-4(3c+1)+3c=3(2c+6)+5

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Solution for -4(3c+1)+3c=3(2c+6)+5 equation: 



-4(3c+1)+3c=3(2c+6)+5

We move all terms to the left:

-4(3c+1)+3c-(3(2c+6)+5)=0

We add all the numbers together, and all the variables

3c-4(3c+1)-(3(2c+6)+5)=0

We multiply parentheses

3c-12c-(3(2c+6)+5)-4=0



We calculate terms in parentheses: -(3(2c+6)+5), so:

3(2c+6)+5

We multiply parentheses

6c+18+5

We add all the numbers together, and all the variables

6c+23

Back to the equation:

-(6c+23)



We add all the numbers together, and all the variables

-9c-(6c+23)-4=0

We get rid of parentheses

-9c-6c-23-4=0

We add all the numbers together, and all the variables

-15c-27=0

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

-15c=27

c=27/-15

c=-1+4/5

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