What is 4/(x-2)+x/(x+2)<= (8x)/(x^2-4) ?

The solution to 4/(x-2)+x/(x+2)<= (8x)/(x^2-4) is -2<x<2\lor 2<x<= 4

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4/(x-2)+x/(x+2)<= (8x)/(x^2-4)

\frac{4}{x - 2} + \frac{x}{x + 2} \leq \frac{8 x}{x ^{2} - 4}

- 2 < x < 2 or 2 < x \leq 4

Solution steps

\frac{4}{x - 2} + \frac{x}{x + 2} \leq \frac{8 x}{x ^{2} - 4}

Rewrite in standard form

\frac{x ^{2} - 6 x + 8}{( x - 2 ) ( x + 2 )} \leq 0

Factor

\frac{x ^{2} - 6 x + 8}{( x - 2 ) ( x + 2 )} : \frac{( x - 2 ) ( x - 4 )}{( x - 2 ) ( x + 2 )}

\frac{( x - 2 ) ( x - 4 )}{( x - 2 ) ( x + 2 )} \leq 0

Identify the intervals

- 2 < x < 2 or 2 < x \leq 4

\frac{300}{( 1 + 0.06 ) ^{10}}

2 x + 4 = 8 x

3 (1 - x) = 4 (1 + 2 x) + 5

7 (- 3) + 9 (2) > - 3

(1 + 5) (1 - 5)

What is 4/(x-2)+x/(x+2)<= (8x)/(x^2-4) ?
The solution to 4/(x-2)+x/(x+2)<= (8x)/(x^2-4) is -2<x<2\lor 2<x<= 4