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Question and solution

How do the factors not cancel? And how is this possible by not multiplying both sides by (4x-1)? I don't see what rules have been used here.

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closed as off-topic by Did, Qwerty, naslundx, Daniel W. Farlow, user223391 Sep 4 '16 at 16:37

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  • $\begingroup$ but the whole computation can you find at the paper $\endgroup$ – Dr. Sonnhard Graubner Sep 4 '16 at 12:27
  • $\begingroup$ what should I do? $\endgroup$ – Dr. Sonnhard Graubner Sep 4 '16 at 12:28
  • $\begingroup$ When it says "instead proceed as follows" I don't understand what is happening. Because normally, I would just multiply both sides by the denominator. $\endgroup$ – james Sep 4 '16 at 12:29
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@james Huge mistake. Multiplying/dividing in inequalities is most of the time a huge mistake unless you know the fact you're multiplying/dividing by is positive. This is precisely what the proposed solution avoids, and then you only sum $\;-x\;$ to both sides of the inequality (and this is done exactly as with equations) and etc. I think the rest is clear:

$$0<\frac{4x^2-x-3}{4x-1}=\frac{(x-1)(4x+3)}{4x-1}=\frac{(x-1)\left(x+\frac43\right)}{x-\frac14}$$

and now multiply both sides by $\;\left(x-\frac14\right)^2\;$ ... observe that, taking into account that you already know that it must be $\;x\neq\frac14\;$ , you multiply by a positive quantity!:

$$(x-1)\left(x+\frac34\right)\left(x-\frac14\right)>0...etc.$$

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you want to solve the inequality $$\frac{3}{4x-1}-x<0$$ first of all we remark that $4x-1\ne 0$ and then we write $$\frac{3-x(4x-1)}{4x-1}<0$$ after simplifying we have $$\frac{-4x^2+x+3}{4x-1}<0$$ multiplying numerator by $-1$ we get $$\frac{4x^2-x-3}{4x-1}>0$$ after factorization we have to solve $$\frac{(4x+3)(x-1)}{4x-1}>0$$ the solution is given by $$-3/4<x<1/4$$ or $$x>1$$

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