What is ((x-3)(4-2x)(1-3x))/((1-4x)(5-3x))<= 0 ?

The solution to ((x-3)(4-2x)(1-3x))/((1-4x)(5-3x))<= 0 is x< 1/4 \lor 1/3 <= x< 5/3 \lor 2<= x<= 3

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((x-3)(4-2x)(1-3x))/((1-4x)(5-3x))<= 0

\frac{( x - 3 ) ( 4 - 2 x ) ( 1 - 3 x )}{( 1 - 4 x ) ( 5 - 3 x )} \leq 0

x < \frac{1}{4} or \frac{1}{3} \leq x < \frac{5}{3} or 2 \leq x \leq 3

Interval Notation

(- \infty, \frac{1}{4}) \cup [\frac{1}{3}, \frac{5}{3}) \cup [2, 3]

Decimal Notation

x < 0.25 or 0.33333 \dots \leq x < 1.66666 \dots or 2 \leq x \leq 3

Solution steps

\frac{( x - 3 ) ( 4 - 2 x ) ( 1 - 3 x )}{( 1 - 4 x ) ( 5 - 3 x )} \leq 0

Factor

\frac{( x - 3 ) ( 4 - 2 x ) ( 1 - 3 x )}{( 1 - 4 x ) ( 5 - 3 x )} : \frac{2 ( x - 3 ) ( x - 2 ) ( 3 x - 1 )}{( 1 - 4 x ) ( 5 - 3 x )}

\frac{2 ( x - 3 ) ( x - 2 ) ( 3 x - 1 )}{( 1 - 4 x ) ( 5 - 3 x )} \leq 0

Divide both sides by

2

\frac{\frac{2 ( x - 3 ) ( x - 2 ) ( 3 x - 1 )}{( 1 - 4 x ) ( 5 - 3 x )}}{2} \leq \frac{0}{2}

Simplify

\frac{( x - 3 ) ( x - 2 ) ( 3 x - 1 )}{( - 4 x + 1 ) ( - 3 x + 5 )} \leq 0

Identify the intervals

x < \frac{1}{4} or \frac{1}{3} \leq x < \frac{5}{3} or 2 \leq x \leq 3

x^{4} - 8 x^{2} - 9

14 x \geq 56

(\frac{12 x y ^{7}}{4 x ^{4} y ^{3}})^{2}

3 x^{2} - 22 x + 39 = 0

- 4 T + 8 - 6 T = - 4 - 10 T + 3

What is ((x-3)(4-2x)(1-3x))/((1-4x)(5-3x))<= 0 ?
The solution to ((x-3)(4-2x)(1-3x))/((1-4x)(5-3x))<= 0 is x< 1/4 \lor 1/3 <= x< 5/3 \lor 2<= x<= 3