Let's solve the equation \( \frac{1}{n-8} - 1 = \frac{7}{n-8} \).

### Step 1: Simplify the equation

We start by moving everything to a common denominator:

\[
\frac{1}{n-8} - \frac{n-8}{n-8} = \frac{7}{n-8}
\]

Simplify the left side:

\[
\frac{1 - (n-8)}{n-8} = \frac{7}{n-8}
\]

Distribute the negative sign:

\[
\frac{1 - n + 8}{n-8} = \frac{7}{n-8}
\]

Combine like terms:

\[
\frac{9 - n}{n-8} = \frac{7}{n-8}
\]

### Step 2: Set the numerators equal

Since the denominators are the same, we can set the numerators equal to each other:

\[
9 - n = 7
\]

### Step 3: Solve for \( n \)

Subtract 9 from both sides:

\[
-n = 7 - 9
\]

\[
-n = -2
\]

Multiply both sides by -1:

\[
n = 2
\]

### Step 4: Check the solution

Substitute \( n = 2 \) back into the original equation to verify:

\[
\frac{1}{2-8} - 1 = \frac{7}{2-8}
\]

Simplify:

\[
\frac{1}{-6} - 1 = \frac{7}{-6}
\]

Convert \( 1 \) to a fraction with a denominator of \(-6\):

\[
\frac{1}{-6} - \frac{6}{-6} = \frac{7}{-6}
\]

Simplify the left side:

\[
\frac{1 - 6}{-6} = \frac{7}{-6}
\]

\[
\frac{-5}{-6} = \frac{7}{-6}
\]

Since \(\frac{-5}{-6} \neq \frac{7}{-6}\), there was an error in reasoning. Upon re-evaluating, it turns out no value of \( n \) satisfies the original equation exactly.

Therefore, **there is no solution** for this equation.

Would you like to know more about why this result came about?

---

Here are 5 related questions for further exploration:

1. How do you solve equations involving variables in the denominator?
2. What are extraneous solutions, and how can they arise?
3. Can you explain the concept of "no solution" in algebraic equations?
4. How do you verify solutions for rational equations?
5. What is the difference between a consistent and inconsistent equation?

**Tip:** When solving rational equations, always check your solutions by substituting them back into the original equation to ensure they satisfy all terms.

Learn how to solve the equation 1/(n-8) - 1 = 7/(n-8) step-by-step. This problem involves rational equations and equations with variables in the denominator. Suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation