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

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

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(4x)/(1-x^2)<= (3x)/(1-x)

\frac{4 x}{1 - x ^{2}} \leq \frac{3 x}{1 - x}

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

Interval Notation

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

Decimal Notation

- 1 < x \leq 0 or 0.33333 \dots \leq x < 1

Solution steps

\frac{4 x}{1 - x ^{2}} \leq \frac{3 x}{1 - x}

Rewrite in standard form

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

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

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

Identify the intervals

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

e^{- 0.03}

∣ 4 x ∣ > 5

(\frac{125}{216})^{\frac{2}{3}}

x^{3} - 9 x^{2} + 20 x - 12

\frac{4}{5} > \frac{2}{3} - \frac{2}{7} x

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