What is (x^2-x-6)/(x^2-5x+4)>0 ?

The solution to (x^2-x-6)/(x^2-5x+4)>0 is x 4

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(x^2-x-6)/(x^2-5x+4)>0

\frac{x ^{2} - x - 6}{x ^{2} - 5 x + 4} > 0

x < - 2 or 1 < x < 3 or x > 4

Interval Notation

(- \infty, - 2) \cup (1, 3) \cup (4, \infty)

Solution steps

\frac{x ^{2} - x - 6}{x ^{2} - 5 x + 4} > 0

Factor

\frac{x ^{2} - x - 6}{x ^{2} - 5 x + 4} : \frac{( x + 2 ) ( x - 3 )}{( x - 1 ) ( x - 4 )}

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

Identify the intervals

x < - 2 or 1 < x < 3 or x > 4

- 5 (- x + 1) - 3 x - 1 = - 12

\frac{14 x ^{6} y ^{3} - 49 x ^{5} y ^{9}}{- 7 x ^{4} y}

2 x^{3} - 6 x^{2} - 56 x

∣ w ∣ > 2

154000 (1 + \frac{0.2}{1})^{5}

What is (x^2-x-6)/(x^2-5x+4)>0 ?
The solution to (x^2-x-6)/(x^2-5x+4)>0 is x<-2\lor 1<x<3\lor x>4