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I have the following inequality:

$ \frac{x-1}{x+2} \geq 0.$

I solved it pretty fast:

$\begin{align} \frac{x-1}{x+2} +1 & \geq 1\\\\ \left(\frac{x-1}{x+2} + 1\right)\cdot(x+2) & \geq 1 \cdot (x+2)\\\\ x-1 + 1\cdot(x+2) & \geq 1\cdot (x+2)\\\\ 2x + 1 & \geq x+2\\\\ x + 1 & \geq 2\\\\ x & \geq 1 \end{align}$

But that is not the only solution, the other solution is $x < -2$. How do I get to this solution?

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    Note, you don't really need the "add 1" trick, just multiply. Then if x+2>0, then you need $x-1\geq 0$ or $x\geq 1$. And if x+2< 0 you need $x-1\leq 1$ or $x\leq 1$. So either $x\geq 1$ or x<-2.2012-12-27

3 Answers 3

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A geometric idea applied to algebra: multiply the inequality by a positive quantity, so that you won't have problems with the inequality sign's direction:

$\frac{x-1}{x+2}\geq 0\Longleftrightarrow \frac{x-1}{x+2}\cdot(x+2)^2\geq 0\cdot(x+2)^2\Longrightarrow (x-1)(x+2)\geq 0$

The left side is a parabola that opens upwards and vanishes at $\,x=-2\,,\,1\,$ , and from the geometric picture we can see the parabola lies above the $\,x-$axis, which is what we want, precisely whenever $\,x<-2\,\,\;\vee\;\,\,x>1\,$.

Finally, we just add the point $\,x=1\,$ to the above as we had a weak inequality.

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First note that when you multiply by a negative number the inequality changes in sign. In general, if $\dfrac{a}{b} > 0$, we have $a>0, b>0$ or $a<0, b<0$. Hence, we get that $(x-1) > 0, \,\,\,\,\, (x+2) > 0 \text{ i.e. } x>1$ or $(x-1) < 0, \,\,\,\,\, (x+2) < 0 \text{ i.e. }x < -2$ If $\dfrac{x-1}{x+2} = 0$, then $x=1$. Hence, the solution is $x \in (-\infty,-2) \cup [1,\infty)$

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    @MaikKlein Split $\dfrac{a}b \geq 0$ into two disjoint cases i.e. \dfrac{a}b > 0 and $\dfrac{a}b = 0$. For the former, we know that a,b > 0 or a,b<0. For the latter, we have $a=0$, provided $b \neq 0$.2012-12-27
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The $=0$ part is easy. So let's worry about the $\gt 0$ part.

Our expression can only change sign when the top changes sign, or when the bottom changes sign. Thus the only "sign change" candidates are $x=1$ and $x=-2$.

It follows that our function is uniform in sign in $(-\infty,-2)$, also in $(-2,1)$, also in $(1,\infty)$.

In each of these regions, sind a test point: any point will do.

For the region $(-\infty,-2)$, use say the test point $x=-12$. Our function is $\frac{-13}{-10}$ at this point, positive. So our function is positive in all of $(-\infty,-2)$.

For the interval $(-2,1)$, use the test point $x=0$. At $0$, our function is clearly negative, so it is negative in the whole interval.

Finally, use a test point in $(1,\infty)$ to conclude our function is positive in that interval.

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    Of course, knowing that the numerator changes sign at $1$ (while the denominator does not) and the denominator changes sign at $-2$ (while the numerator does not) and those are the only points where a sign changes, you really only need one test point.2012-12-27