128=4/9(n+42)

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Solution for 128=4/9(n+42) equation: 



128=4/9(n+42)

We move all terms to the left:

128-(4/9(n+42))=0



Domain of the equation: 9(n+42))!=0

n∈R



We multiply all the terms by the denominator


-(4+128*9(n+42))=0



We calculate terms in parentheses: -(4+128*9(n+42)), so:

4+128*9(n+42)

determiningTheFunctionDomain
128*9(n+42)+4

Wy multiply elements

1152n(n+4

Back to the equation:

-(1152n(n+4)





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