# Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$

I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.

$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$

$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0$$

$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$

I don't know how to get any further, and I'm starting to get too high values to handle.

The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$.

But how do I get the last step? Or am I already dead wrong?

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I think that after the first step you can start determining the intervals where $x$ satisfies the inequalities. Draw a number line, mark the points where the denominator is 0 and then test points in between. – echoone Aug 25 '12 at 14:41

Hint: From here: $$\frac{2x^2 - 43x - 45}{2x(x + \frac{3}{2})} \lt 0$$

you could say that the fraction is less than zero if and only if either (1) the numerator is positive and the denominator is negative or (2) the numerator is negative and the denominator is positive. So for case (2), you solve:

$$2x^2 - 43x - 45 <0 \quad \text{and}\quad 2x(x +\frac{3}{2}) > 0.$$

You might already know this, but the way you solve each of these two inequalities is by finding possible roots. So for example for the $2x(x+ \frac{3}{2}) > 0$, the roots are $x = 0$ and $x = -\frac{3}{2}$. Now you just need to find the (constant) signs of the expression $2x(x+\frac{3}{2})$ on the intervals $(-\infty, -\frac{3}{2})$, $(-\frac{3}{2}, 0)$, and $(0, \infty)$.

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Thank you! Calculating for case (2) I find the root $-1$ for $2x^2 - 43x - 45$. Adding this to the roots of $2x(x+\frac{3}{2})$, this should mean that the entire expression is negative when $\frac{-3}{2} \lt x \lt -1$ or $0 \lt x$. However, when testing higher numbers (approximately where $x \gt 22$), $2x^2 - 43x - 45$ becomes positive and so the entire expression becomes positive. Why is this? – Quispiam Aug 25 '12 at 21:55
Note that you also have the root $22.5$ for $2x^2-43x-45$. – Thomas Aug 25 '12 at 21:57
Sorry, that was me being too excited after suddenly getting things right for once. Should have kept checking for roots after finding the obvious one. Thank you. – Quispiam Aug 25 '12 at 22:00
@Quispiam: :) Just ask again if you get stuck. – Thomas Aug 25 '12 at 22:00
It seems I figured everything out now. It was part of a bigger problem (uniting it with another inequality) and it seems like everything goes together perfectly now. Thanks a lot for the help. – Quispiam Aug 25 '12 at 23:41

One approach is to find the (potential) "switchover" points. Note that one rational function can only become bigger than another after either intersecting or after a hole or vertical asymptote (of one or both). (Why?) There our no holes, and our asymptotes occur at $x=0$ and $x=-\frac32$. Intersection occurs when $x\neq 0$, $x\neq-\frac32$ and $$\frac{2x-13}{2x+3}=\frac{15}x$$ $$2x^2-13x=30x+45$$ $$2x^2-43x-45=0$$ $$2x^2-45x+2x-45=0$$ $$(2x-45)(x+1)=0.$$ Thus, our other possible switchover points are $x=-1$ and $x=\frac{45}2$.

Now, check the inequality on the intervals $\left(-\infty,-\frac32\right)$, $\left(-\frac32,-1\right)$, $(-1,0)$, $\left(0,\frac{45}2\right)$, and $\left(\frac{45}2,\infty\right)$ by using test points in each interval. Note that for the first interval a sign chart would readily tell us that the desired inequality fails (without need for a test point), since $\cfrac{2x-13}{2x+3}$ is positive and $\cfrac{15}x$ is negative.

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you have to resolve this inequality depending $x$ : $$\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$$

that is equivalent with: $\displaystyle \frac{x(2x-13)-15(2x+3)}{x(2x+3)} \lt 0$ so: $\displaystyle \frac{2x^2-13x-30x-45}{2x^2+3x} \lt 0$ and now the simple form is : $$\frac{2x^2-43x-45}{2x^2+3x} \lt 0$$ Now we have two equation . first: $2x^2-43x-45=0$ with : $\displaystyle x_{1,2}=-1 \mbox{and} \frac{45}{2}$ and second $2x^2+3x=0$ with $\displaystyle x_{1,2}=-\frac{3}{2} \mbox{and} 0$.

For that inequality to be satisfied it must: $\displaystyle 2x^2-43x-45 \lt0$ and $\displaystyle 2x^2+3x \gt 0$ (1) OR $2x^2-43x-45 \gt 0$ and $2x^2+3x \lt 0$ (2).

I attach a draw. it is easy now to view the solutions

for first case: $2x^2-43x-45 \lt 0$ when $\displaystyle x \in (-1, \frac{45}{2})$ and $2x^2+3x \gt 0$ equivalent with $x\in (-\infty, -\frac{3}{2}) \cup (0, \infty)$. Now the solution for this case is the intersection : $$(-1, \frac{45}{2}) \cap( (-\infty, -\frac{3}{2}) \cup (0, \infty))=(0, \frac{45}{2})$$ and now the answer for second case is: $x^2-43x-45 \gt 0$ and $2x^2+3x \lt 0$. We have for first inequality $x\in (-\infty, -1) \cup (\frac{45}{2}, \infty)$ and for second inequality we have $x \in (-\frac{3}{2},0)$. So the answer for second case is : $$(-\frac{3}{2},0) \cap((-\infty, -1) \cup (\frac{45}{2}, \infty))=(-\frac{3}{2},-1)$$
And the final answer is $$x \in (-\frac{3}{2}, -1) \cup (0, \frac{45}{2}).$$

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I had a mistake but now it is correct. Thanks to @Fly by Night to tell me that I was wrong. – Iuli Aug 25 '12 at 21:53

After simplifying you have $\frac{2x^2-43x-45}{x(2x+3)}=\frac{(2x-45)(x+1)}{x(2x+3)}<0$. So we consider the function on the intervals $x<-\frac{3}{2}$, $x \in (-3/2,-1)$, $x \in (-1,0)$, $x \in (0,\frac{45}{2})$, $x>\frac{45}{2}$. At $x=-2$ the function is positive and the sign alternates in each successive region.

Thus the function is negative for $x \in (-3/2,-1)$ and $x \in (0,\frac{45}{2})$.

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Thank you. Correction, though - the $45$ should be negative. – Quispiam Aug 25 '12 at 21:16

There is at least one incorrect "solution" above.

The usual trick is to multiply both sides by something positive. So multiply both sides by $x^2(2x+3)^2$. Next, eliminate common factors in the numerators and denominators, bring everything over to the left hand side and then factorise. The inequality reduces to

$x(2x-45)(2x+3)(x+1) < 0 \, .$

You need to find the values of $x$ for which an odd number of $x$, $2x-45$, $2x+3$ and $x+1$ are negative. So clearly $x = 0$, $x = 45/2$, $x = -3/2$ and $x = -1$ play an important role: these are the values for which the factors change sign. The final result, contrary to Iuli's "solution", is: $-3/2 < x < -1$ or $0 < x < 45/2$.

To see why Iuli's "solution" is incorrect, let $x = -1/2 \in (-3/2,45/2) \backslash \{0\}$, yet

$-7 \not< -30 \, .$

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Yes, you are right . I will correct my mistake but my method is a good one. – Iuli Aug 25 '12 at 21:41
Having re-read my answer, I realise it sounded quite harsh. I hope I didn't offend you. Your method is a good one, I agree. – Fly by Night Aug 25 '12 at 22:05
@ Fly by Night It's ok :) have a good night – Iuli Aug 25 '12 at 22:06