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

The solution to 1/(x-4)>= 3/(x+1) is x<-1\lor 4<x<= 13/2

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1/(x-4)>= 3/(x+1)

\frac{1}{x - 4} \geq \frac{3}{x + 1}

x < - 1 or 4 < x \leq \frac{13}{2}

Interval Notation

(- \infty, - 1) \cup (4, \frac{13}{2}]

Decimal Notation

x < - 1 or 4 < x \leq 6.5

Solution steps

\frac{1}{x - 4} \geq \frac{3}{x + 1}

Rewrite in standard form

\frac{- 2 x + 13}{( x - 4 ) ( x + 1 )} \geq 0

Identify the intervals

x < - 1 or 4 < x \leq \frac{13}{2}

2 x^{2} + 8 x - 5 = 0

\frac{6 x + 1}{4 x + 1} + \frac{4 x + 3}{2 x - 3} \leq 0

\frac{125 y ^{2}}{10 x y}

x^{2} - 2 x - 1

What is 1/(x-4)>= 3/(x+1) ?
The solution to 1/(x-4)>= 3/(x+1) is x<-1\lor 4<x<= 13/2