What is 1/(|x-1|)< 1/(|x+2|) ?

The solution to 1/(|x-1|)< 1/(|x+2|) is x<-2\lor-2<x<-1/2

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1/(|x-1|)< 1/(|x+2|)

\frac{1}{∣ x - 1 ∣} < \frac{1}{∣ x + 2 ∣}

x < - 2 or - 2 < x < - \frac{1}{2}

Solution steps

\frac{1}{∣ x - 1 ∣} < \frac{1}{∣ x + 2 ∣}

Find positive and negative intervals

x < - 2, - 2 \leq x < 1, x \geq 1

Solve the inequality for each interval

x < - 2 or - 2 < x < - \frac{1}{2} or No Solution

Merge Overlapping Intervals

x < - 2 or - 2 < x < - \frac{1}{2}

220 - 20 + 320 - 245

\frac{4}{3 ^{3}}

\frac{\frac{1}{y} + \frac{1}{x}}{\frac{x}{y} - \frac{y}{x}}

\frac{x ^{2} - y ^{2}}{x - y}

2 x^{2} - 13 x + 15 = 0

What is 1/(|x-1|)< 1/(|x+2|) ?
The solution to 1/(|x-1|)< 1/(|x+2|) is x<-2\lor-2<x<-1/2