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I've been doing some calculations in fractions, and found this equation pop up to calculate my answer:

$$\frac{1-x}{1+x}=x$$

the initial equation is

$$\frac{2(x-1)}{\frac{4(x+1)}{2}}+x=4x+9(-4x-2)-2(-17x+34)+61+6$$

(I used a random number generator)I started tackling it by solving the right side

$$ \begin{align} \cdots&=4x+9(-4x+2)-2(-17x+34)+61+6\\ &=4x+(-36x)+18-(-34x)-68+61+6\\ &=4x+(-36x)+18-(-34x)-68+61+6\\ &=4x-36x+1+34x-68+61+6\\ &=2x\\ \end{align} $$

then i simplified it even further using the other side as well, getting:

$$ \begin{align} \frac{2(x-1)}{\frac{4(x+1)}{2}}+x&=2x\\ \frac{2(x-1)}{2(x+1)}+x&=2x\\ \frac{x-1}{x+1}+x&=2x\\ \frac{x-1}{x+1}&=x\\ \end{align} $$

This is my problem. so what is $x???$ also, did I do this correctly? if not, could you solve the equation for me, and then still solve this annoying equation?

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    $\begingroup$ The lines after "I started tackling it by solving the right side" are wrong as well — 61(x+1) should expand to 61x+61 $\endgroup$
    – Toby Mak
    Apr 27, 2017 at 23:56
  • $\begingroup$ @TobyMak i made an error, and changed it back to normal. thanks for pointing it out $\endgroup$ Apr 27, 2017 at 23:57
  • $\begingroup$ $4x+9(-4x-2)-2(-17x+34)+61+6=2x-19$ $\endgroup$ Apr 28, 2017 at 6:30

2 Answers 2

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$$\frac{x-1}{x+1} = x$$ Multiply both sides by $x+1:$ $$ x-1 = x(x+1) $$ $$ x-1 = x^2 + x $$ $$ -1 = x^2 $$ $$\text{etc.} $$

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  • $\begingroup$ great, great!! my soultion is not even real!! i now gotta change my 'real analysis' tag $\endgroup$ Apr 27, 2017 at 23:44
  • $\begingroup$ If you replace "etc." with "$x=\pm i$", you have a complete answer lol $\endgroup$
    – mrnovice
    Apr 27, 2017 at 23:46
  • $\begingroup$ also, good call on the \cdots idea. i didn't know it existed! $\endgroup$ Apr 27, 2017 at 23:50
  • $\begingroup$ @AlexanderDay : Actually this is neither real analysis nor complex analysis; rather it is algebra. Real analysis deals with limits and continuity and integration and differentiation, and complex analysis deals with consequences of the holomorphic nature of complex-valued functions of a complex variable. $$ \S $$ As for typography, I use \ldots between commas, as in $1,\ldots, n,$ and \cdots between binary relation or binary operation signs as in $A+\cdots+Z$ or $A < B < \cdots < Z.$ \vdots and \ddots also exist and are sometimes used in matrices. $\endgroup$ Apr 28, 2017 at 0:35
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The right hand side of the equation is $$x-19$$ not $2x $ as you wrote.

the equation is

$$\frac {x-1}{x+1}=x-19$$

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  • $\begingroup$ sorry, and thanks. but this is a comment, not an answer $\endgroup$ Apr 27, 2017 at 23:42
  • $\begingroup$ you right member seems not correct. $\endgroup$ Apr 28, 2017 at 0:04
  • $\begingroup$ $1-68+61+6=67-67=0$, which gets $2x\pm 0$ $\endgroup$ Apr 28, 2017 at 0:31

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