What is (2x-8)/((x-3)(x-7))<0 ?

The solution to (2x-8)/((x-3)(x-7))<0 is x<3\lor 4<x<7

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(2x-8)/((x-3)(x-7))<0

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

x < 3 or 4 < x < 7

Interval Notation

(- \infty, 3) \cup (4, 7)

Solution steps

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

Factor

\frac{2 x - 8}{( x - 3 ) ( x - 7 )} : \frac{2 ( x - 4 )}{( x - 3 ) ( x - 7 )}

\frac{2 ( x - 4 )}{( x - 3 ) ( x - 7 )} < 0

Divide both sides by

2

\frac{\frac{2 ( x - 4 )}{( x - 3 ) ( x - 7 )}}{2} < \frac{0}{2}

Simplify

\frac{x - 4}{( x - 3 ) ( x - 7 )} < 0

Identify the intervals

x < 3 or 4 < x < 7

(5 \frac{1}{16})^{\frac{1}{4}}

9 x^{3} - 7 x - 2

- 2 x^{2} + 16 x - 14 = 0

15 = \frac{7}{5} (4 x + 15)

- 1 \leq \frac{3 x - 7}{2} \leq 14

What is (2x-8)/((x-3)(x-7))<0 ?
The solution to (2x-8)/((x-3)(x-7))<0 is x<3\lor 4<x<7