Useful property: \(\frac{a-b}{b-a} = -1\)

Here's why the property works:
Take: \(\frac{a-b}{b-a}\)
Factor \(-1\) from the numerator to get: \(\frac{-1(-a+b)}{b-a}\)
Rearrange the numerator as follows: \(\frac{-1(b-a)}{b-a}\) , which simplifies to become \(-1\)

APPROACH #1: Apply the above property to the given equation

So, our equation becomes: \(-1=\frac{1}{x}\), which means \(x = -1\)

So, we have:
QUANTITY A: \(-1\)
QUANTITY B: \(-\frac{1}{2}\)

Answer: B


APPROACH #2: Algebra + matching operation (see video below)

If \(\frac{1-x}{x-1}=\frac{1}{x}\), we can also say that \(\frac{x-1}{1-x}=\frac{x}{1}\)

In other words: \(\frac{x-1}{1-x}=x\), which means we can substitute this equivalent value into quantity A to get:

QUANTITY A: \(\frac{x-1}{1-x}\)

QUANTITY B: \(-\frac{1}{2}\)

Add \(\frac{1}{2}\) both sides to get:

QUANTITY A: \(\frac{x-1}{1-x}+\frac{1}{2}\)
QUANTITY B: \(0\)

Rewrite the two fractions with a common denominator:
QUANTITY A: \(\frac{2(x-1)}{2(1-x)}+\frac{1(1-x)}{2(1-x)}\)
QUANTITY B: \(0\)

Expand the numerators:
QUANTITY A: \(\frac{2x-2}{2(1-x)}+\frac{1-x}{2(1-x)}\)
QUANTITY B: \(0\)

Combine fractions:
QUANTITY A: \(\frac{x-1}{2(1-x)}\)
QUANTITY B: \(0\)

Apply the above property:
QUANTITY A: \(\frac{x-1}{2(1-x)} = \frac{1}{2}(\frac{x-1}{1-x})= \frac{1}{2}(-1)= -\frac{1}{2}\)
QUANTITY B: \(0\)

Answer: B

APPROACH #3: Different algebra
Take the given equation: \(\frac{1-x}{x-1}=\frac{1}{x}\)

Cross multiply to get: \(x - x^2 = x - 1\)
Subtract \(x\) from both sides: \(-x^2 = -1\)
Multiply both sides by \(-1\) to get: \(x^2 = 1\), which means either \(x = 1\) or \(x = -1\)
Important: At this point it certainly seems as though \(x\) can have two different values. HOWEVER, when we plug \(x = 1\) and \(x = -1\) back into the original equation, we see that \(x = 1\) is NOT a solution.
That is, when we plug \(x = 1\) into the original equation we get: \(\frac{1-1}{1-1}=1\), which becomes \(\frac{0}{0}=1\) upon simplification.


Since \(x = -1\) is the only possible solution to the given equation, we get:
QUANTITY A: \(-1\)
QUANTITY B: \(\frac{-1}{2}\)

Answer: B

RELATED VIDEO

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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.
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(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

Alternate approach:

1-x and x-1 are OPPOSITES:
If 1-x = 2, then x-1 = -2
If 1-x = -10, then x-1 = 10

To generalize:
If 1-x = k, then x-1 = -k
Thus:
\(\frac{1-x}{x-1} = \frac{k}{-k} = -1\)

Substituting \(\frac{1-x}{x-1} = -1\) into \(\frac{1-x}{x-1}=\frac{1}{x}\), we get:
\(-1 = \frac{1}{x}\)
\(x = -1\)

Since \(-\frac{1}{2}\) is greater than -1, Quantity B wins.


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