To solve the equation:

\[
\frac{x - 14}{x + 1} - \frac{x}{x - 1} = 0
\]

we will follow these steps:

### Step 1: Combine the two fractions
To combine the fractions, find a common denominator. The common denominator between \(x + 1\) and \(x - 1\) is \((x + 1)(x - 1)\).

Thus, rewrite the expression with the common denominator:

\[
\frac{(x - 14)(x - 1)}{(x + 1)(x - 1)} - \frac{x(x + 1)}{(x + 1)(x - 1)} = 0
\]

### Step 2: Simplify the numerators
Expand the numerators:

1. For the first term: \((x - 14)(x - 1)\)
\[
(x - 14)(x - 1) = x^2 - x - 14x + 14 = x^2 - 15x + 14
\]

2. For the second term: \(x(x + 1)\)
\[
x(x + 1) = x^2 + x
\]

### Step 3: Combine the two terms
Now we can combine the numerators, keeping the denominator the same:

\[
\frac{x^2 - 15x + 14 - (x^2 + x)}{(x + 1)(x - 1)} = 0
\]

Simplify the numerator:
\[
x^2 - 15x + 14 - x^2 - x = -16x + 14
\]

So the expression becomes:
\[
\frac{-16x + 14}{(x + 1)(x - 1)} = 0
\]

### Step 4: Solve for \(x\)
For a fraction to be zero, the numerator must be zero. So we set the numerator equal to zero:

\[
-16x + 14 = 0
\]

Solve for \(x\):

\[
-16x = -14
\]
\[
x = \frac{-14}{-16} = \frac{7}{8}
\]

### Step 5: Check for restrictions
The original denominators were \(x + 1\) and \(x - 1\), so \(x \neq 1\) and \(x \neq -1\). Since \(x = \frac{7}{8}\) does not violate these restrictions, the solution is valid.

Thus, the solution is:
\[
x = \frac{7}{8}
\]

Would you like any additional explanation or details?

### Related Questions:
1. How do you find a common denominator in rational expressions?
2. What is the process for expanding binomials?
3. Why is it necessary to check for restrictions in rational equations?
4. How can you verify a solution to a rational equation?
5. What are other methods to solve rational equations?

### Tip:
Always check for any restrictions on the variable (i.e., values that make the denominator zero) when solving rational equations.

Learn how to solve the rational equation (x - 14) / (x + 1) − x / (x − 1) = 0 by finding a common denominator, simplifying, and solving the resulting linear equation. This solution covers algebraic concepts such as combining fractions and solving linear equations, suitable for students in Grades 9-11.

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