What is ((5-2x))/((3x^2-2x-16))<1 ?

The solution to ((5-2x))/((3x^2-2x-16)) 8/3

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((5-2x))/((3x^2-2x-16))<1

\frac{( 5 - 2 x )}{( 3 x ^{2} - 2 x - 16 )} < 1

x < - 7 or - 2 < x < 7 or x > \frac{8}{3}

Interval Notation

(- \infty, - 7) \cup (- 2, 7) \cup (\frac{8}{3}, \infty)

Decimal Notation

x < - 2.64575 \dots or - 2 < x < 2.64575 \dots or x > 2.66666 \dots

Solution steps

\frac{5 - 2 x}{3 x ^{2} - 2 x - 16} < 1

Rewrite in standard form

\frac{- 3 x ^{2} + 21}{3 x ^{2} - 2 x - 16} < 0

Factor

\frac{- 3 x ^{2} + 21}{3 x ^{2} - 2 x - 16} : \frac{- 3 ( x + 7 ) ( x - 7 )}{( x + 2 ) ( 3 x - 8 )}

\frac{- 3 ( x + 7 ) ( x - 7 )}{( x + 2 ) ( 3 x - 8 )} < 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 3 ( x + 7 ) ( x - 7 ) ) ( - 1 )}{( x + 2 ) ( 3 x - 8 )} > 0 \cdot (- 1)

Simplify

\frac{3 ( x + 7 ) ( x - 7 )}{( x + 2 ) ( 3 x - 8 )} > 0

Divide both sides by

3

\frac{\frac{3 ( x + 7 ) ( x - 7 )}{( x + 2 ) ( 3 x - 8 )}}{3} > \frac{0}{3}

Simplify

\frac{( x + 7 ) ( x - 7 )}{( x + 2 ) ( 3 x - 8 )} > 0

Identify the intervals

x < - 7 or - 2 < x < 7 or x > \frac{8}{3}

(x^{2} - 4 x + 4) (1 - 2 x) (x^{2} + 3 x - 1) x > 0

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

\frac{( 3 x - 2 )}{( 2 x - 3 )} < 3

\frac{( x ^{2} - 5 ) ( x ^{2} - 4 )}{( x - 1 )} < 0

3 x^{2} - 8 x + 2 < 0

What is ((5-2x))/((3x^2-2x-16))<1 ?
The solution to ((5-2x))/((3x^2-2x-16))<1 is x<-sqrt(7)\lor-2<x<sqrt(7)\lor x> 8/3