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Ok so I have the following two inequalities:

\begin{equation} \left| \frac{x+6}{x-2}\right| \leq 4 \end{equation} and \begin{equation} \frac{x^2-1}{\left| x+2\right|} \leq 3(1-x) \end{equation} Also for the value of $x$ it just holds $x\ \in \mathbb{R}$.

For the first one what I have done by now is the following:

\begin{equation} \frac{1}{\left| \frac{x+6}{x-2}\right|} \geq \frac{1}{4} \Leftrightarrow \left| \frac{x-2}{x+6}\right| \geq \frac{1}{4} \end{equation} which is equivalent to

\begin{equation} \left( \frac{x-2}{x+6} \leq -\frac{1}{4}\right) \vee \left( \frac{x-2}{x+6} \geq \frac{1}{4}\right) \end{equation} After this point I am not sure how to proceed. For the second one, I do not even know how to start. Thanks a ton!

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  • $\begingroup$ Note that for all real numbers $x$ different from $2$ one has $$\dfrac{x+6}{x-2}=\dfrac{(x-2)+6}{x-2}=1+\dfrac{6}{x-2}.$$ If you want to use what you did until this point, note instead that $$\forall x\in \mathbb R\setminus \{6\}\left(\dfrac{x-2}{x+6}=1-\dfrac{8}{x+6}\right).$$ $\endgroup$ – Git Gud Jan 23 '16 at 23:47
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    $\begingroup$ @GitGud: I think you might mean $\dfrac{x+6}{x-2}=\dfrac{(x-2)+8}{x-2} = 1+\dfrac{8}{x-2}$. $\endgroup$ – Frentos Jan 24 '16 at 0:02
  • $\begingroup$ @Frentos Indeed, thanks. $\endgroup$ – Git Gud Jan 24 '16 at 0:07
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The way that I would go about doing this problem (at least for the first inequality) is by squaring both sides and then cross-multiplying the denominator over.

This should give you an inequality between two quadratics:

$x^2 + 12x + 36 ≤ 16(x^2 + 4 - 4x)$

You can solve this by setting one side of the inequality equal to zero:

$0≤15x^2 - 76x + 28$

Factor the quadratic on the right and use a number line or similar procedure to identify when the inequality holds.

For the second inequality (the bottom one), multiply the denominator over (it must be positive):

$x^2+1 ≤3*|x-2|*(1-x)$

Now consider cases. If $x<1$, then $1-x$ is positive, and the following division may be conducted without flipping the inequality sign:

$\frac{x^2+1}{1-x} ≤3|x-2|$

The left side reduces by factoring the numerator as a difference of squares:

$-(x+1) ≤ 3|x-2|$

Now, square both sides and solve the resulting quadratic-inequality in the method we used for inequality (1).

We must now consider the contrary case, that $x>1$. All this involves is flipping the inequality sign when we divide the term $1-x$. Follow the same procedure that we did in the case $x<1$.

The final case we need to consider is that $x=1$. If this is the case it is readily apparent that the inequality holds because it is not strict.

The final step to do is to find the intersection between the solution sets for inequalities 1 and 2. This will be the solution set x to the system of inequalities.

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  • $\begingroup$ By the way, the first quadratic inequality, though it looks incredibly ugly, has remarkably simple roots: 2/5 and 14/3. (Found via application of the quadratic formula). Between those two values, the quadratic is negative, so the first inequality is met when $x≥\frac{14}{3}$ or when $x≤\frac{2}{5}$. $\endgroup$ – KR136 Jan 24 '16 at 0:05
  • $\begingroup$ The reason the roots are all rational is because you don't need to use quadratics to solve the equation. It can all be solved with linear inequalities using different cases for the absolute value. $\endgroup$ – Noble Mushtak Jan 24 '16 at 0:19
  • $\begingroup$ @NobleMushtak Yes linear inequalities definitely apply here (and perhaps are easier/what the OP was looking for). This is just the solution method that occurred first to me when looking at the problem. $\endgroup$ – KR136 Jan 24 '16 at 0:39
  • $\begingroup$ I'm not sure if it's easier because it took me a few hours to finally get the answer, but I personally like different cases and sub-cases more than quadratics. $\endgroup$ – Noble Mushtak Jan 24 '16 at 1:12
  • $\begingroup$ Thank you guys! Appreciate it! :D $\endgroup$ – Mitscaype Jan 24 '16 at 14:27
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Here's a solution without quadratics, but with a lot of sub-cases. I like this solution because I like sub-cases more than factoring/solving quadratics. However, if you just want the answer to this and now how to get the answer, enter your inequalities into here. It will give you the same solution set as my answer.

For the first inequality, rewrite the inequality so you can get rid of the absolute value: \begin{equation} -4 \leq \frac{x+6}{x-2} \leq 4 \end{equation} Multiply all sides by $x-2$. There are now two cases: The case where $x > 2$, in which case we multiply normally, and the case in which $x < 2$, in which case we switch the signs.

Case $x > 2$: \begin{equation} -4x+8 \leq x+6 \leq 4x-8 \end{equation} Now, split this into two linear inequalities: \begin{equation} -4x+8 \leq x+6 \wedge x+6 \leq 4x-8 \end{equation} Solve both inequalities: \begin{equation} x \geq 2/5 \wedge x \geq 14/3 \end{equation} Now, compound this with the fact that $x > 2$. Since $2/5 \leq 2 < 14/3 \leq x$, the solution is simply $x \geq 14/3$.

