Who is right? (Complex fraction problem)

Hello everyone, that's my first discussion (open-ended post), so if there are any improvements to be done, then please don't hesitate in informing about it.

Getting to the point, there's an equation which I've seen in some wiki from Brilliant, and it's really bothering me because I can't understand it at all. I tried to solve it myself, and I got a completely different result. I also entered it in some math computing websites (such as WolframAlpha), but they also gave me completely different results from the equation I've seen around here. I'm going to post both ways (the way I've seen, Computation A, and my way, Computation B) so someone might be able to point which of both are right (or explain why none are right).

Computation A

x+1x−1+x−1x+1x+1x−1+x−1x+1=(x+1)²+(x−1)²(x−1)(x+1)(x+1)²−(x−1)²(x−1)(x+1)=x²+2x+1+x²−2x+1x²−1x²+2x+1−(x²−2x+1)x²−1=2x²+2x²−1×x²−14x=x²+12x \LARGE \frac{ \frac{ x+1 }{ x-1 } + \frac{ x-1 }{ x+1} }{ \frac{ x+1 }{ x-1 } + \frac{ x-1 }{ x+1} } = \frac{ \frac{(x+1)²+(x-1)²}{(x-1)(x+1)} }{ \frac{(x+1)²-(x-1)²}{(x-1)(x+1)} } = \frac{ \frac{x²+2x+1+x²-2x+1}{x²-1} }{ \frac{x²+2x+1-(x²-2x+1)}{x²-1} } = \frac{2x²+2}{x²-1} \times \frac{x²-1}{4x} = \frac{x²+1}{2x}

x≠0,x≠±1 \LARGE x \neq 0, x \neq \pm1

I'm still in high-school level on math, but in either way I wasn't able to follow that math. Not completely, but I'll keep an ordered list on how I interpreted the steps in case anyone can clarify anything to me. Each number from the list refers to its respective step in the Computation A equation.

  1. Okay, that was given.
  2. That's where I lost track. I mean, he probably just multiplied both fractions from the numerator and denominator by their respective LCMs, but here's what I find strange there: if the numerator and denominator in the largest fraction are the same, then why did they turned into different products when simplified? And if they're the same, then why isn't the fracion't quotient 1? Anyway, from that result we know that x≠±1 x \neq \pm 1.
  3. Further simplification. I get that.
  4. Ok... A bunch of steps were skipped from the third to the fourth step, but if you simplify the third step yourself you'll get that result (I hope so). From this step we can infer that x≠0 x \neq 0 as well.

  5. Simplification from previous step, but it makes sense to me.

    Now here's my way of solving it when considering that aa=1 \frac{a}{a} = 1

Computation B

x+1x−1+x−1x+1x+1x−1+x−1x+1 \LARGE \frac{\frac{x+1}{x-1}+\frac{x-1}{x+1}}{\frac{x+1}{x-1}+\frac{x-1}{x+1}}

a=x+1x−1+x−1x+1 \LARGE a = \frac{x+1}{x-1}+\frac{x-1}{x+1}

∴x+1x−1+x−1x+1x+1x−1+x−1x+1=aa=1 \LARGE \therefore \frac{\frac{x+1}{x-1}+\frac{x-1}{x+1}}{\frac{x+1}{x-1}+\frac{x-1}{x+1}} = \frac{a}{a} = 1

And by the way, I don't trust my own calculations and methods, so I threw that problem into some online math-computing programs. There are the links:

#Algebra

Note by Daniel Maia
4 years, 11 months ago

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Comments

The answer is indeed 1. After taking L.C.M in the denominator, you did subtraction instead of addition.

A Former Brilliant Member - 4 years, 11 months ago

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That's what I thought. Actually the "Computation A" I got from a wiki page from Brilliant itself, but I thought it was really weird. But thanks for the input anyway!

Daniel Maia - 4 years, 11 months ago

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Seems to be a typo. It'll be better if someone corrects it to prevent more people from being confused.

Do you remember which wiki page?

Siddhartha Srivastava - 4 years, 11 months ago

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@Siddhartha Srivastava – Yep. It was the fractions wiki, but don't worry, I'll change the plus sign from the denominator to a minus sign, so it should get corrected.

Daniel Maia - 4 years, 11 months ago
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