What is ((2x+3))/((x^2+x-12))<= 1/2 ?

The solution to ((2x+3))/((x^2+x-12)) = 6

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((2x+3))/((x^2+x-12))<= 1/2

\frac{( 2 x + 3 )}{( x ^{2} + x - 12 )} \leq \frac{1}{2}

x < - 4 or - 3 \leq x < 3 or x \geq 6

Interval Notation

(- \infty, - 4) \cup [- 3, 3) \cup [6, \infty)

Solution steps

\frac{2 x + 3}{x ^{2} + x - 12} \leq \frac{1}{2}

Rewrite in standard form

\frac{- x ^{2} + 3 x + 18}{( x - 3 ) ( x + 4 )} \leq 0

Factor

\frac{- x ^{2} + 3 x + 18}{( x - 3 ) ( x + 4 )} : \frac{- ( x + 3 ) ( x - 6 )}{( x - 3 ) ( x + 4 )}

\frac{- ( x + 3 ) ( x - 6 )}{( x - 3 ) ( x + 4 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 3 ) ( x - 6 ) ) ( - 1 )}{( x - 3 ) ( x + 4 )} \geq 0 \cdot (- 1)

Simplify

\frac{( x + 3 ) ( x - 6 )}{( x - 3 ) ( x + 4 )} \geq 0

Identify the intervals

x < - 4 or - 3 \leq x < 3 or x \geq 6

\frac{3}{2 x - 1} \geq \frac{5}{x - 1}

\frac{1}{x} - 2 - \frac{1}{x} \leq \frac{2}{x} + 2

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

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

x^{3} < 7 x + 6

What is ((2x+3))/((x^2+x-12))<= 1/2 ?
The solution to ((2x+3))/((x^2+x-12))<= 1/2 is x<-4\lor-3<= x<3\lor x>= 6