Ok so I've got an exam in 4 hours and I can't ever figure out these problems.

Here's the problem I'm struggling with now:

$$\frac{x}{3x - 5} \leq \frac{2}{x - 1}.$$

I've learned this so many times but my brain just cannot intuitively understand how these work. I can't cross-multiply because I don't know if the binomials are negative or not. If someone could just give a quick explanation of how to solve this sort of thing I'd really appreciate it.

  • $\begingroup$ You could start by bringing everything to one side: ${x\over 3x-5}-{2\over x-1}\le 0$. Can you take it from here? $\endgroup$ – David Mitra Sep 13 '12 at 16:09
  • $\begingroup$ Then I've just got two binomials in the numerator and two in the denominator. How do you derive a solution from that? $\endgroup$ – user39398 Sep 13 '12 at 16:12

$1.$ Find out where you have equality. This involves simplifying to a quadratic equation. In your case you will I think end up with $x^2-7x+10=0$, which has the roots $x=2$ and $x=5$.

$2.$ Find the singularities, where your expression is not defined. These are $x=1$ and $x=5/3$.

$3.$ The points $x=1$, $x=5/3$, $x=2$, and $x=5$ are the only places where the inequality might reverse. These $4$ points divide the number line into $5$ regions. For each region, pick a convenient test point and check whether the inequality holds at that point. If it holds at the test point, it holds for the whole region. If it fails, then it fails for the whole region.

$4.$ Write down your conclusions, paying attention to the fact that we will have equality at $2$ and $5$, and that the inequality makes no sense at $x=1$ and $x=5/3$.


$\frac{x}{3x-5}-\frac{2}{x-1} ≤0$



(1)$(x-2)(x-5)≤0$ and $(3x-5)(x-1)>0$

or (2) $(x-2)(x-5)≥0$ and $(3x-5)(x-1)<0$

(1)$\implies 2≤x≤5$ and ($x<1$ or $x>\frac{5}{3})\implies 2≤x≤5$.

(2)$\implies (x≥5$ or $x≤2)$ and $1<x<\frac{5}{3}\implies 1<x<\frac{5}{3}$.

Taking union of regions, we get $1<x≤5$

The formula used :

$(x-a)(x-b)≤0$ where $a≤b\implies a≤x≤b$

and $(x-a)(x-b)≥0$ where $a≤b\implies x≤a$ or $x≥b$


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