What is 3|x-3|+|x+4|>= 12 ?

The solution to 3|x-3|+|x+4|>= 12 is x = 17/4

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3|x-3|+|x+4|>= 12

3 ∣ x - 3 ∣ + ∣ x + 4 ∣ \geq 12

x \leq \frac{1}{2} or x \geq \frac{17}{4}

Interval Notation

(- \infty, \frac{1}{2}] \cup [\frac{17}{4}, \infty)

Decimal Notation

x \leq 0.5 or x \geq 4.25

Solution steps

3 ∣ x - 3 ∣ + ∣ x + 4 ∣ \geq 12

Find positive and negative intervals

x < - 4, - 4 \leq x < 3, x \geq 3

Solve the inequality for each interval

x < - 4 or - 4 \leq x \leq \frac{1}{2} or x \geq \frac{17}{4}

Merge Overlapping Intervals

x \leq \frac{1}{2} or x \geq \frac{17}{4}

x^{2} - 5 x + 4 - x^{2} - 4 < 0

∣ x + 2 ∣ < 4 ∣ 2 x - 3 ∣

x^{2} - ∣ x + 2 ∣ + x => 0

∣ 4 - x ∣ < - 11

\frac{( x - 2 )}{( x - 2 )} \cdot (\frac{- 1}{2}) \leq 0

What is 3|x-3|+|x+4|>= 12 ?
The solution to 3|x-3|+|x+4|>= 12 is x<= 1/2 \lor x>= 17/4