Question

1. \\(\frac{9n - 45}{n^2 + 6n} = \frac{1}{n} - 1\\)

Explanation:

Step1: Find the LCD

The denominators are \(n^2 + 6n=n(n + 6)\), \(n\). So the least common denominator (LCD) is \(n(n + 6)\). Multiply each term by \(n(n + 6)\) to eliminate the fractions:

$$ \frac{9n-45}{n(n + 6)}\times n(n + 6)=\frac{1}{n}\times n(n + 6)-1\times n(n + 6) $$

Simplify each term:
\(9n - 45=(n + 6)-n(n + 6)\)

Step2: Simplify the right - hand side

Expand the right - hand side:
\(9n-45=n + 6-n^{2}-6n\)
Combine like terms on the right - hand side:
\(9n-45=-n^{2}-5n + 6\)

Step3: Rearrange into standard quadratic form

Bring all terms to one side to get a quadratic equation:
\(n^{2}+9n + 5n-45 - 6 = 0\)
\(n^{2}+14n-51 = 0\)

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(- 51\) and add up to \(14\). The numbers are \(17\) and \(-3\).
So, \(n^{2}+14n - 51=(n + 17)(n-3)=0\)

Step5: Solve for n

Set each factor equal to zero:
\(n+17 = 0\) or \(n - 3=0\)
\(n=-17\) or \(n = 3\)

Step6: Check for extraneous solutions

We need to check if these solutions make the original denominators zero.
For \(n=-17\): The denominators \(n^2 + 6n=(-17)^2+6\times(-17)=289 - 102 = 187
eq0\) and \(n=-17
eq0\).
For \(n = 3\): The denominators \(n^2+6n=9 + 18=27
eq0\) and \(n = 3
eq0\).

Answer:

The solutions are \(n=-17\) and \(n = 3\)