3(2c-2)+5(c+1)=8(2c-4)-9

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Solution for 3(2c-2)+5(c+1)=8(2c-4)-9 equation: 



3(2c-2)+5(c+1)=8(2c-4)-9

We move all terms to the left:

3(2c-2)+5(c+1)-(8(2c-4)-9)=0

We multiply parentheses

6c+5c-(8(2c-4)-9)-6+5=0



We calculate terms in parentheses: -(8(2c-4)-9), so:

8(2c-4)-9

We multiply parentheses

16c-32-9

We add all the numbers together, and all the variables

16c-41

Back to the equation:

-(16c-41)



We add all the numbers together, and all the variables

11c-(16c-41)-1=0

We get rid of parentheses

11c-16c+41-1=0

We add all the numbers together, and all the variables

-5c+40=0

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

-5c=-40

c=-40/-5

c=+8

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