What is (6x+1)/(4x+1)+(4x+3)/(2x-3)<= 0 ?

The solution to (6x+1)/(4x+1)+(4x+3)/(2x-3)<= 0 is -1/4 <x< 3/2

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(6x+1)/(4x+1)+(4x+3)/(2x-3)<= 0

\frac{6 x + 1}{4 x + 1} + \frac{4 x + 3}{2 x - 3} \leq 0

- \frac{1}{4} < x < \frac{3}{2}

Interval Notation

(- \frac{1}{4}, \frac{3}{2})

Decimal Notation

- 0.25 < x < 1.5

Solution steps

\frac{6 x + 1}{4 x + 1} + \frac{4 x + 3}{2 x - 3} \leq 0

Rewrite in standard form

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

Identify the intervals

- \frac{1}{4} < x < \frac{3}{2}

\frac{125 y ^{2}}{10 x y}

x^{2} - 2 x - 1

x^{2} - 18 x + 14 < 0

3 r^{5} + 3 r^{4} - 90 r^{3}

\frac{( - \frac{42}{13} )}{( - \frac{7}{26} )}

What is (6x+1)/(4x+1)+(4x+3)/(2x-3)<= 0 ?
The solution to (6x+1)/(4x+1)+(4x+3)/(2x-3)<= 0 is -1/4 <x< 3/2