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

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

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

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

x \leq - \frac{11}{3} or - \frac{9}{7} \leq x < \frac{1}{2} or x > \frac{1}{2}

Interval Notation

(- \infty, - \frac{11}{3}] \cup [- \frac{9}{7}, \frac{1}{2}) \cup (\frac{1}{2}, \infty)

Decimal Notation

x \leq - 3.66666 \dots or - 1.28571 \dots \leq x < 0.5 or x > 0.5

Solution steps

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

Find positive and negative intervals

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

Solve the inequality for each interval

x \leq - \frac{11}{3} or - \frac{9}{7} \leq x < \frac{1}{2} or x > \frac{1}{2}

2 x + 12 x^{3}

\frac{a ^{2} - 25}{a ^{3} + 125}

8 x^{3} y + 38 x^{2} y - 10 x y

x + 3 = x - 2

(2 y - 1)^{2} + (2 y + 1)^{2} = (2 y)^{2} + 6

What is | 5/(2x-1)|>= | 1/(x+2)| ?
The solution to | 5/(2x-1)|>= | 1/(x+2)| is x<=-11/3 \lor-9/7 <= x< 1/2 \lor x> 1/2