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

The solution to (2x+2)/(x^2+4x+3)<= (3x+3)/(3x^2+2x-1) is x<-3\lor 1/3 <x<= 11/3

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

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

x < - 3 or \frac{1}{3} < x \leq \frac{11}{3}

Interval Notation

(- \infty, - 3) \cup (\frac{1}{3}, \frac{11}{3}]

Decimal Notation

x < - 3 or 0.33333 \dots < x \leq 3.66666 \dots

Solution steps

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

Rewrite in standard form

\frac{3 x ^{2} - 8 x - 11}{( x + 1 ) ( x + 3 ) ( 3 x - 1 )} \leq 0

Factor

\frac{3 x ^{2} - 8 x - 11}{( x + 1 ) ( x + 3 ) ( 3 x - 1 )} : \frac{( x + 1 ) ( 3 x - 11 )}{( x + 1 ) ( x + 3 ) ( 3 x - 1 )}

\frac{( x + 1 ) ( 3 x - 11 )}{( x + 1 ) ( x + 3 ) ( 3 x - 1 )} \leq 0

Identify the intervals

x < - 3 or \frac{1}{3} < x \leq \frac{11}{3}

(- 64)^{\frac{1}{3}}

\frac{( - 3 c ^{- 1} ) ^{4}}{- 12 c ^{7}}

3 x + 2 \geq 0

a^{2} + 4 - 4 a - 9 b^{2}

\frac{7 x + 3}{- 2} \leq 7 x + 5

What is (2x+2)/(x^2+4x+3)<= (3x+3)/(3x^2+2x-1) ?
The solution to (2x+2)/(x^2+4x+3)<= (3x+3)/(3x^2+2x-1) is x<-3\lor 1/3 <x<= 11/3