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

How do I solve this math problem?

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[tex] \frac{1}{n-8} -1= \frac{7}{n-8} [/tex]

[tex] \frac{1-(n-8)}{n-8}= \frac{7}{n-8} [/tex]

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

[tex] \frac{1-(n-8)}{n-8}*({n-8})= \frac{7}{n-8} *({n-8})[/tex]

[tex]1-(n-8) = 7[/tex]

[tex]1-n+8 = 7[/tex]

[tex]-n = 7-1-8[/tex]

[tex]-n = -2[/tex]

dividing both sides by -1, we get,

[tex]n =2[/tex]
[tex]\frac{1}{n-8}-1=\frac{7}{n-8};\ D:n\neq8\\\\\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}[/tex]
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