9(c+2c)-11=5(2c+3)-2c

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Solution for 9(c+2c)-11=5(2c+3)-2c equation: 



9(c+2c)-11=5(2c+3)-2c

We move all terms to the left:

9(c+2c)-11-(5(2c+3)-2c)=0

We add all the numbers together, and all the variables

9(+3c)-(5(2c+3)-2c)-11=0

We multiply parentheses

27c-(5(2c+3)-2c)-11=0



We calculate terms in parentheses: -(5(2c+3)-2c), so:

5(2c+3)-2c

We add all the numbers together, and all the variables

-2c+5(2c+3)

We multiply parentheses

-2c+10c+15

We add all the numbers together, and all the variables

8c+15

Back to the equation:

-(8c+15)



We get rid of parentheses

27c-8c-15-11=0

We add all the numbers together, and all the variables

19c-26=0

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

19c=26

c=26/19

c=1+7/19

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