2(5c+2)+2c=3(3c+3)+7

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Solution for 2(5c+2)+2c=3(3c+3)+7 equation: 



2(5c+2)+2c=3(3c+3)+7

We move all terms to the left:

2(5c+2)+2c-(3(3c+3)+7)=0

We add all the numbers together, and all the variables

2c+2(5c+2)-(3(3c+3)+7)=0

We multiply parentheses

2c+10c-(3(3c+3)+7)+4=0



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

3(3c+3)+7

We multiply parentheses

9c+9+7

We add all the numbers together, and all the variables

9c+16

Back to the equation:

-(9c+16)



We add all the numbers together, and all the variables

12c-(9c+16)+4=0

We get rid of parentheses

12c-9c-16+4=0

We add all the numbers together, and all the variables

3c-12=0

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

3c=12

c=12/3

c=4

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