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

The solution to 1/(x-3)+1/(x+2)>= (2x)/(x^2+x-2) is x 3

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1/(x-3)+1/(x+2)>= (2x)/(x^2+x-2)

\frac{1}{x - 3} + \frac{1}{x + 2} \geq \frac{2 x}{x ^{2} + x - 2}

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

Interval Notation

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

Decimal Notation

x < - 2 or - 0.33333 \dots \leq x < 1 or x > 3

Solution steps

\frac{1}{x - 3} + \frac{1}{x + 2} \geq \frac{2 x}{x ^{2} + x - 2}

Rewrite in standard form

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

Identify the intervals

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

x^{3} - x^{2} - 14 x + 24

4 h + 36 = - 20

10 - x \leq 35

3 a^{2} - 48

2 x^{2} - 1 x - 15 = 0

What is 1/(x-3)+1/(x+2)>= (2x)/(x^2+x-2) ?
The solution to 1/(x-3)+1/(x+2)>= (2x)/(x^2+x-2) is x<-2\lor-1/3 <= x<1\lor x>3