What is |x-1|+|x+2|>= 3 ?

The solution to |x-1|+|x+2|>= 3 is True for all x

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|x-1|+|x+2|>= 3

∣ x - 1 ∣ + ∣ x + 2 ∣ \geq 3

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

∣ x - 1 ∣ + ∣ x + 2 ∣ \geq 3

Find positive and negative intervals

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

Solve the inequality for each interval

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

Merge Overlapping Intervals

True for all x

\frac{1}{( 2 - ∣ x ∣ )} \geq 1

x^{2} - 3 x - 3 > x^{2} + 7 x - 13

∣ x - 1 ∣ + ∣ x - 2 ∣ + ∣ x - 3 ∣ \geq 6

\frac{( ∣ x ∣ + 4 )}{( ∣ x ∣ + 3 )} \geq 0

∣ x - 1 ∣ < ∣ x ∣