What is (x^2-2x-2}{x^2-3x}+(7x-19)/(x-3)<= \frac{8x+1)/x ?

The solution to (x^2-2x-2}{x^2-3x}+(7x-19)/(x-3)<= \frac{8x+1)/x is x<=-1/2 \lor 0<x<3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2-2x-2}{x^2-3x}+(7x-19)/(x-3)<= \frac{8x+1)/x

\frac{x ^{2} - 2 x - 2}{x ^{2} - 3 x} + \frac{7 x - 19}{x - 3} \leq \frac{8 x + 1}{x}

x \leq - \frac{1}{2} or 0 < x < 3

Interval Notation

(- \infty, - \frac{1}{2}] \cup (0, 3)

Decimal Notation

x \leq - 0.5 or 0 < x < 3

Solution steps

\frac{x ^{2} - 2 x - 2}{x ^{2} - 3 x} + \frac{7 x - 19}{x - 3} \leq \frac{8 x + 1}{x}

Rewrite in standard form

\frac{2 x + 1}{x ( x - 3 )} \leq 0

Identify the intervals

x \leq - \frac{1}{2} or 0 < x < 3

\frac{5}{6} v < \frac{8}{5}

p^{2} - 4 p + 4 = 0

∣ 2 b - 23 ∣ \geq - 15

5 < x \leq 7

(\frac{x ^{3}}{2 y ^{6}})^{- 3}

What is (x^2-2x-2}{x^2-3x}+(7x-19)/(x-3)<= \frac{8x+1)/x ?
The solution to (x^2-2x-2}{x^2-3x}+(7x-19)/(x-3)<= \frac{8x+1)/x is x<=-1/2 \lor 0<x<3