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When I was reading on the topic of equations that can be written as quadratic equations I encountered this example (without showing the steps leading to the quadratic equation).

$$\frac{1}{x+2} - \frac{4}{x} - 1 = 0$$

My first attempts at trying to recreate the process did not work. The only time it worked was when I took the $x$ in $\frac{-4}{x}$ as $(-x)$ and multiplied the equation with $-x$ (to get rid of the denominator). Then I managed to have the quadratic equation $$x^{2} + 5x + 8 = 0.$$

The problem is in this example I knew what to expect, thus I also knew that my previous solutions were wrong. So I fixed it accordingly until I got the expected result (as stated in the book). Yet in a previous example in the book the solution was given and the $x$ part of a fraction having a negative sign before it was evaluated as having a positive value (unlike this example the numerator part had the minus sign accordingly after the multiplication - which I would also automatically assume). I am still not sure about the way I solved it: so I would appreciate a demonstration.

In case what I did was right: How can one decide when to take the denominator $x$ as positive or negative when they don't have the result at hand? How is it possible for me to see that the negative sign before the fraction applies to the numerator or denominator when multiplying with a denominator to get a quadratic equation?

Frankly speaking, this kind of manipulations which seem to be haphazard when given without any explanations as if it is the duty of anyone to automatically decide which manipulation to conduct are among my biggest problems in math. Is there a general rule?

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    $\begingroup$ Just clear all the denominators. In your original equation you have two denominators, namely $x+1$ and $x$. So multiply by $x(x+2)$ to see that $x-4(x+2)-x(x+2)=0\implies x-4x-8-x^2-2x=0\implies -x^2-5x-8=0\implies$ $ x^2+5+8=0$ $\endgroup$ – lulu Oct 21 '18 at 20:49
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    $\begingroup$ Multiply through by $-1$. $\endgroup$ – lulu Oct 21 '18 at 21:02
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    $\begingroup$ @JavaMan Yes, thank you. The denominators are $x+2$ and $x$ and so you multiply through by $x(x+2)$. $\endgroup$ – lulu Oct 21 '18 at 21:03
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    $\begingroup$ @serratusmagnus or add $x^2 + 5x + 8$ to both sides of the equation giving: $0 = x^2 + 5x + 8$. By the symmetric property of equality, this is the same as $x^2 + 5x + 8 = 0$ $\endgroup$ – David Diaz Oct 21 '18 at 21:06
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    $\begingroup$ No...multiplying by $-1$ isn't at all standard, but getting the lead coefficient to be $1$ is. Here, the first version you (and I) got had $-x^2$ as the lead term. To get rid of that you divide by $-1$, which is the same as multiplying by $-1$. Note: it is absolutely not necessary to get the lead coefficient to be $1$. But it is a fairly standard convention. $\endgroup$ – lulu Oct 21 '18 at 23:54
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It so happens that $$-\frac{x}{4} = \frac{-x}{4} = \frac{x}{-4}$$ It may be helpful to think of a negative fraction as $$(-1)\frac{x}{4}$$ and notice that the negative one can be multiplied into the fraction: $$\frac{(-1)x}{4}$$ additionally, any fraction can be multiplied by $\frac{a}{a}, a\not=0$ without changing the value of the fraction. Try multiplying $\frac{-x}{4}$by $\frac{-1}{-1}$ $$=\frac{(-1)(-1)x}{(-1)4}=\frac{x}{(-1)4}$$

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