Questions: Find all real solutions of the following. Check your results. 1/(x-5) - 5/(x^2 - 5x) = 1 Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The solution(s) is/are and the extraneous solution(s) is/are. B. The solution(s) is/are and there are no extraneous solutions. C. There are no real solutions, but an extraneous solution of. D. There are no real solutions.

Find all real solutions of the following. Check your results.

1/(x-5) - 5/(x^2 - 5x) = 1

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
A. The solution(s) is/are and the extraneous solution(s) is/are.
B. The solution(s) is/are and there are no extraneous solutions.
C. There are no real solutions, but an extraneous solution of.
D. There are no real solutions.

Transcript text: Find all real solutions of the following. Check your results. \[ \frac{1}{x-5}-\frac{5}{x^{2}-5 x}=1 \] Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The solution(s) is/are $\square$ and the extraneous solution(s) is/are $\square$ . B. The solution(s) is/are $\square$ and there are no extraneous solutions. C. There are no real solutions, but an extraneous solution of $\square$ . D. There are no real solutions.

Solution

Solution Steps

To solve the equation $\frac{1}{x-5} - \frac{5}{x^2-5x} = 1$, we first need to find a common denominator for the fractions on the left-hand side. The common denominator is $x(x-5)$. We then rewrite each fraction with this common denominator and combine them. After simplifying, we solve the resulting equation for $x$. Finally, we check for any extraneous solutions by substituting back into the original equation.

Step 1: Rewrite the Equation

We start with the equation: \[ \frac{1}{x-5} - \frac{5}{x^2 - 5x} = 1 \] The denominator $x^2 - 5x$ can be factored as $x(x-5)$. Thus, we rewrite the equation as: \[ \frac{1}{x-5} - \frac{5}{x(x-5)} = 1 \]

Step 2: Find a Common Denominator

The common denominator for the left-hand side is $x(x-5)$. We rewrite the fractions: \[ \frac{x}{x(x-5)} - \frac{5}{x(x-5)} = 1 \] This simplifies to: \[ \frac{x - 5}{x(x-5)} = 1 \]

Step 3: Cross-Multiply and Simplify

Cross-multiplying gives us: \[ x - 5 = x(x - 5) \] Expanding the right-hand side: \[ x - 5 = x^2 - 5x \] Rearranging the equation leads to: \[ x^2 - 6x + 5 = 0 \]

Step 4: Solve the Quadratic Equation

We can factor the quadratic: \[ (x - 1)(x - 5) = 0 \] Thus, the solutions are: \[ x = 1 \quad \text{and} \quad x = 5 \]

Step 5: Check for Extraneous Solutions

We substitute $x = 1$ and $x = 5$ back into the original equation. For $x = 1$: \[ \frac{1}{1-5} - \frac{5}{1^2 - 5 \cdot 1} = \frac{1}{-4} - \frac{5}{-4} = \frac{1 + 5}{-4} = \frac{6}{-4} = -\frac{3}{2} \neq 1 \] Thus, $x = 1$ is a valid solution.

For $x = 5$: \[ \frac{1}{5-5} - \frac{5}{5^2 - 5 \cdot 5} \text{ is undefined.} \] Thus, $x = 5$ is an extraneous solution.