What is |4x-1|+4x<|x+2| ?

The solution to |4x-1|+4x<|x+2| is x<-3\lor-1<x< 3/7

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|4x-1|+4x<|x+2|

∣ 4 x - 1 ∣ + 4 x < ∣ x + 2 ∣

x < - 3 or - 1 < x < \frac{3}{7}

Interval Notation

(- \infty, - 3) \cup (- 1, \frac{3}{7})

Decimal Notation

x < - 3 or - 1 < x < 0.42857 \dots

Solution steps

∣ 4 x - 1 ∣ + 4 x < ∣ x + 2 ∣

Find positive and negative intervals

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

Solve the inequality for each interval

x < - 3 or - 1 < x < \frac{1}{4} or \frac{1}{4} \leq x < \frac{3}{7}

Merge Overlapping Intervals

x < - 3 or - 1 < x < \frac{3}{7}

- 3.56 = 2.7 (4 - y)

5 x \cdot 5 x

2 x (x - 3) = 4 x^{2} - 7 x

2 x + 7 = 3

(\frac{4 a ^{2} b}{a ^{3} b ^{2}}) (\frac{5 a ^{2} b}{2 b ^{4}})