3(c-3)=21+6(2c-4)

Solution for 3(c-3)=21+6(2c-4) equation:

3(c-3)=21+6(2c-4)

We move all terms to the left:

3(c-3)-(21+6(2c-4))=0

We multiply parentheses

3c-(21+6(2c-4))-9=0


We calculate terms in parentheses: -(21+6(2c-4)), so:

21+6(2c-4)

determiningTheFunctionDomain
6(2c-4)+21

We multiply parentheses

12c-24+21

We add all the numbers together, and all the variables

12c-3

Back to the equation:

-(12c-3)


We get rid of parentheses

3c-12c+3-9=0

We add all the numbers together, and all the variables

-9c-6=0

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

-9c=6

c=6/-9

c=-2/3