Mr. Knotts found the difference of the following expression. Which statement is true about Mr. Knotts's work?

[tex]\[
\frac{x}{x^2-1}-\frac{1}{x-1}
\][/tex]

Step 1: [tex]\[\frac{x}{(x+1)(x-1)} - \frac{1}{x-1}\][/tex]

Step 2: [tex]\[\frac{x}{(x+1)(x-1)} - \frac{1(x+1)}{(x+1)(x-1)}\][/tex]

Step 3: [tex]\[\frac{x - (x + 1)}{(x+1)(x-1)}\][/tex]

Step 4: [tex]\[\frac{-1}{(x+1)(x-1)}\][/tex]

A. Mr. Knotts correctly simplified the expression.

B. Mr. Knotts made an error in Step 1.

C. Mr. Knotts made an error in Step 2.

D. Mr. Knotts made an error in Step 3.

E. Mr. Knotts made an error in Step 4.



Answer :

Let's examine Mr. Knotts's work step-by-step to verify whether it’s correct.

We start with:
[tex]\[ \frac{x}{x^2 - 1} - \frac{1}{x - 1} \][/tex]

### Step 1
[tex]\[ \frac{x}{(x+1)(x-1)} - \frac{1}{x-1} \][/tex]

This factorization of [tex]\( x^2 - 1 \)[/tex] as [tex]\((x + 1)(x - 1)\)[/tex] is correct.

### Step 2
[tex]\[ \frac{x}{(x+1)(x-1)} - \frac{1(x + 1)}{(x+1)(x-1)} \][/tex]

Mr. Knotts found a common denominator, which is [tex]\((x + 1)(x - 1)\)[/tex]. He then adjusted the second fraction to have this common denominator. To do this properly, he multiplied the numerator and the denominator of the second term by [tex]\( x + 1 \)[/tex]. This is correct.

### Step 3
[tex]\[ \frac{x - (x + 1)}{(x+1)(x-1)} \][/tex]

However, there is a mistake in this step. The term should be:
[tex]\[ \frac{x - (x + 1)}{(x + 1)(x - 1)} = \frac{x - x - 1}{(x + 1)(x - 1)} \][/tex]

The correction should lead to:
[tex]\[ \frac{-1}{(x + 1)(x - 1)} \][/tex]

### Step 4
Mr. Knotts's work shows:
[tex]\[ \frac{1}{(x+1)(x-1)} \][/tex]

However, the correct simplified form is:
[tex]\[ \frac{-1}{(x+1)(x-1)} \][/tex]

Thus, upon review, it appears Mr. Knotts made a mistake in the sign while simplifying the numerator in Step 3. Therefore, the correct form should have been [tex]\( \frac{-1}{(x+1)(x-1)} \)[/tex].

### Conclusion
The correct statement is that Mr. Knotts should have obtained:
[tex]\[ \frac{-1}{(x + 1)(x - 1)} \][/tex]

Hence, his final step should reflect this negative sign. Therefore, Mr. Knotts’s work in Step 3 and Step 4 is incorrect due to the sign error in simplifying the numerator.