-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+13

Back to the equation:

-(6c+13)



We add all the numbers together, and all the variables

-9c-(6c+13)-4=0

We get rid of parentheses

-9c-6c-13-4=0

We add all the numbers together, and all the variables

-15c-17=0

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

-15c=17

c=17/-15

c=-1+2/15

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