41=5/9(f-32)

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Solution for 41=5/9(f-32) equation: 



41=5/9(f-32)

We move all terms to the left:

41-(5/9(f-32))=0



Domain of the equation: 9(f-32))!=0

f∈R



We multiply all the terms by the denominator


-(5+41*9(f-32))=0



We calculate terms in parentheses: -(5+41*9(f-32)), so:

5+41*9(f-32)

determiningTheFunctionDomain
41*9(f-32)+5

Wy multiply elements

369f(f+5

Back to the equation:

-(369f(f+5)





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369f(f+5





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369f(f+5





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369f(f+5





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