To solve the equation
\[ \frac{36}{3x - 15} + \frac{1}{x - 5} = \frac{13}{3}, \]
let's first simplify it step by step.
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Simplify \(3x - 15\): \[ 3x - 15 = 3(x - 5). \] Therefore, we can rewrite the first fraction: \[ \frac{36}{3x - 15} = \frac{36}{3(x - 5)} = \frac{12}{x - 5}. \]
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Rewrite the equation with the simplified fraction: \[ \frac{12}{x - 5} + \frac{1}{x - 5} = \frac{13}{3}. \] Combine the fractions on the left: \[ \frac{12 + 1}{x - 5} = \frac{13}{x - 5}. \] Thus, the equation becomes: \[ \frac{13}{x - 5} = \frac{13}{3}. \]
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Cross-multiply: \[ 13 \cdot 3 = 13 \cdot (x - 5). \] Canceling out \(13\) (assuming \(x - 5 \neq 0\)): \[ 3 = x - 5. \]
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Solve for \(x\): \[ x = 3 + 5 = 8. \]
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Check for extraneous solutions: We need to verify that \(x = 8\) does not make any denominators zero. The original denominators were \(3x - 15\) and \(x - 5\). For \(x = 8\): \[ 3(8) - 15 = 24 - 15 = 9 \quad (\text{not zero}), \] \[ 8 - 5 = 3 \quad (\text{not zero}). \]
Thus, the solution is valid.
The final answer is: \[ \boxed{8}. \]