What is |[ n/((2-3))]-1|< 1/10 ?

The solution to |[ n/((2-3))]-1|< 1/10 is -11/10 <n<-9/10

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|[ n/((2-3))]-1|< 1/10

[\frac{n}{( 2 - 3 )}] - 1 < \frac{1}{10}

- \frac{11}{10} < n < - \frac{9}{10}

Interval Notation

(- \frac{11}{10}, - \frac{9}{10})

Decimal Notation

- 1.1 < n < - 0.9

Solution steps

[\frac{n}{2 - 3}] - 1 < \frac{1}{10}

Apply absolute rule: If

∣ u ∣ < a, a > 0

then

- a < u < a

- \frac{1}{10} < [\frac{n}{2 - 3}] - 1 < \frac{1}{10}

[\frac{n}{2 - 3}] - 1 > - \frac{1}{10} and [\frac{n}{2 - 3}] - 1 < \frac{1}{10}

n < - \frac{9}{10} and n > - \frac{11}{10}

Merge Overlapping Intervals

- \frac{11}{10} < n < - \frac{9}{10}

∣ x - 1 ∣ > ∣ x - 2 ∣

∣ 4 x - 5 ∣ < 8

∣ 2 x + 1 ∣ + ∣ 2 x - 1 ∣ > 5

\frac{( x - 4 )}{( x ^{2} - 36 )} \leq 0

3 ∣ x - 3 ∣ + ∣ x + 4 ∣ \geq 12

What is |[ n/((2-3))]-1|< 1/10 ?
The solution to |[ n/((2-3))]-1|< 1/10 is -11/10 <n<-9/10