To solve the equation
\[ \frac{3}{x - 1} - \frac{1}{5} = \frac{4}{x + 1}, \]
we will first eliminate the fractions by finding a common denominator.
The common denominator for the fractions is \(5(x - 1)(x + 1)\). We can multiply the entire equation by this common denominator:
\[ 5(x - 1)(x + 1)\left(\frac{3}{x - 1}\right) - 5(x - 1)(x + 1)\left(\frac{1}{5}\right) = 5(x - 1)(x + 1)\left(\frac{4}{x + 1}\right). \]
This simplifies to:
\[ 15(x + 1) - (x - 1)(x + 1) = 20(x - 1). \]
Now expand each term:
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For \(15(x + 1)\): \[ 15x + 15. \]
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For \((x - 1)(x + 1)\) (difference of squares): \[ x^2 - 1. \]
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For \(20(x - 1)\): \[ 20x - 20. \]
Putting it all together, the equation becomes:
\[ 15x + 15 - (x^2 - 1) = 20x - 20. \]
Simplifying gives:
\[ 15x + 15 - x^2 + 1 = 20x - 20, \] \[ 15x + 16 - x^2 = 20x - 20. \]
Rearranging all terms to one side:
\[ -x^2 + 15x + 16 - 20x + 20 = 0, \] \[ -x^2 - 5x + 36 = 0. \]
Multiplying through by -1 to make the equation standard:
\[ x^2 + 5x - 36 = 0. \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \(a = 1\), \(b = 5\), and \(c = -36\). Plugging in these values:
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1}. \]
Calculating the discriminant:
\[ b^2 - 4ac = 25 + 144 = 169. \]
Thus,
\[ x = \frac{-5 \pm \sqrt{169}}{2}, \] \[ x = \frac{-5 \pm 13}{2}. \]
This gives us two solutions:
- \(x = \frac{8}{2} = 4,\)
- \(x = \frac{-18}{2} = -9.\)
We have two potential solutions: \(x = 4\) and \(x = -9\).
Finally, let's check for any restrictions in the original equation. Since \(x\) cannot be \(1\) or \(-1\) (due to denominators), both solutions \(4\) and \(-9\) are valid.
Thus, the solutions to the equation are
\[ \boxed{4} \quad \text{and} \quad \boxed{-9}. \]