-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

-3c+19=0

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

-3c=-19

c=-19/-3

c=6+1/3

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