What is 1/(x+2)+3/(x-4)>0 ?

The solution to 1/(x+2)+3/(x-4)>0 is -2 4

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1/(x+2)+3/(x-4)>0

\frac{1}{x + 2} + \frac{3}{x - 4} > 0

- 2 < x < - \frac{1}{2} or x > 4

Interval Notation

(- 2, - \frac{1}{2}) \cup (4, \infty)

Decimal Notation

- 2 < x < - 0.5 or x > 4

Solution steps

\frac{1}{x + 2} + \frac{3}{x - 4} > 0

Rewrite in standard form

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

Factor

\frac{4 x + 2}{( x + 2 ) ( x - 4 )} : \frac{2 ( 2 x + 1 )}{( x + 2 ) ( x - 4 )}

\frac{2 ( 2 x + 1 )}{( x + 2 ) ( x - 4 )} > 0

Divide both sides by

2

\frac{\frac{2 ( 2 x + 1 )}{( x + 2 ) ( x - 4 )}}{2} > \frac{0}{2}

Simplify

\frac{2 x + 1}{( x + 2 ) ( x - 4 )} > 0

Identify the intervals

- 2 < x < - \frac{1}{2} or x > 4

2 p^{2} + 2 p - 4

3 x^{2} - 6 x + 9

\frac{x - 2}{3} = \frac{x + 5}{4}

125 - 64 c^{3}

(2 y)^{2}

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