8=1/5(c+2)

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Solution for 8=1/5(c+2) equation: 



8=1/5(c+2)

We move all terms to the left:

8-(1/5(c+2))=0



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

c∈R



We multiply all the terms by the denominator


-(1+8*5(c+2))=0



We calculate terms in parentheses: -(1+8*5(c+2)), so:

1+8*5(c+2)

determiningTheFunctionDomain
8*5(c+2)+1

Wy multiply elements

40c(c+1

Back to the equation:

-(40c(c+1)





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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40c(c+1





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