-4(11c+9)+40c=6(c-8)+3c

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Solution for -4(11c+9)+40c=6(c-8)+3c equation: 



-4(11c+9)+40c=6(c-8)+3c

We move all terms to the left:

-4(11c+9)+40c-(6(c-8)+3c)=0

We add all the numbers together, and all the variables

40c-4(11c+9)-(6(c-8)+3c)=0

We multiply parentheses

40c-44c-(6(c-8)+3c)-36=0



We calculate terms in parentheses: -(6(c-8)+3c), so:

6(c-8)+3c

We add all the numbers together, and all the variables

3c+6(c-8)

We multiply parentheses

3c+6c-48

We add all the numbers together, and all the variables

9c-48

Back to the equation:

-(9c-48)



We add all the numbers together, and all the variables

-4c-(9c-48)-36=0

We get rid of parentheses

-4c-9c+48-36=0

We add all the numbers together, and all the variables

-13c+12=0

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

-13c=-12

c=-12/-13

c=12/13

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