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The question says to find all the integral values of x for which the inequality holds. the question is

$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$

My Solution

\begin{align} & \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt] & \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\[6pt] & \frac{14x^2-64x-9x^2+21x+30}{(x+1)(x-4)} < 0 \\[6pt] & \frac{5x^2-43x+30}{(x+1)(x-4)}<0 \end{align}

using quadratic formula, roots of $5x^2-43x+30$ comes $7.83$(approx.) and $0.763$(approx.)

so rewriting the equation as

$$\frac{(x-7.83)(x-0.763)}{(x+1)(x-4)}<0$$

I then did the plotting of zeroes and poles on number line for finding the values for $x$ but I donot get 2 integral values (which is the answer). Can anyone tell where I did wrong?

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  • $\begingroup$ Well take their intersection with $\mathbb Z$ $\endgroup$ – Oussama Boussif Jul 9 '16 at 17:54
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    $\begingroup$ You made mistake, you should get $\frac{5x^2=35x+30}{(x+1)(-4)}<0$ or equivalently $\frac{5(x-1)(x-6)}{(x+1)(x-4)}<0$. $\endgroup$ – alans Jul 9 '16 at 17:57
  • $\begingroup$ I took but i get 4 integral values 0,5,6,7 but the answer says it has only two possible integral values $\endgroup$ – danny Jul 9 '16 at 17:57
  • $\begingroup$ @alans i also thought that i made a mistake there but i am not able to figure out what is the mistake i did. $\endgroup$ – danny Jul 9 '16 at 18:05
  • $\begingroup$ I deleted all the inappropriate "equals" signs. Those should be used ONLY when you're saying things are equal. $\qquad$ $\endgroup$ – Michael Hardy Jul 9 '16 at 18:06
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Inequality is equivalent with $$\frac{(x-1)(x-6)}{(x+1)(x-4)}<0.$$ Function $f(x)=\frac{(x-1)(x-6)}{(x+1)(x-4)}$ has sign $+$ on intervals $(-\infty,-1)$, $(1,4)$, $(6,\infty)$ and sign $-$ on intervals $(-1,1)$ and $(4,6)$.

Therefore, only integer solutions that satisfy inequality are $0$ and $5$.

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Basic approach. Perhaps easier would be to rewrite the original inequality as

$$ 14 - \frac{14}{x+1} < 9 + \frac{6}{x-4} $$

This leads to

$$ \frac{14}{x+1} + \frac{6}{x-4} > 5 $$

which becomes

$$ \frac{4x-10}{(x+1)(x-4)} > 1 $$

You can now (a) consider the cases $x = 0, 1, 2, 3$ separately, and then otherwise, (b) for $x < -1$ or $x > 4$, we have

$$ (x+1)(x-4) < 4x-10 $$

which becomes

$$ x^2-7x+6 < 0 $$

which you should be able to handle. Keep in mind that this inequality is only valid for the subcase (b) $x < -1$ or $x > 4$.

(There's probably a simpler way, incidentally. See alans' comment, for instance. This is just what I wrote up.)

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  • $\begingroup$ $x^2-7x+6<0$ is equivalent with $1<x<6$, which is not solution (take $x=2$ as eample). You shouldn't multiply with $(x+1)(x-4)$, that caused mistake. $\endgroup$ – alans Jul 9 '16 at 18:16
  • $\begingroup$ @alans: I left it a little unclear, but that section is prefaced with "otherwise, for $x < -1$ or $x > 4$." In conjunction with $1 < x < 6$, that leaves $x = 5$. The other solution, $x = 0$, is found by direct inspection. ¶ Did you downvote? $\endgroup$ – Brian Tung Jul 9 '16 at 21:36
  • $\begingroup$ Yes, I downvoted, sorry, it was unclear. I'll fix that now. $\endgroup$ – alans Jul 10 '16 at 10:46
  • $\begingroup$ @alans: It was worth noting that it was unclear. +1 to you too! :-) $\endgroup$ – Brian Tung Jul 10 '16 at 20:49
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One easy way to solve is to consider the corresponding equation. A change of sings can only occur due to continuity at the equation's zeros or where the equation is undefined.

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Correcting your initial mistake will lead to the inequality

\begin{equation} \dfrac{(x+1)(x-6)}{(x+1)(x-4)}<0 \end{equation}

which is positive on the interval $(6,\infty)$ since the product of $x$ coefficients is positive, and the expression changes sign at each zero and pole as you move left along the $x$-axis since each of the four linear factors is raised to an odd power. Thus it is negative only on the intervals $(-1,1)$ and $(4,6)$.

So the only integral solutions are $x=0$ and $x=5$.

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