Question

Solve.
StartFraction 3 Over x minus 1 EndFraction
minusone fifth
equalsStartFraction 4 Over x plus 1 EndFraction

Answer 1

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:

1. For \(15(x + 1)\):
\[
15x + 15.
\]

2. For \((x - 1)(x + 1)\) (difference of squares):
\[
x^2 - 1.
\]

3. 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:

1. \(x = \frac{8}{2} = 4,\)
2. \(x = \frac{-18}{2} = -9.\)

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

\[
\boxed{4} \quad \text{and} \quad \boxed{-9}.
\]