What is 1/((3x-8))<= 1/((8x-3)) ?

The solution to 1/((3x-8))<= 1/((8x-3)) is x<=-1\lor 3/8 <x< 8/3

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1/((3x-8))<= 1/((8x-3))

\frac{1}{( 3 x - 8 )} \leq \frac{1}{( 8 x - 3 )}

x \leq - 1 or \frac{3}{8} < x < \frac{8}{3}

Interval Notation

(- \infty, - 1] \cup (\frac{3}{8}, \frac{8}{3})

Decimal Notation

x \leq - 1 or 0.375 < x < 2.66666 \dots

Solution steps

\frac{1}{3 x - 8} \leq \frac{1}{8 x - 3}

Rewrite in standard form

\frac{5 x + 5}{( 3 x - 8 ) ( 8 x - 3 )} \leq 0

Factor

\frac{5 x + 5}{( 3 x - 8 ) ( 8 x - 3 )} : \frac{5 ( x + 1 )}{( 3 x - 8 ) ( 8 x - 3 )}

\frac{5 ( x + 1 )}{( 3 x - 8 ) ( 8 x - 3 )} \leq 0

Divide both sides by

5

\frac{\frac{5 ( x + 1 )}{( 3 x - 8 ) ( 8 x - 3 )}}{5} \leq \frac{0}{5}

Simplify

\frac{x + 1}{( 3 x - 8 ) ( 8 x - 3 )} \leq 0

Identify the intervals

x \leq - 1 or \frac{3}{8} < x < \frac{8}{3}

x (x + 3) (x - 1) < 0

\frac{( ( x - 4 ) ^{2} )}{x} > 0

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

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

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

What is 1/((3x-8))<= 1/((8x-3)) ?
The solution to 1/((3x-8))<= 1/((8x-3)) is x<=-1\lor 3/8 <x< 8/3