Solving $|\frac{x+1}{x}|< 1$

Question

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.) $\displaystyle\frac{|x+1|}{|x|} < 1$

Answer 1

$$\left|\frac{x+1}{x}\right|< 1$$

$$\left|1 + \frac{1}{x}\right|<1\\-1<1 + \frac{1}{x}<1\\ -2<\frac{1}{x}<0 \\\left(\frac{1}{x} >-2\right) \wedge \left(\frac{1}{x}<0\right)$$

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.

Answer 2

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?

Answer 3

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 $x^2$....

Answer 4

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 $a=\frac{1}{x}$.

Answer 5

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 $x<-0.5$.

Answer 6

Since both sides of the inequality are positive,you can get them squared..and solve it like a normal equation..