What is (2x^2)/(3x-2)<=-1 ?

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

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(2x^2)/(3x-2)<=-1

\frac{2 x ^{2}}{3 x - 2} \leq - 1

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

Interval Notation

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

Decimal Notation

x \leq - 2 or 0.5 \leq x < 0.66666 \dots

Solution steps

\frac{2 x ^{2}}{3 x - 2} \leq - 1

Rewrite in standard form

\frac{2 x ^{2} + 3 x - 2}{3 x - 2} \leq 0

Factor

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

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

Identify the intervals

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

\frac{3}{5} = \frac{3}{10} d - 12

\frac{x - 2}{4} \geq 5

(x - 2) (1 - 5 (x - \frac{1}{5})) < - 5 (x^{2} - 2)

(\frac{a}{b})^{- 1}

x^{2} = 225

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