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

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(x-3}{x^2}<= \frac{x+3)/x

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

x < 0 or x > 0

Interval Notation

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

Solution steps

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

Rewrite in standard form

\frac{- x ^{2} - 2 x - 3}{x ^{2}} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x ^{2} + 2 x + 3 ) ) ( - 1 )}{x ^{2}} \geq 0 \cdot (- 1)

Simplify

\frac{x ^{2} + 2 x + 3}{x ^{2}} \geq 0

Identify the intervals

x < 0 or x > 0

(x + 6)^{2} = 3 (x + 12) - x^{2}

\frac{3 x - 4}{2} < \frac{4 x + 7}{5}

28 = 5 x - 7

2 m^{2} - 12 m - 17 = - 5

x^{2} + 17 x + 16 = 0

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