What is 1/(x+3)+1/(x+4)<= 0 ?

The solution to 1/(x+3)+1/(x+4)<= 0 is x<-4\lor-7/2 <= x<-3

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1/(x+3)+1/(x+4)<= 0

\frac{1}{x + 3} + \frac{1}{x + 4} \leq 0

x < - 4 or - \frac{7}{2} \leq x < - 3

Interval Notation

(- \infty, - 4) \cup [- \frac{7}{2}, - 3)

Decimal Notation

x < - 4 or - 3.5 \leq x < - 3

Solution steps

\frac{1}{x + 3} + \frac{1}{x + 4} \leq 0

Rewrite in standard form

\frac{2 x + 7}{( x + 3 ) ( x + 4 )} \leq 0

Identify the intervals

x < - 4 or - \frac{7}{2} \leq x < - 3

x y - 5 y - 2 x + 10

1300

- 13 \geq \frac{x}{- 4} - 4

(x + 2) (x - 2) = 5

14 x + 14 = 4 (2 x + 2)

What is 1/(x+3)+1/(x+4)<= 0 ?
The solution to 1/(x+3)+1/(x+4)<= 0 is x<-4\lor-7/2 <= x<-3