What is |(6x-4)/(3+x)|>= 1/2 ?

The solution to |(6x-4)/(3+x)|>= 1/2 is x = 1

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|(6x-4)/(3+x)|>= 1/2

\frac{6 x - 4}{3 + x} \geq \frac{1}{2}

x < - 3 or - 3 < x \leq \frac{5}{13} or x \geq 1

Interval Notation

(- \infty, - 3) \cup (- 3, \frac{5}{13}] \cup [1, \infty)

Decimal Notation

x < - 3 or - 3 < x \leq 0.38461 \dots or x \geq 1

Solution steps

\frac{6 x - 4}{3 + x} \geq \frac{1}{2}

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

\frac{6 x - 4}{3 + x} \leq - \frac{1}{2} or \frac{6 x - 4}{3 + x} \geq \frac{1}{2}

\frac{6 x - 4}{3 + x} \leq - \frac{1}{2} or \frac{6 x - 4}{3 + x} \geq \frac{1}{2}

- 3 < x \leq \frac{5}{13} or (x < - 3 or x \geq 1)

Merge Overlapping Intervals

x < - 3 or - 3 < x \leq \frac{5}{13} or x \geq 1

x - 3 (x - 2) = 6 x - 2

x < \frac{1}{3} and x > - \frac{3}{4}

7147

1.5 = x^{2}

3 + 4 (x + 1) = 159

What is |(6x-4)/(3+x)|>= 1/2 ?
The solution to |(6x-4)/(3+x)|>= 1/2 is x<-3\lor-3<x<= 5/13 \lor x>= 1