x is replaced by 1 - x everywhere in the expression 1/x - 1/(1 - x)

I am getting D as the answer.
After replacing x by 1-x the expression is
1/(1-x) - 1/(1-(1-x) = 1/(1-x) - 1/x
now if i simplify this expression by taking lcm i get ---> (2x-1)/x(x-1)
upon multiplying it with x(x-1) i get 2x-1.

Initial equation is

\(\frac{1}{x}\) - \(\frac{1}{(1 - x)}\)

Replace \(x\) with \(1-x\)

\(\frac{1}{(1 - x)}\) - \(\frac{1}{x}\) = \(\frac{(x-1+x)}{(1-x)(x)}\) =\(\frac{2x-1}{(1-x)(x)}\)

Multiply above equation with \(x^2 - x\) i.e. \((x-1)x\)

\(\frac{(2x-1)*(x-1)(x)}{(1-x)(x)}\) = \(1-2x\)

Answer: E

Regards,
Ammu

As per the question let's replace the value of x

\(\frac{1}{1-x}\) - \(\frac{1}{1-(1-x)}\)

\(\frac{1}{1-x}\) - \(\frac{1}{x}\)

\(\frac{1-2x}{x^2 - x}\)

Multiply above equation by \(x^{2}\) - x

= 1 - 2x

Answer E

Official Solution:

\(x\) is replaced by \(1 - x\) everywhere in the expression \(\frac{1}{x} - \frac{1}{1 - x}\), with \(x \neq 0\) and \(x \neq 1\). If the result is then multiplied by \(x^2 - x\), the outcome equals

A. \(x + 1\)
B. \(x - 1\)
C. \(1 - x^2\)
D. \(2x - 1\)
E. \(1 - 2x\)

First, carry out the replacement. Literally replace every \(x\) in the expression with \(1 - x\), putting parentheses around the \(1 - x\) in order to preserve proper order of operations:

Original: \(\frac{1}{x} - \frac{1}{1 - x}\)

Replacement:
\(\frac{1}{(1 - x)} - \frac{1}{1 - (1 - x)}\)

Now simplify the second denominator: \((1 - (1 - x)) = (1 - 1 + x) = x\)

So the replacement expression becomes this:
\(\frac{1}{(1 - x)} - \frac{1}{x}\)

This should make sense. If we replace \(x\) by \(1 - x\), then it turns out that we are also replacing \(1 - x\) by \(x\) (since \(1 - (1 - x) = x\)). Thus, the denominators of the original expression are simply swapped.

Now we can either combine these fractions first (by finding a common denominator) or go ahead and multiply by \(x^2 - x,\) as we are instructed to. Let’s take the latter approach.
\(( \frac{1}{1 - x} - \frac{1}{x}) (x^2 - x)\)

Instead of FOILing this product right away, we should factor the expression \(x^2 - x\) first. If we do so, we will be able to cancel denominators quickly.

\(x^2 - x\) factors into \((x - 1)x\). We can now rewrite the product:
\(( \frac{1}{1 - x} - \frac{1}{x} ) (x - 1)x = \frac{(x - 1)x}{(1 - x)} - \frac{(x - 1)x}{x}\)

The second term, \(\frac{(x - 1)x}{x}\), becomes just \(x - 1\) after we cancel the \(x\)'s.

Since \((x - 1) = -(1 - x)\), we can rewrite the first term as \(\frac{-(1 - x)x}{(1 - x)}\) and then cancel the \((1 - x)\)’s, leaving \(-x\).

So, the final result is \(-x - (x - 1) = -x - x + 1 = 1 - 2x\). This is the answer.

Separately, since this is a Variables In Choices problem, we could instead pick a number and calculate a target. Since 0 and 1 are disallowed, let's pick \(x = 2\). We are told that \(x\) should be replaced by \(1 - x\), so we calculate \(1 - x = -1\) and put in -1 wherever \(x\) is in the original expression.
\(\frac{1}{x} - \frac{1}{(1 - x)} = \frac{1}{(-1)} - \frac{1}{(1 - (-1))} = -1 - \frac{1}{2} = -\frac{3}{2}\)

Now multiply this number by \(x^2 - x = 2^2 - 2 = 2\). We get -3 as our target number.

Finally, we plug \(x = 2\) into the answer choices and look for -3:

(A) \(x + 1 = 2 + 1 = 3\)

(B) \(x - 1 = 2 - 1 = 1\)

(C) \(1 - x^2 = 1 - 2^2 = -3\)

(D) \(2x - 1 = 2(2) - 1 = 3\)

(E) \(1 - 2x = 1 - 2(2) = -3\)

We can eliminate choices A, B, and D, but to choose between C and E, we would need to pick another number.

Answer: E.

After replacing
1/(1-x) - 1/(1-(1-x)
= 1/(1-x) - 1/x

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

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