What is (3x+3)/(4x-1)-9/(2x-3)<= 0 ?

The solution to (3x+3)/(4x-1)-9/(2x-3)<= 0 is 0<= x< 1/4 \lor 3/2 <x<= 13/2

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(3x+3)/(4x-1)-9/(2x-3)<= 0

\frac{3 x + 3}{4 x - 1} - \frac{9}{2 x - 3} \leq 0

0 \leq x < \frac{1}{4} or \frac{3}{2} < x \leq \frac{13}{2}

Interval Notation

[0, \frac{1}{4}) \cup (\frac{3}{2}, \frac{13}{2}]

Decimal Notation

0 \leq x < 0.25 or 1.5 < x \leq 6.5

Solution steps

\frac{3 x + 3}{4 x - 1} - \frac{9}{2 x - 3} \leq 0

Rewrite in standard form

\frac{6 x ^{2} - 39 x}{( 4 x - 1 ) ( 2 x - 3 )} \leq 0

Factor

\frac{6 x ^{2} - 39 x}{( 4 x - 1 ) ( 2 x - 3 )} : \frac{3 x ( 2 x - 13 )}{( 4 x - 1 ) ( 2 x - 3 )}

\frac{3 x ( 2 x - 13 )}{( 4 x - 1 ) ( 2 x - 3 )} \leq 0

Divide both sides by

3

\frac{\frac{3 x ( 2 x - 13 )}{( 4 x - 1 ) ( 2 x - 3 )}}{3} \leq \frac{0}{3}

Simplify

\frac{x ( 2 x - 13 )}{( 4 x - 1 ) ( 2 x - 3 )} \leq 0

Identify the intervals

0 \leq x < \frac{1}{4} or \frac{3}{2} < x \leq \frac{13}{2}

15 n^{2} + n - 2

\frac{k ^{3}}{k + 5} + \frac{7 k ^{3} + 3}{k + 5}

5 + 9 x - 3 x = 2 + 6 x

0 = 3 x^{2} - 12 x + 9

625 - 16 t^{2}

What is (3x+3)/(4x-1)-9/(2x-3)<= 0 ?
The solution to (3x+3)/(4x-1)-9/(2x-3)<= 0 is 0<= x< 1/4 \lor 3/2 <x<= 13/2