What is |((9x)/2-1/3)/(2/3)|>= 4 ?

The solution to |((9x)/2-1/3)/(2/3)|>= 4 is x = 2/3

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|((9x)/2-1/3)/(2/3)|>= 4

\frac{\frac{9 x}{2} - \frac{1}{3}}{\frac{2}{3}} \geq 4

x \leq - \frac{14}{27} or x \geq \frac{2}{3}

Interval Notation

(- \infty, - \frac{14}{27}] \cup [\frac{2}{3}, \infty)

Decimal Notation

x \leq - 0.51851 \dots or x \geq 0.66666 \dots

Solution steps

\frac{\frac{9 x}{2} - \frac{1}{3}}{\frac{2}{3}} \geq 4

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

\frac{\frac{9 x}{2} - \frac{1}{3}}{\frac{2}{3}} \leq - 4 or \frac{\frac{9 x}{2} - \frac{1}{3}}{\frac{2}{3}} \geq 4

\frac{\frac{9 x}{2} - \frac{1}{3}}{\frac{2}{3}} \leq - 4 or \frac{\frac{9 x}{2} - \frac{1}{3}}{\frac{2}{3}} \geq 4

x \leq - \frac{14}{27} or x \geq \frac{2}{3}

x^{2} + 8 \geq 6 x

9 a^{4} + 11 a^{3} b + 2 a^{2} b^{2}

\frac{3 ^{327}}{3 ^{323}}

3 x^{3} + 5 x^{2} - 2 x

(\frac{1}{3})^{6}

What is |((9x)/2-1/3)/(2/3)|>= 4 ?
The solution to |((9x)/2-1/3)/(2/3)|>= 4 is x<=-14/27 \lor x>= 2/3