5=1/2(c+2)

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



5=1/2(c+2)

We move all terms to the left:

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



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

c∈R



We multiply all the terms by the denominator


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



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

1+5*2(c+2)

determiningTheFunctionDomain
5*2(c+2)+1

Wy multiply elements

10c(c+1

Back to the equation:

-(10c(c+1)





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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10c(c+1





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