Question

Solve the equation for x: 1/x-5 = 1/x+3 + x-1/x^2-2x-15

Answer 1

To solve the equation

\[
\frac{1}{x} - 5 = \frac{1}{x + 3} + \frac{x - 1}{x^2 - 2x - 15},
\]

we first factor the quadratic in the denominator on the right side. The expression \( x^2 - 2x - 15 \) factors to

\[
(x - 5)(x + 3).
\]

Therefore, we can rewrite the equation as

\[
\frac{1}{x} - 5 = \frac{1}{x + 3} + \frac{x - 1}{(x - 5)(x + 3)}.
\]

Next, we multiply both sides of the equation by \( x(x + 3)(x - 5) \) to eliminate the denominators. This gives us:

\[
(x - 5)(x)(x + 3) \left(\frac{1}{x} - 5\right) = (x - 5)(x) + (x - 1)(x).
\]

The left side becomes:

\[
(x - 5)(x + 3) - 5x(x - 5)(x + 3).
\]

Distributing the terms:

\[
(x^2 - 2x - 15) - 5x^2 + 25x,
\]
which simplifies to:
\[
-4x^2 + 23x - 15.
\]

Reorganizing gives us:

\[
-4x^2 + 23x - 15 = x^2 - 5x.
\]

Now, moving all terms to one side:

\[
-4x^2 + 23x - 15 - x^2 + 5x = 0.
\]
This simplifies to:

\[
-5x^2 + 28x - 15 = 0.
\]

Multiplying through by -1 gives:

\[
5x^2 - 28x + 15 = 0.
\]

Next, we apply the quadratic formula where \( a = 5 \), \( b = -28 \), and \( c = 15 \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 5 \cdot 15}}{2 \cdot 5}.
\]

Calculating the discriminant:

\[
28^2 - 4 \cdot 5 \cdot 15 = 784 - 300 = 484.
\]

Now substituting this back into the quadratic formula:

\[
x = \frac{28 \pm \sqrt{484}}{10} = \frac{28 \pm 22}{10}.
\]

Calculating the two cases:

1. For the positive root:
\[
x = \frac{28 + 22}{10} = \frac{50}{10} = 5.
\]

2. For the negative root:
\[
x = \frac{28 - 22}{10} = \frac{6}{10} = \frac{3}{5}.
\]

Therefore, the potential solutions are \( x = 5 \) and \( x = \frac{3}{5} \).

Next, we need to check for restrictions from the original equation denominators, specifically:

1. \( x \neq 0 \)
2. \( x + 3 \neq 0 \), thus \( x \neq -3 \)
3. \( x - 5 \neq 0 \), thus \( x \neq 5 \)

Since \( x = 5 \) results in a division by zero, it is not a valid solution.

Thus, the only valid solution is

\[
\boxed{\frac{3}{5}}.
\]