What is (6x)/(x-1)<= x/(x+3)+6 ?

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(6x)/(x-1)<= x/(x+3)+6

\frac{6 x}{x - 1} \leq \frac{x}{x + 3} + 6

x < - 3 or - 2 \leq x < 1 or x \geq 9

Interval Notation

(- \infty, - 3) \cup [- 2, 1) \cup [9, \infty)

Solution steps

\frac{6 x}{x - 1} \leq \frac{x}{x + 3} + 6

Rewrite in standard form

\frac{- x ^{2} + 7 x + 18}{( x - 1 ) ( x + 3 )} \leq 0

Factor

\frac{- x ^{2} + 7 x + 18}{( x - 1 ) ( x + 3 )} : \frac{- ( x + 2 ) ( x - 9 )}{( x - 1 ) ( x + 3 )}

\frac{- ( x + 2 ) ( x - 9 )}{( x - 1 ) ( x + 3 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 2 ) ( x - 9 ) ) ( - 1 )}{( x - 1 ) ( x + 3 )} \geq 0 \cdot (- 1)

Simplify

\frac{( x + 2 ) ( x - 9 )}{( x - 1 ) ( x + 3 )} \geq 0

Identify the intervals

x < - 3 or - 2 \leq x < 1 or x \geq 9

\frac{x}{8} = 8

x^{3} - x^{2} - 21 x + 45

- 0.04 x^{2} + 2000 x

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

∣ 4 x - 1 ∣ + 4 x < ∣ x + 2 ∣

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