6(3+2c)=(7c-2)-3c-4

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Solution for 6(3+2c)=(7c-2)-3c-4 equation: 



6(3+2c)=(7c-2)-3c-4

We move all terms to the left:

6(3+2c)-((7c-2)-3c-4)=0

We add all the numbers together, and all the variables

6(2c+3)-((7c-2)-3c-4)=0

We multiply parentheses

12c-((7c-2)-3c-4)+18=0



We calculate terms in parentheses: -((7c-2)-3c-4), so:

(7c-2)-3c-4

We add all the numbers together, and all the variables

-3c+(7c-2)-4

We get rid of parentheses

-3c+7c-2-4

We add all the numbers together, and all the variables

4c-6

Back to the equation:

-(4c-6)



We get rid of parentheses

12c-4c+6+18=0

We add all the numbers together, and all the variables

8c+24=0

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

8c=-24

c=-24/8

c=-3

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