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

The solution to |3x-2|>= |x+1| is x = 3/2

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

∣ 3 x - 2 ∣ \geq ∣ x + 1 ∣

x \leq \frac{1}{4} or x \geq \frac{3}{2}

Interval Notation

(- \infty, \frac{1}{4}] \cup [\frac{3}{2}, \infty)

Decimal Notation

x \leq 0.25 or x \geq 1.5

Solution steps

∣ 3 x - 2 ∣ \geq ∣ x + 1 ∣

Find positive and negative intervals

x < - 1, - 1 \leq x < \frac{2}{3}, x \geq \frac{2}{3}

Solve the inequality for each interval

x < - 1 or - 1 \leq x \leq \frac{1}{4} or x \geq \frac{3}{2}

Merge Overlapping Intervals

x \leq \frac{1}{4} or x \geq \frac{3}{2}

\frac{( x ^{2} - 5 x + 4 )}{( x ^{2} - 4 )} \leq 1

∣ x - 2 ∣ + ∣ x - 3 ∣ < 6

x^{2} - 1 \leq ∣ 2 x - 1 ∣

\frac{x}{( x ^{2} - x - 1 )} < 2

∣ 4 x - 3 ∣ + 2 < 10