Aside: from the given equation, we know that x ≠ 0 and we know that x-1 ≠ 0
Here's why:
If x = 0, we get: (1-0)/(0-1) = 1/0
Simplify to get: -1 = 1/0 DOESN'T WORK
So, x ≠ 0

Likewise, if x-1 = 0, then x = 1
If we plug in 1 for x, we get: (1-1)/(1-1) = 1/1
Simplify to get: 0/0 = 1/1 DOESN'T WORK
So, x-1 ≠ 0

--------------------------------
Okay, now for a different approach....

Notice that 1-x = -1(-1 + x) = -1(x-1)
So, let's take (1-x)/(x-1) = 1/x and replace (1-x) with -1(x-1)

We get: -1(x-1)/(x-1) = 1/x
Simplify to get: -1 = 1/x
Solve to get: x = -1

So, we have:
Quantity A: -1
Quantity B: -1/2

Answer: B
_________________

Solution

Since the denominator must be different from zero we do have that x is different from 1 and zero

Multiplying both sides we do have: \(x-x^2=x-1\) from which \(x^2=1\)

Our roots are \(x=1\) or \(x=-1\) but we already know that x must be different from 1. So the only value we do have is \(x=-1\)

\(-1 < \frac{1}{2}\)

The answer is \(B\)
_________________

"we already know that x must be different from " from where you already known that?

Hey,

The relation \(\frac{1-x}{x-1}=\frac{1}{x}\) is already told to be true.

Now if you put x=1 the LHS becomes undefined and and if x=0 then RHS becomes undefined.

Hence X cannot be 0 or 1.
_________________

(1-x)/(x-1)=1/x
-1(x-1)/(x-1)=1/x
therefore, x=-1 then qty b=-0.5 so qty b is greater

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