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How and why did the sign + and - switch places in the solution?

Question:

$$\frac{x-6}x - \frac6x = \frac{x-6}{x+6}$$

Answer:

Step one

$$\frac{x-6-6}x = \frac{x+6}{x-6}$$

equation from textbook

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    $\begingroup$ probably a typo. $\endgroup$
    – Spectre
    Oct 31, 2020 at 15:31
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    $\begingroup$ Ok so how do you solve the original equation? $\endgroup$ Oct 31, 2020 at 15:38
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    $\begingroup$ @JessicaHam Do the same process by plugging the correct signs. $\endgroup$
    – 19aksh
    Oct 31, 2020 at 15:41
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    $\begingroup$ I tried... I don't think it works.. Can someone pls try it out $\endgroup$ Oct 31, 2020 at 15:42
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    $\begingroup$ To the downvoters: Yes, people should post equations using MathJax. But for suspected typo questions it's very helpful to see an image of the text. It would be nice to have a MathJax transcription as well, but I don't think it's strictly necessary for typos. $\endgroup$
    – PM 2Ring
    Oct 31, 2020 at 15:53

2 Answers 2

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The original solution is like this :

$$\dfrac{x - 6}{x} - \dfrac 6x = \dfrac{x - 6}{x + 6} \implies \dfrac{x - 12}{x} = \dfrac{x - 6}{x + 6}\\ \implies x(x -6) = (x - 12)(x +6)\\ \implies x^2 - 6x = x^2 +6x-12x-72\\ \implies -6x = -6x - 72\\ \implies 6x = 6x + 72$$

Now that is unsolvable, since $6x-6x = 0 \neq 72$.

Therefore, the actual question must be $\dfrac{x - 6}{x} - \dfrac 6x = \dfrac{x + 6}{x - 6} $.

If this is the question, the solution in the picture is the right solution.

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  • $\begingroup$ It looks like you intended to place implies rather than equals signs at the beginning of the third, fourth, and fifth lines of your calculations. $\endgroup$ Oct 31, 2020 at 18:00
  • $\begingroup$ @N.F.Taussig, no... I did it deliberately... $\endgroup$
    – Spectre
    Nov 1, 2020 at 5:05
  • $\begingroup$ That does not make sense since $x^2 - 6x$ does not equal $-6x$ unless $x = 0$. Similarly, $-6x$ does not equal $6x$ unless $x = 0$. $\endgroup$ Nov 1, 2020 at 11:05
  • $\begingroup$ @N.F.Taussig , at school I usually do it like that. But you may edit it if you want it to be more clear ':) My mistake.. sorry 'bout it ! $\endgroup$
    – Spectre
    Nov 1, 2020 at 11:06
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Here's another way to solve the equation $$\frac{x - 12}{x} = \frac{x - 6}{x + 6}$$ by corrensponding addition: subtract the numerator from the denominator on both fractions to get $$\frac{x - 12}{x-(x-12)} = \frac{x - 6}{x + 6-(x-6)}\iff\frac{x - 12}{12} = \frac{x - 6}{12}$$ which is impossible.

NB: The given solution solves $$\frac{x - 12}{x} = \frac{x + 6}{x - 6}.$$ Using the method above gives $$\frac{x - 12}{12} = \frac{x + 6}{-12} \iff x-12=-6-x$$ that is $x=3$.

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