120-(11c+11)=6-(3c-4c+7)

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Solution for 120-(11c+11)=6-(3c-4c+7) equation: 



120-(11c+11)=6-(3c-4c+7)

We move all terms to the left:

120-(11c+11)-(6-(3c-4c+7))=0

We add all the numbers together, and all the variables

-(11c+11)-(6-(-1c+7))+120=0

We get rid of parentheses

-11c-(6-(-1c+7))-11+120=0



We calculate terms in parentheses: -(6-(-1c+7)), so:

6-(-1c+7)

determiningTheFunctionDomain
-(-1c+7)+6

We get rid of parentheses

1c-7+6

We add all the numbers together, and all the variables

c-1

Back to the equation:

-(c-1)



We add all the numbers together, and all the variables

-11c-(c-1)+109=0

We get rid of parentheses

-11c-c+1+109=0

We add all the numbers together, and all the variables

-12c+110=0

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

-12c=-110

c=-110/-12

c=9+1/6

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