180=2c+6+3c+4+(1/2)c+5

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Solution for 180=2c+6+3c+4+(1/2)c+5 equation: 



180=2c+6+3c+4+(1/2)c+5

We move all terms to the left:

180-(2c+6+3c+4+(1/2)c+5)=0



Domain of the equation: 2)c+5)!=0

c!=0/1

c!=0

c∈R



We add all the numbers together, and all the variables

-(2c+6+3c+4+(+1/2)c+5)+180=0

We multiply all the terms by the denominator


-(2c+6+3c+4+(+1+180*2)c+5)=0



We calculate terms in parentheses: -(2c+6+3c+4+(+1+180*2)c+5), so:

2c+6+3c+4+(+1+180*2)c+5

determiningTheFunctionDomain
2c+3c+(+1+180*2)c+6+4+5

We add all the numbers together, and all the variables

2c+3c+361c+6+4+5

We add all the numbers together, and all the variables

366c+15

Back to the equation:

-(366c+15)



We get rid of parentheses

-366c-15=0

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

-366c=15

c=15/-366

c=-5/122

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