What is -(9x)/(x^2-5x+6)>((x^2+8))/(x^2-5x+6) ?

The solution to -(9x)/(x^2-5x+6)>((x^2+8))/(x^2-5x+6) is -8<x<-1\lor 2<x<3

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-(9x)/(x^2-5x+6)>((x^2+8))/(x^2-5x+6)

- \frac{9 x}{x ^{2} - 5 x + 6} > \frac{( x ^{2} + 8 )}{x ^{2} - 5 x + 6}

- 8 < x < - 1 or 2 < x < 3

Interval Notation

(- 8, - 1) \cup (2, 3)

Solution steps

- \frac{9 x}{x ^{2} - 5 x + 6} > \frac{x ^{2} + 8}{x ^{2} - 5 x + 6}

Rewrite in standard form

\frac{- 9 x - x ^{2} - 8}{x ^{2} - 5 x + 6} > 0

Factor

\frac{- 9 x - x ^{2} - 8}{x ^{2} - 5 x + 6} : \frac{- ( x + 1 ) ( x + 8 )}{( x - 2 ) ( x - 3 )}

\frac{- ( x + 1 ) ( x + 8 )}{( x - 2 ) ( x - 3 )} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 1 ) ( x + 8 ) ) ( - 1 )}{( x - 2 ) ( x - 3 )} < 0 \cdot (- 1)

Simplify

\frac{( x + 1 ) ( x + 8 )}{( x - 2 ) ( x - 3 )} < 0

Identify the intervals

- 8 < x < - 1 or 2 < x < 3

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

2 x^{2} + 8 x - 10

x^{2} - 9 x + 10 = 0

6 x + 7 = \frac{4 x + 8}{5}

lo g_{3} (\frac{1}{9}) = y

What is -(9x)/(x^2-5x+6)>((x^2+8))/(x^2-5x+6) ?
The solution to -(9x)/(x^2-5x+6)>((x^2+8))/(x^2-5x+6) is -8<x<-1\lor 2<x<3