quation: \( \frac{5 s}{s^{2}-81}+\frac{4}{s-9}=\frac{1}{s+9} \)
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To solve the equation \( \frac{5 s}{s^{2}-81}+\frac{4}{s-9}=\frac{1}{s+9} \), we first recognize that \( s^2 - 81 \) can be factored as \( (s - 9)(s + 9) \). This gives us a common denominator: \[ \frac{5s}{(s - 9)(s + 9)} + \frac{4(s + 9)}{(s - 9)(s + 9)} = \frac{1(s - 9)}{(s - 9)(s + 9)} \] Now, we can combine the left side over the common denominator: \[ \frac{5s + 4(s + 9)}{(s - 9)(s + 9)} = \frac{s - 9}{(s - 9)(s + 9)} \] Next, let's simplify the left side: \[ 5s + 4s + 36 = 9s + 36 \] This gives us: \[ \frac{9s + 36}{(s - 9)(s + 9)} = \frac{s - 9}{(s - 9)(s + 9)} \] Now, equate the numerators (as the denominators are the same): \[ 9s + 36 = s - 9 \] Moving all terms involving \( s \) to one side and constant terms to the other: \[ 9s - s = -9 - 36 \] \[ 8s = -45 \] \[ s = -\frac{45}{8} \] So, the solution to the equation is \( s = -\frac{45}{8} \).