Question

1. \\(\frac{1}{n^2 + 11n + 18} = \frac{3}{2n^2 + 22n + 36} + \frac{3}{2n + 4}\\)

Explanation:

Step1: Factor Denominators

Factor each quadratic denominator:

• \( n^2 + 11n + 18 = (n + 2)(n + 9) \)
• \( 2n^2 + 22n + 36 = 2(n^2 + 11n + 18) = 2(n + 2)(n + 9) \)
• \( 2n + 4 = 2(n + 2) \)

The equation becomes:
\( \frac{1}{(n + 2)(n + 9)} = \frac{3}{2(n + 2)(n + 9)} + \frac{3}{2(n + 2)} \)

Step2: Multiply by LCM

The least common multiple (LCM) of the denominators \( (n + 2)(n + 9) \) and \( 2(n + 2) \) is \( 2(n + 2)(n + 9) \). Multiply each term by \( 2(n + 2)(n + 9) \) to eliminate denominators:
\( 2(n + 2)(n + 9) \cdot \frac{1}{(n + 2)(n + 9)} = 2(n + 2)(n + 9) \cdot \frac{3}{2(n + 2)(n + 9)} + 2(n + 2)(n + 9) \cdot \frac{3}{2(n + 2)} \)

Simplify each term:

• Left term: \( 2 \)
• Middle term: \( 3 \)
• Right term: \( 3(n + 9) \)

Equation: \( 2 = 3 + 3(n + 9) \)

Step3: Solve for \( n \)

Simplify the right-hand side:
\( 2 = 3 + 3n + 27 \)
\( 2 = 3n + 30 \)

Subtract \( 30 \) from both sides:
\( 2 - 30 = 3n \)
\( -28 = 3n \)

Divide by \( 3 \):
\( n = -\frac{28}{3} \)

Step4: Check Excluded Values

Denominators cannot be zero:

• \( n + 2

eq 0 \implies n
eq -2 \)

• \( n + 9

eq 0 \implies n
eq -9 \)

\( -\frac{28}{3} \approx -9.33 \), which is not \( -2 \) or \( -9 \), so it is valid.

Answer:

\( n = \boldsymbol{-\frac{28}{3}} \)