6(c+1)-9(c-5)=3(4c+2)

Solution for 6(c+1)-9(c-5)=3(4c+2) equation:

6(c+1)-9(c-5)=3(4c+2)

We move all terms to the left:

6(c+1)-9(c-5)-(3(4c+2))=0

We multiply parentheses

6c-9c-(3(4c+2))+6+45=0


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

3(4c+2)

We multiply parentheses

12c+6

Back to the equation:

-(12c+6)


We add all the numbers together, and all the variables

-3c-(12c+6)+51=0

We get rid of parentheses

-3c-12c-6+51=0

We add all the numbers together, and all the variables

-15c+45=0

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

-15c=-45

c=-45/-15

c=+3