2(1+c)+4c=5(c+1)

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Solution for 2(1+c)+4c=5(c+1) equation: 



2(1+c)+4c=5(c+1)

We move all terms to the left:

2(1+c)+4c-(5(c+1))=0

We add all the numbers together, and all the variables

2(c+1)+4c-(5(c+1))=0

We add all the numbers together, and all the variables

4c+2(c+1)-(5(c+1))=0

We multiply parentheses

4c+2c-(5(c+1))+2=0



We calculate terms in parentheses: -(5(c+1)), so:

5(c+1)

We multiply parentheses

5c+5

Back to the equation:

-(5c+5)



We add all the numbers together, and all the variables

6c-(5c+5)+2=0

We get rid of parentheses

6c-5c-5+2=0

We add all the numbers together, and all the variables

c-3=0

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

c=3

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