I'm trying to solve $\frac{3x-5}{8x-2}<6$ ?

I'm not sure which first step to take. I mean if I multiply both sides by $8x-2$ then I'm not sure if the sign would switch, as this could be positive or negative depending on $x$.

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    $\begingroup$ Make a case distinction. $\endgroup$ – Daniel Fischer Jun 13 '14 at 14:56
  • $\begingroup$ Look at where the denominator is $<0,=0, >0$. $\endgroup$ – copper.hat Jun 13 '14 at 14:57


You can't multiply by $8x-2$ without discuss on its sign. The best way to answer the question is:

$$\frac{3x-5}{8x-2}<6\iff\frac{3x-5}{8x-2}-6=\frac{-45x+7}{8x-2}<0$$ and now draw a signs table for this quotient.

Edit The sign table is enter image description here

so the answer is $$\left(-\infty,\frac7{45}\right)\cup \left(\frac14,+\infty\right)$$

  • $\begingroup$ The other answer shows the way to solve the problem of multiplying: multiply by a non-negative factor! Much faster than checking signs cases... $\endgroup$ – DonAntonio Jun 13 '14 at 15:02
  • $\begingroup$ The other answer is similar and he should reason on the sign of each factor! $\endgroup$ – user63181 Jun 13 '14 at 15:04
  • $\begingroup$ Why "should" he reason about those signs beyond the fact that knowing that multiplying an inequality by a non-negative factor leaves the inequality sign steady, @user63181 ? $\endgroup$ – DonAntonio Jun 13 '14 at 15:06
  • $\begingroup$ In what cases we have $ab>0$? Didn't you think on the sign of $a$ and $b$? Have you an alternative method? $\endgroup$ – user63181 Jun 13 '14 at 15:10
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    $\begingroup$ I understand all of this. And I think lab's answer is actually very good. I just know that when you introduce new methods when students are still wrapping their heads around the old ones, that it can introduce confusion. Much later, I would introduce other methods. That is when students have a firm grasp on the old method. It is a matter of taste really. $\endgroup$ – Joel Jun 13 '14 at 15:37

Check if the denominator $=0$


$$\frac{3x-5}{8x-2}<6\iff \frac{3x-5}{8x-2}-6<0\iff\frac{45x-7}{8x-2}>0\iff (45x-7)(8x-2)>0$$

as for real $x,(8x-2)^2>0$


If $\displaystyle(x-a)(x-b)>0,$ where $a<b$ can you prove that either $x<a$ or $x>b$ for real $a,b$

  • $\begingroup$ Except for $x={1 \over 4}$? $\endgroup$ – copper.hat Jun 13 '14 at 14:58
  • $\begingroup$ The easiest, fastest and simplest way +1 $\endgroup$ – DonAntonio Jun 13 '14 at 14:58
  • $\begingroup$ Never learned anything like this. Is this the "standard" way to solve those kind of inequalities ? $\endgroup$ – 90intuition Jun 13 '14 at 14:59
  • $\begingroup$ In fact @copper.hat, $\;x=\frac14\;$ doesn't satisfay the rightmost, last inequality....but I agree: it should be clearly stated from the beginning what the definition set is (as, for example, if the inequality was weak then...) $\endgroup$ – DonAntonio Jun 13 '14 at 14:59
  • $\begingroup$ @90intuition, it is probably the best, easiest way. Observe the last inequality is reached after multiplying by $\;(8x-2)^2\;$ , which is always non-negative and thus we have no problems with the inequality sign. $\endgroup$ – DonAntonio Jun 13 '14 at 15:00

Multiply both sides of $\frac{3x-5}{8x-2}<6$ by $(8x-2)^2$ will make you free from the worry (if the sign would switch).

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    $\begingroup$ This is exactly the same method shown in lab's answer 17 minutes ago... $\endgroup$ – DonAntonio Jun 13 '14 at 15:12
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    $\begingroup$ @DonAntonio Not quite. He did that after moving terms and simplifications. I suggest performing it right at the beginning. $\endgroup$ – Mick Jun 13 '14 at 15:15
  • $\begingroup$ Mick, and that's an improvement on something that gives exactly the same because...? $\endgroup$ – DonAntonio Jun 13 '14 at 15:16
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    $\begingroup$ @DonAntonio The OP says “…not sure which first step to take. ……”; and “not sure if the sign would switch”. I am showing him the worry-free initial step. After that, it is a go with full-throttle. $\endgroup$ – Mick Jun 13 '14 at 15:24
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    $\begingroup$ @DonAntonio Because the 'annoying' denominator is gone after the first step allready? The sooner the worry of the student is gone, the better, I would say. $\endgroup$ – drhab Jun 13 '14 at 15:25

Assume $8x-2>0$ (i.e. $x>\frac14$) and multiply both sides: $3x-5<48x-12$, or $x>\frac7{45}$.

Symmetrically, when $x<\frac14$ you get $x<\frac7{45}$.

Combined, $x>\frac14\lor x<\frac7{45}$.


$$\frac{3x-5}{8x-2}<6\iff\frac{3x-5}{8x-2}-6<0\iff\frac{3x-5-48x+12}{8x-2}<0\iff$$ $$\iff\frac{7-45x}{2(4x-1)}<0\iff\frac{7-45x}{4x-1}<0\iff$$ $$\left(7-45x>0\text{and}4x-1<0\iff x<\frac{7}{45}\text{and} x<\frac{1}{4}\iff x<\frac{1}{4}\right)$$ or $$\left(7-45x<0\text{and}4x-1>0\iff x>\frac{7}{45}\text{and} x>\frac{1}{4}\iff x>\frac{7}{45}\right)\iff$$ $$\iff x\in\mathbb R-\ [1/4,7/45\ ]$$


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