Question

solve the equation. remember to check for extraneous solutions.
\frac{n + 6}{n}=\frac{n^{2}-2n - 15}{n^{2}}+\frac{5}{n}

o n = 4 and n = - 1
o n = - 4 and n = 1
o n = 5
o n = - 5

question #2
solve the equation. remember to check for extraneous solutions.
\frac{9}{x}=\frac{3x - 9}{x}

o x = \frac{2}{3}
o x = 6
o x = 1
o x = 3

Explanation:

Step1: Clear the denominators

Multiply each term by $n^2$ (for the first - equation) and $x$ (for the second - equation).
For the first equation $\frac{n + 6}{n}=\frac{n^2-2n - 15}{n^2}+\frac{5}{n}$, we get $(n + 6)n=n^2-2n - 15+5n$.
Expand the left - hand side: $n^2+6n=n^2-2n - 15+5n$.
Subtract $n^2$ from both sides: $6n=-2n - 15+5n$.
Combine like terms: $6n - 5n+2n=-15$, $3n=-15$, so $n=-5$.
We need to check: when $n = - 5$, the original equation's denominators $n$ and $n^2$ are non - zero.
For the second equation $\frac{9}{x}=\frac{3x - 9}{x}$, multiply both sides by $x$ (where $x
eq0$), we get $9 = 3x-9$.
Add 9 to both sides: $3x=9 + 9=18$.
Divide both sides by 3: $x = 6$.
We need to check: when $x = 6$, the original equation's denominator $x$ is non - zero.

Answer:

For the first equation: $n=-5$
For the second equation: $x = 6$