Gregg simplified the rational expression $$\frac{x}{x-1}-\frac{1}{1-x}$$ in the following steps:
(1) He determined the condition $x\neq1$ right at the beginning.
(2) He modified the second fraction as follows: $$\frac{x}{x-1}-\frac{1}{1-x}=\frac{x}{x-1}+\frac{1}{x-1}$$
(3) Then, he eliminated the fractions by multiplying the rational expresion by $x-1$:
$$\frac{x}{x-1}+\frac{1}{x-1}=x+1$$
Is his solution correct? If not, identify all his mistakes.
Yes. The whole solution is perfectly fine.
No, his solution is not correct. The mistake is in step (2). He should have modified the second fraction as follows: $$\frac{x}{x-1}-\frac{1}{1-x}=\frac{x}{x-1}-\frac{1}{x-1}$$ Then he should have eliminated the fractions by multiplying by $x-1$ and obtained the result $$\frac{x}{x-1}-\frac{1}{x-1}=x-1$$
No, his solution is not correct. The mistake is in step (3). Multiplying the expression by $x-1$ changes its value. The correct step (3) is:
$$\frac{x}{x-1}+\frac{1}{x-1}=\frac{x+1}{x-1}$$
No, his solution is not correct. The mistakes are in both steps (2) and (3). Gregg should have modified the second fraction as follows: $$\frac{x}{x-1}-\frac{1}{1-x}=\frac{x}{x-1}-\frac{1}{x-1}$$
The next step should then have been:$\frac{x}{x-1}-\frac{1}{x-1}=\frac{x-1}{x-1}=1$.
Correct solution:
$$\frac{x}{x-1}-\frac{1}{1-x}=\frac{x}{x-1}+\frac{1}{x-1}=\frac{x+1}{x-1},\quad x\neq1$$