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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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3  
Without restrictions on $x$, you don't know the sign of $x+2$. This is dangerous when you multiply both sides of your inequality by $x+2$. –  1015 Dec 27 '12 at 20:13
    
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 $x<-2$. –  Thomas Andrews Dec 27 '12 at 20:36

3 Answers 3

up vote 1 down vote accepted

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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okay thanks, but how do you know that x < -2? I mean in this sample it is kinda obvious. But if it gets more complex I need to know how I would calculate both solutions. (See my solution above). –  Maik Klein Dec 27 '12 at 20:14
    
I think my main issue is how I get from a >= to a <. I mean multiplying with a negative number only changes the sign. –  Maik Klein Dec 27 '12 at 20:21
    
@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 $b \neq 0$. –  user17762 Dec 27 '12 at 20:22
    
(+1) Like it ${}~~~~~~~~$ –  Nancy Rutkowskie Jul 8 '13 at 12:27

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. –  Robert Israel Dec 27 '12 at 20:53

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