I need some help/suggestions solving the following math problem. I don't know how to continue from step 2. Find x.
1.) $\displaystyle\left|\frac{x+1}{x}\right|< 1$
2.) $\displaystyle\frac{|x+1|}{|x|} < 1$
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But... what is the problem to solve? – ajotatxe May 21 '14 at 14:11
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$|a|<b <=> -b < a < b $, then solve for the two separate inequality($a<b$ and $-b<a$) and take the intersection of the solutions. – 05storm26 May 21 '14 at 14:11
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3Multiply with $|x|$, to get $|x + 1| < |x|$. Now draw some pictures. – Arthur May 21 '14 at 14:11
6 Answers
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$$\left|\frac{x+1}{x}\right|< 1$$
$$\left|1 + \frac{1}{x}\right|<1\\-1<1 + \frac{1}{x}<1\\ -2<\frac{1}{x}<0 \\\left(\frac{1}{x} >-2\right) \wedge \left(\frac{1}{x}<0\right)$$
Dismiss $\displaystyle x>0$ because of the second inequality.
Hence $\displaystyle x<0$.
From first inequality, $\displaystyle 1 < -2x\\\displaystyle x<-\frac{1}{2}$
Therefore $\displaystyle x<-\frac{1}{2}$ is the only solution.
Credit to Zarrax who helped me realise what I had originally missed.
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I got the same solution. I assume that x must not be zero, because that is where the fraction is not defined. Why does it say in the solutions:$(x< -\frac{1}{2}) V (x>0); x<0$ ? – user114141 May 21 '14 at 14:23
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$1/x < 0$ is relevant here; it forces $x$ to be negative. Note that $-2 < {1 \over x}$ is solved by any $x > 0$. – Zarrax May 21 '14 at 14:27
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@Zarrax. No that inequality is not relevant. All solutions of $x < -\frac{1}{2}$ are already negative. – Deepak May 21 '14 at 14:29
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1@Deepak $-2 < {1 \over x}$ is solved by any $x > 0$, so that inequality would not be enough to get the correct answer. This sort of annoyance is why I suggest squaring both sides in this type of problem. – Zarrax May 21 '14 at 14:30
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@Zarrax. Good point. I missed that. Yes, squaring does seem neater. But one must be careful to exclude redundant roots introduced in squaring. In this case, no redundant roots are introduced, though. – Deepak May 21 '14 at 14:33
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When you have an equation only involving absolute values and other positive quantities, you can square both sides. Sometimes this makes it easier to solve, as in this case. $$\bigg({x +1 \over x}\bigg)^2 < 1$$ Multiply both sides by $x^2$....
Zarrax
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Hint:
$\left|\frac{x+1}{x}\right|=\left|1+\frac{1}{x}\right|$ and in general $\left|1+a\right|<1\iff a\in\left(-2,0\right)$. So apply this on $a=\frac{1}{x}$.
drhab
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For which $x$ is this fraction defined?
Can we multiply both sides by $|x|$? What will happen with the sign of the inequality?
What are the points of sign-changing of left and right hand sides of the new inequality?
Can you solve it case by case?
TZakrevskiy
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Since $|x|\ge 0$, and we may assume $x\neq 0$ for the fraction to be defined, we multiply both sides by $|x|$ to get $$|x+1|<|x|$$ We now complete the problem with geometry, rewriting as $$|x-(-1)|<|x-0|$$ The expression $|x-a|$ measures the distance between $x$ and $a$, so we want all points $x$ that are closer to $-1$ than they are to $0$. This is exactly all points $x<-0.5$.
vadim123
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Since both sides of the inequality are positive,you can get them squared..and solve it like a normal equation..
user152456
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