What is |z+1|+|z-1|<= 3 ?

The solution to |z+1|+|z-1|<= 3 is -3/2 <= z<= 3/2

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|z+1|+|z-1|<= 3

∣ z + 1 ∣ + ∣ z - 1 ∣ \leq 3

- \frac{3}{2} \leq z \leq \frac{3}{2}

Interval Notation

[- \frac{3}{2}, \frac{3}{2}]

Decimal Notation

- 1.5 \leq z \leq 1.5

Solution steps

∣ z + 1 ∣ + ∣ z - 1 ∣ \leq 3

Find positive and negative intervals

z < - 1, - 1 \leq z < 1, z \geq 1

Solve the inequality for each interval

- \frac{3}{2} \leq z < - 1 or - 1 \leq z < 1 or 1 \leq z \leq \frac{3}{2}

Merge Overlapping Intervals

- \frac{3}{2} \leq z \leq \frac{3}{2}

\frac{6 a ^{5} b ^{2}}{4 a b ^{3}}

1 < \frac{4 x - 4}{2 x + 3} < 3

\frac{x ^{9}}{x ^{7}}

- 6 (4 - 6 i)

4 (x + 3) = 1