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

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

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

x < - 4 or - 2 \leq x < 1 or x \geq 6

Interval Notation

(- \infty, - 4) \cup [- 2, 1) \cup [6, \infty)

Solution steps

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

Rewrite in standard form

\frac{- x ^{2} + 4 x + 12}{( x - 1 ) ( x + 4 )} \leq 0

Factor

\frac{- x ^{2} + 4 x + 12}{( x - 1 ) ( x + 4 )} : \frac{- ( x + 2 ) ( x - 6 )}{( x - 1 ) ( x + 4 )}

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

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 2 ) ( x - 6 ) ) ( - 1 )}{( x - 1 ) ( x + 4 )} \geq 0 \cdot (- 1)

Simplify

\frac{( x + 2 ) ( x - 6 )}{( x - 1 ) ( x + 4 )} \geq 0

Identify the intervals

x < - 4 or - 2 \leq x < 1 or x \geq 6

x^{2} + 7 x y - 18 y^{2}

5^{6}

- 4 x > - 32

5^{l o g_{5} (17)}

12 x^{2} - 104 x + 17

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