What is ((x+3))/3*4/((x+2))> x/3 ?

The solution to ((x+3))/3*4/((x+2))> x/3 is x<1-sqrt(13)\lor-2<x<1+sqrt(13)

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((x+3))/3*4/((x+2))> x/3

\frac{( x + 3 )}{3} \cdot \frac{4}{( x + 2 )} > \frac{x}{3}

x < 1 - 13 or - 2 < x < 1 + 13

Interval Notation

(- \infty, 1 - 13) \cup (- 2, 1 + 13)

Decimal Notation

x < - 2.60555 \dots or - 2 < x < 4.60555 \dots

Solution steps

\frac{x + 3}{3} \cdot \frac{4}{x + 2} > \frac{x}{3}

Rewrite in standard form

\frac{- x ^{2} + 2 x + 12}{x + 2} > 0

Factor

- x^{2} + 2 x + 12 : - (x - 1 - 13) (x - 1 + 13)

\frac{- ( x - 1 - 13 ) ( x - 1 + 13 )}{x + 2} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - 1 - 13 ) ( x - 1 + 13 ) ) ( - 1 )}{x + 2} < 0 \cdot (- 1)

Simplify

\frac{( x - 1 - 13 ) ( x - 1 + 13 )}{x + 2} < 0

Identify the intervals

x < 1 - 13 or - 2 < x < 1 + 13

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

\frac{x}{( x + 1 )} > \frac{1}{x}

\frac{6 x ( 12 + 8 )}{5} > 2 x^{2}

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

\frac{( x ^{2} - 5 x + 6 )}{( x - 2 )} < 0

What is ((x+3))/3*4/((x+2))> x/3 ?
The solution to ((x+3))/3*4/((x+2))> x/3 is x<1-sqrt(13)\lor-2<x<1+sqrt(13)