-1)

We move all terms to the left:

3(c+1)+2c-(7+3(c-1))=0

We add all the numbers together, and all the variables

2c+3(c+1)-(7+3(c-1))=0

We multiply parentheses

2c+3c-(7+3(c-1))+3=0



We calculate terms in parentheses: -(7+3(c-1)), so:

7+3(c-1)

determiningTheFunctionDomain
3(c-1)+7

We multiply parentheses

3c-3+7

We add all the numbers together, and all the variables

3c+4

Back to the equation:

-(3c+4)



We add all the numbers together, and all the variables

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

We get rid of parentheses

5c-3c-4+3=0

We add all the numbers together, and all the variables

2c-1=0

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

2c=1

c=1/2

c=1/2

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