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$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$

I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?

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    $\begingroup$ I feel immature for laughing at the title... $\endgroup$ – thunderbolt Oct 5 '16 at 3:07
  • $\begingroup$ Solve what? I don't see any equation. What does the problem ask you to do? $\endgroup$ – bof Oct 5 '16 at 3:11
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    $\begingroup$ Catchy title. Next, if you'd put the minimal effort in to format your question properly using MathJax, then that would increase your chances at good answers dramatically. $\endgroup$ – dxiv Oct 5 '16 at 3:12
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    $\begingroup$ @thunderbolt his profile pic is awesome too $\endgroup$ – yoyostein Oct 5 '16 at 3:13
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    $\begingroup$ Thanks for formatting my equation for me. (my title was removed sadly) $\endgroup$ – bigjuicywatermelons Oct 5 '16 at 3:13
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Multiply the numerator and denominator by $x(1+x)(1-x)$. This clears away all the "inner" denominators, leaving $${xx(1+x)+(1+x)(1+x)(1-x)\over (1-x)(1-x)(1+x)+xx(1-x)}={(1+x)\over (1-x)}{x^2+1-x^2\over x^2+1-x^2}={1+x\over 1-x}.$$

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Multiply both the numerator and the denominator by $x(1-x)(1+x)$ and then use the fact that $(1-x)(1+x)=1-x^2$ to see that

\begin{align} \frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}&=\left(\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}\right)\frac{x(1-x)(1+x)}{x(1-x)(1+x)} \\&=\left(\frac{x^2+(1+x)(1-x)}{(1-x)(1+x)+x^2}\right)\frac{1+x}{1-x} \\&=\left(\frac{1}{1}\right)\frac{1+x}{1-x} \\&=\frac{1+x}{1-x}. \end{align}

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  • $\begingroup$ I feel like you could just skip the whole $(1-x)(1+x)=1-x^2$ part and see through additive and multiplicative commutation that the left half of the third equation has the same numerator and denominator. Alternately, it might be helpful to show an intermediate step with ${x^2+1-x^2\over 1-x^2+x^2}$ to see where the identity is being used. But sometimes I'm just stupid so it may not help. $\endgroup$ – MichaelS Oct 5 '16 at 3:49
  • $\begingroup$ @MichaelS You're right; that may be a bit easier. $\endgroup$ – Mitchell Spector Oct 5 '16 at 3:50
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If you note that the expression, which is of form $$\cfrac{a+b}{\frac 1a +\frac 1b}$$ simplifies easily to $ab$ (multiply through by $\frac {ab}{ab}$) you can avoid a lot of potentially confusing/error inducing algebra, and easily get the result that others have given.

It is sometimes useful to identify the form of an expression and simplify that, especially when it looks likely to get confusing. Sometimes it is a distraction, but here it would be a quick try and you could revert to a more hands-on approach if it doesn't work.

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Let me try.

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

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