Solve: \( \frac{5}{a-1}+\frac{2}{a-7}=\frac{a^{2}+4 a-65}{(a-1)(a-7)} \) \( a=\square \) (Enter your answer(s), separated by a comma, or DNE if there is no solution)
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To solve the equation \[ \frac{5}{a-1}+\frac{2}{a-7}=\frac{a^{2}+4 a-65}{(a-1)(a-7)}, \] we can start by finding a common denominator for the left side. The common denominator is \((a-1)(a-7)\). Rewriting the left-hand side gives: \[ \frac{5(a-7) + 2(a-1)}{(a-1)(a-7)} = \frac{5a - 35 + 2a - 2}{(a-1)(a-7)} = \frac{7a - 37}{(a-1)(a-7)}. \] Now, we set the numerators equal to each other since the denominators are the same: \[ 7a - 37 = a^2 + 4a - 65. \] Rearranging gives: \[ 0 = a^2 - 3a - 28. \] Factoring this quadratic equation: \[ 0 = (a - 7)(a + 4). \] Setting each factor to zero gives us the solutions: \[ a - 7 = 0 \quad \Rightarrow \quad a = 7, \] \[ a + 4 = 0 \quad \Rightarrow \quad a = -4. \] However, we must check for any extraneous solutions by substituting back into the original denominators: - For \(a = 7\), the term \(\frac{5}{a-1} = \frac{5}{6}\) is valid but \(\frac{2}{a-7}\) becomes undefined since the denominator is zero. - For \(a = -4\), both denominators are valid: \(\frac{5}{-5} = -1\) and \(\frac{2}{-11} = -\frac{2}{11}\), hence valid. Thus, the only valid solution is: \[ a = -4. \] So the answer is \( a = -4 \).
