68=1/3(20c+50)+2

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Solution for 68=1/3(20c+50)+2 equation: 



68=1/3(20c+50)+2

We move all terms to the left:

68-(1/3(20c+50)+2)=0



Domain of the equation: 3(20c+50)+2)!=0

c∈R



We multiply all the terms by the denominator


-(1+68*3(20c+50)+2)=0



We calculate terms in parentheses: -(1+68*3(20c+50)+2), so:

1+68*3(20c+50)+2

determiningTheFunctionDomain
68*3(20c+50)+1+2

We add all the numbers together, and all the variables

68*3(20c+50)+3

Wy multiply elements

204c(2+3

Back to the equation:

-(204c(2+3)



We add all the numbers together, and all the variables

-(204c5=0

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