What is (x-5)/(4x-4)+2/(x^2-1)>= 0 ?

The solution to (x-5)/(4x-4)+2/(x^2-1)>= 0 is x = 3

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(x-5)/(4x-4)+2/(x^2-1)>= 0

\frac{x - 5}{4 x - 4} + \frac{2}{x ^{2} - 1} \geq 0

x < - 1 or x \geq 3

Interval Notation

(- \infty, - 1) \cup [3, \infty)

Solution steps

\frac{x - 5}{4 x - 4} + \frac{2}{x ^{2} - 1} \geq 0

Rewrite in standard form

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

Factor

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

\frac{( x - 1 ) ( x - 3 )}{( x - 1 ) ( x + 1 )} \geq 0

Identify the intervals

x < - 1 or x \geq 3

12 n^{2} - 7 n + 1

69 = 1 \cdot 0.037 \cdot (x - 23.45)

3 x^{2} + 17 x + 4 = 9 x

∣ 2 x + 3 ∣ - 5 = 9

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