What is |(6-5x)/(3+x)|>= 1/2 ?

The solution to |(6-5x)/(3+x)|>= 1/2 is x = 5/3

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|(6-5x)/(3+x)|>= 1/2

\frac{6 - 5 x}{3 + x} \geq \frac{1}{2}

x < - 3 or - 3 < x \leq \frac{9}{11} or x \geq \frac{5}{3}

Interval Notation

(- \infty, - 3) \cup (- 3, \frac{9}{11}] \cup [\frac{5}{3}, \infty)

Decimal Notation

x < - 3 or - 3 < x \leq 0.81818 \dots or x \geq 1.66666 \dots

Solution steps

\frac{6 - 5 x}{3 + x} \geq \frac{1}{2}

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

\frac{6 - 5 x}{3 + x} \leq - \frac{1}{2} or \frac{6 - 5 x}{3 + x} \geq \frac{1}{2}

\frac{6 - 5 x}{3 + x} \leq - \frac{1}{2} or \frac{6 - 5 x}{3 + x} \geq \frac{1}{2}

(x < - 3 or x \geq \frac{5}{3}) or - 3 < x \leq \frac{9}{11}

Merge Overlapping Intervals

x < - 3 or - 3 < x \leq \frac{9}{11} or x \geq \frac{5}{3}

- 81 + 100 x^{2} = 0

f (y) = \frac{4 y - 5}{2} - \frac{2 y - 1}{6}

∣ 4 x + 5 ∣ < 7

5 - 16 + 3 - 81

432 c^{3} d^{2}

What is |(6-5x)/(3+x)|>= 1/2 ?
The solution to |(6-5x)/(3+x)|>= 1/2 is x<-3\lor-3<x<= 9/11 \lor x>= 5/3