Verified Answer :

rabinshrestha41

\(\frac{1}{n-8} -1= \frac{7}{n-8}\)

\(\frac{1-(n-8)}{n-8}= \frac{7}{n-8}\)

Multiplying by (n-8) on both sides, we get,

\(\frac{1-(n-8)}{n-8}*({n-8})= \frac{7}{n-8} *({n-8})\)

\(1-(n-8) = 7\)

\(1-n+8 = 7\)

\(-n = 7-1-8\)

\(-n = -2\)

dividing both sides by -1, we get,

\(n =2\)

\(\frac{1}{n-8}-1=\frac{7}{n-8};\ D:n eq8\\\\\frac{1}{n-8}-\frac{n-8}{n-8}=\frac{7}{n-8}\\\\\frac{1-(n-8)}{n-8}=\frac{7}{n-8}\\\\\frac{1-n+8}{n-8}=\frac{7}{n-8}\\\\\frac{9-n}{n-8}=\frac{7}{n-8}\iff9-n=7\ \ \ \ \ \ |subtract\ 9\ from\ both\ sides\\\\-n=-2\\\\\boxed{n=2\in D}\)

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