Case $x < 2$:

We get the same answer as the above sub-case, except with the inequality signs switched: \begin{equation} x \leq 2/5 \wedge x \leq 14/3 \end{equation} Now, compound this with the fact that $x < 2$. Since $x \leq 2/5 < 2 \leq 14/3$, the solution is simply $x \leq 2/5$.

Thus, for the first inequality, by looking at all of our cases, the solution is $x \leq 2/5 \vee 14/3 \leq x$.


For the second inequality, just divide by $1-x$. However, there are three cases: $x=1$, in which case the inequality holds true, $x < 1$, in which case $1-x$ is positive and one only needs to divide by $1-x$, and $x > 1$, in which case $1-x$ is negative and one needs to switch the inequality sign after dividing by $1-x$.

Case $x < 1$: \begin{equation} \frac{-x-1}{|x+2|} \leq 3 \end{equation} Multiply both sides by $-1$. Switch the signs accordingly. \begin{equation} \frac{x+1}{|x+2|} \geq -3 \end{equation} Multiply both sides by $|x+2|$: \begin{equation} x+1 \geq -3|x+2| \end{equation} Now, there are two sub cases: $x < -2$, in which case $|x+2|=-x-2$, and $-2 < x < 1$, in which case $|x+2|=x+2$.

Sub-Case $x < -2$: \begin{equation} x+1 \geq -3(-x-2)=3x+6 \end{equation} Solve the linear inequality: \begin{equation} x \leq -5/2 \end{equation} Now, compound this with the fact that $x < -2$. Since $x \leq -5/2 < -2$, the solution for this sub-case is simply $x \leq -5/2$.

Sub-Case $-2 < x < 1$: \begin{equation} x+1 \geq -3(x+2)=-3x-6 \end{equation} Solve the linear inequality: \begin{equation} x \geq -7/4 \end{equation} Now, compound this with the fact that $-2 < x < 1$. Since $-2 < -7/4 \leq x < 1$, the solution is simply $-7/4 \leq x < 1$.

Now, we're done with the case where $x < 1$. Go back to the beginning where we were going to do the case of $x > 1$.

Case $x > 1$: \begin{equation} \frac{-x-1}{|x+2|} \leq 3 \end{equation} Multiply both sides by $|x+2|$: \begin{equation} -x-1 \leq 3|x+2| \end{equation} Since $-2 < 1 < x$, $|x+2|=x+2$: \begin{equation} -x-1 \leq 3(x+2)=3x+6 \end{equation} Solve the linear inequality: \begin{equation} x \leq -7/4 \end{equation} Now, compound this with the fact that $x > 1$. Since $-7/4 \leq x > 1$ and $-7/4 < 1$, this case is impossible and has no solution.

Thus, from this inequality, by looking at all of our cases and sub-cases, the solution is $x \leq -5/2 \vee -7/4 \leq x < 1 \vee x=1$. However, the last two inequalities can be combined to simplify the solution to $x \leq -5/2 \vee -7/4 \leq x \leq 1$.


Now, we just need to compound the solutions from both inequalities: \begin{equation} (x \leq 2/5 \vee 14/3 \leq x) \wedge (x \leq -5/2 \vee -7/4 \leq x \leq 1) \end{equation} Now, notice that we have five numbers in this inequality: $-5/2 < -7/4 < 2/5 < 1 < 14/3$. To figure out when this inequality is true, we need to handle six cases: $x \leq -5/2$, $x$ is in between one of these numbers, or $14/3 < x$.

Case $x \leq -5/2$: The inequality is true because $x \leq 2/5 \wedge x \leq -5/2$.

Case $-5/2 \leq x < 7/4$: The inequality is false because there is no way to make the second statement in the conjunction true in this case.

Case $-7/4 \leq x \leq 2/5$: The inequality is true because $x \leq 2/5 \wedge -7/4 \leq x \leq 1$.

Case $2/5 < x \leq 1$: The inequality is false because there is no way to make the first statement in the conjunction true in this case.

Case $1 < x < 14/3$: The inequality is false because there is no way to make either statement in the conjunction true in this case.

Case $14/3 < x$: The inequality is false because there is no way to make the second statement in the conjunction true in this case.

Thus, by looking at all of the solutions from all of our cases, we find that the solution to this compound inequality -- the solution to this whole problem -- is $x \leq -5/2 \vee -7/4 \leq x \leq 2/5$.

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  • $\begingroup$ Hey thank you a lot! No I do not just care about the outcome of the inequality. What I am interested in is the way to solve it. Again thank you for your time answering this one! $\endgroup$ – Mitscaype Jan 24 '16 at 14:24
  • $\begingroup$ Glad I could help! $\endgroup$ – Noble Mushtak Jan 24 '16 at 14:26
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Thank you Noble Mushtak for that informative analysis but I think that your final solution of $x \leq \frac{−5}2 \vee \frac{−7}4 \leq x\leq \frac{2}5$ to the second inequality was a bit off.

The graph at wolfram alpha tells me that it should be $x \leq \frac{−5}2 \vee \frac{−7}4 \leq x \le1$ as you indicated earlier in your analysis.

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