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\(\frac{1-x}{x-1}=\frac{1}{x}\)
Quantity A
Quantity B
\(x\)
\(-\frac{1}{2}\)
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given.
Practice Questions Question: 8 Page: 339 Difficulty: medium
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:
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: \(\frac{-1}{2}\)
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