LaTeX practice
To solve the equation \(\frac{1}{n-8} - 1 = \frac{7}{n-8}\), we will follow a systematic approach, including checking for extraneous solutions. ### Step 1: Combine the terms on the left side We first want to get a common denominator on the left side. The left side can be rewritten as: \[ \frac{1}{n-8} - 1 = \frac{1}{n-8} - \frac{n-8}{n-8} = \frac{1 - (n-8)}{n-8} = \frac{1 - n + 8}{n-8} = \frac{9 - n}{n-8} \] So, the equation now looks like: \[ \frac{9 - n}{n-8} = \frac{7}{n-8} \] ### Step 2: Eliminate the denominators Since the denominators are the same, we can multiply both sides by \(n - 8\) (assuming \(n \neq 8\)): \[ 9 - n = 7 \] ### Step 3: Solve for \(n\) Now we can solve for \(n\): \[ 9 - n = 7 \] Subtract \(9\) from both sides: \[ -n = 7 - 9 \] \[ -n = -2 \] Multiply both sides by \(-1\): \[ n = 2 \] ### Step 4: Check for extraneous solutions We need to check if our solution \(n = 2\) satisfies the original equation. We substitute \(n = 2\) back into the original equation: The original equation is: \[ \frac{1}{n-8} - 1 = \frac{7}{n-8} \] Substituting \(n = 2\): \[ \frac{1}{2-8} - 1 = \frac{7}{2-8} \] \[ \frac{1}{-6} - 1 = \frac{7}{-6} \] Convert \(1\) into a fraction: \[ \frac{1}{-6} - \frac{6}{6} = \frac{7}{-6} \] Now simplify the left side: \[ \frac{1 - 6}{-6} = \frac{-5}{-6} = \frac{5}{6} \] The right side: \[ \frac{7}{-6} = \frac{-7}{6} \] Since \(\frac{5}{6} \neq \frac{-7}{6}\), \(n = 2\) does not satisfy the original equation. ### Conclusion The solution \(n = 2\) is an extraneous solution. Since our procedure yielded no other possible values, we conclude that **there are no solutions** for the equation \(\frac{1}{n-8} - 1 = \frac{7}{n-8}\).
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