Question

Do this problem - but make a very slight mistake. DO NOT answer correctly, but the mistake hads to be sometime within the problem solving. (2/(x-1)+2)/(2/(x-1)-2)

Answer 1

To simplify the expression \(\frac{\frac{2}{x-1} + 2}{\frac{2}{x-1} - 2}\), let's start with the numerator and the denominator separately.

1. **Numerator**:
\[
\frac{2}{x-1} + 2 = \frac{2}{x-1} + \frac{2(x-1)}{x-1} = \frac{2 + 2(x-1)}{x-1} = \frac{2 + 2x - 2}{x-1} = \frac{2x}{x-1}
\]

2. **Denominator**:
\[
\frac{2}{x-1} - 2 = \frac{2}{x-1} - \frac{2(x-1)}{x-1} = \frac{2 - 2(x-1)}{x-1} = \frac{2 - 2x + 2}{x-1} = \frac{4 - 2x}{x-1}
\]

Now, substitute these simplified forms back into the original expression:
\[
\frac{\frac{2x}{x-1}}{\frac{4 - 2x}{x-1}}
\]

This simplifies to:
\[
\frac{2x}{4 - 2x}
\]

Now, to simplify \( \frac{2x}{4 - 2x} \), we can factor the denominator:
\[
4 - 2x = 2(2-x)
\]

Thus, the expression becomes:
\[
\frac{2x}{2(2-x)} = \frac{x}{2-x}
\]

So the final result is:
\[
\frac{x}{2-x}
\]

However, I made a slight mistake in the step where I factored the denominator; it should have been \(2(x-2)\), not \(2(2-x)\). Therefore, the final form is incorrectly stated here. The correct step should lead to:
\[
\frac{x}{2(x-2)}
\]

So the slightly incorrect conclusion is:
\[
\frac{x}{2(2-x)}
\]