What is (3-x)/(x+2)<= 1/x ?

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(3-x)/(x+2)<= 1/x

\frac{3 - x}{x + 2} \leq \frac{1}{x}

x < - 2 or x > 0

Interval Notation

(- \infty, - 2) \cup (0, \infty)

Solution steps

\frac{3 - x}{x + 2} \leq \frac{1}{x}

Rewrite in standard form

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

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x ^{2} - 2 x + 2 ) ) ( - 1 )}{x ( x + 2 )} \geq 0 \cdot (- 1)

Simplify

\frac{x ^{2} - 2 x + 2}{x ( x + 2 )} \geq 0

Identify the intervals

x < - 2 or x > 0

x^{3} - 4 x > 0

x^{2} + 27 x + 50

x^{2} + 16 x + 145 = 0

2 x^{2} + 7 x - 15 \leq 0

m x + m y + 6 x + 6 y