What is (-2(x-6)+1)/(2*(2-3x))>= (-3+1)/(2*x) ?

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

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(-2(x-6)+1)/(2(2-3x))>= (-3+1)/(2x)

\frac{- 2 ( x - 6 ) + 1}{2 \cdot ( 2 - 3 x )} \geq \frac{- 3 + 1}{2 \cdot x}

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

Interval Notation

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

Decimal Notation

x \leq - 0.5 or 0 < x < 0.66666 \dots or x \geq 4

Solution steps

\frac{- 2 ( x - 6 ) + 1}{2 ( 2 - 3 x )} \geq \frac{- 3 + 1}{2 x}

Rewrite in standard form

\frac{- 2 x ^{2} + 7 x + 4}{x ( - 3 x + 2 )} \geq 0

Factor

\frac{- 2 x ^{2} + 7 x + 4}{x ( - 3 x + 2 )} : \frac{- ( 2 x + 1 ) ( x - 4 )}{x ( - 3 x + 2 )}

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

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

- 8 x^{4} + 6 x^{2} + 5 \leq 0

\frac{( ( x + 7 ) ^{2} )}{( ( 3 + x ) ( 2 - x ) )} \geq 0

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

(x - 5)^{3} + 9 \geq 10

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

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