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

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

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

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

x < - 2 or - \frac{3}{5} \leq x < 1 or x > 5

Interval Notation

(- \infty, - 2) \cup [- \frac{3}{5}, 1) \cup (5, \infty)

Decimal Notation

x < - 2 or - 0.6 \leq x < 1 or x > 5

Solution steps

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

Rewrite in standard form

\frac{5 x + 3}{( x - 5 ) ( x + 2 ) ( x - 1 )} \geq 0

Identify the intervals

x < - 2 or - \frac{3}{5} \leq x < 1 or x > 5

6 d + 21 = 75

\frac{\frac{\frac{y ^{2} - z ^{2}}{yz}}{y - z}}{z}

- \frac{3}{2} + 1

- x < 8

8 x - (2 x - 1) = 2

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