What is 1/(x+4)+1/(x-8)<= 0 ?

The solution to 1/(x+4)+1/(x-8)<= 0 is x<-4\lor 2<= x<8

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1/(x+4)+1/(x-8)<= 0

\frac{1}{x + 4} + \frac{1}{x - 8} \leq 0

x < - 4 or 2 \leq x < 8

Interval Notation

(- \infty, - 4) \cup [2, 8)

Solution steps

\frac{1}{x + 4} + \frac{1}{x - 8} \leq 0

Rewrite in standard form

\frac{2 x - 4}{( x + 4 ) ( x - 8 )} \leq 0

Factor

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

\frac{2 ( x - 2 )}{( x + 4 ) ( x - 8 )} \leq 0

Divide both sides by

2

\frac{\frac{2 ( x - 2 )}{( x + 4 ) ( x - 8 )}}{2} \leq \frac{0}{2}

Simplify

\frac{x - 2}{( x + 4 ) ( x - 8 )} \leq 0

Identify the intervals

x < - 4 or 2 \leq x < 8

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What is 1/(x+4)+1/(x-8)<= 0 ?
The solution to 1/(x+4)+1/(x-8)<= 0 is x<-4\lor 2<= x<8