6(11-c)=2(3+2c)

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Solution for 6(11-c)=2(3+2c) equation: 



6(11-c)=2(3+2c)

We move all terms to the left:

6(11-c)-(2(3+2c))=0

We add all the numbers together, and all the variables

6(-1c+11)-(2(2c+3))=0

We multiply parentheses

-6c-(2(2c+3))+66=0



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

2(2c+3)

We multiply parentheses

4c+6

Back to the equation:

-(4c+6)



We get rid of parentheses

-6c-4c-6+66=0

We add all the numbers together, and all the variables

-10c+60=0

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

-10c=-60

c=-60/-10

c=+6

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