3(3c+5)+1=2(c+28)

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Solution for 3(3c+5)+1=2(c+28) equation: 



3(3c+5)+1=2(c+28)

We move all terms to the left:

3(3c+5)+1-(2(c+28))=0

We multiply parentheses

9c-(2(c+28))+15+1=0



We calculate terms in parentheses: -(2(c+28)), so:

2(c+28)

We multiply parentheses

2c+56

Back to the equation:

-(2c+56)



We add all the numbers together, and all the variables

9c-(2c+56)+16=0

We get rid of parentheses

9c-2c-56+16=0

We add all the numbers together, and all the variables

7c-40=0

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

7c=40

c=40/7

c=5+5/7

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