What is ((x^2-3x+4))/((x+5))< 1/3 ?

The solution to ((x^2-3x+4))/((x+5))< 1/3 is x<-5\lor 1<x< 7/3

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((x^2-3x+4))/((x+5))< 1/3

\frac{( x ^{2} - 3 x + 4 )}{( x + 5 )} < \frac{1}{3}

x < - 5 or 1 < x < \frac{7}{3}

Interval Notation

(- \infty, - 5) \cup (1, \frac{7}{3})

Decimal Notation

x < - 5 or 1 < x < 2.33333 \dots

Solution steps

\frac{x ^{2} - 3 x + 4}{x + 5} < \frac{1}{3}

Rewrite in standard form

\frac{3 x ^{2} - 10 x + 7}{x + 5} < 0

Factor

\frac{3 x ^{2} - 10 x + 7}{x + 5} : \frac{( x - 1 ) ( 3 x - 7 )}{x + 5}

\frac{( x - 1 ) ( 3 x - 7 )}{x + 5} < 0

Identify the intervals

x < - 5 or 1 < x < \frac{7}{3}

4 m^{2} + 4 m - 8 > 0

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

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

\frac{( x ^{2} + 5 x + 6 )}{( x ^{2} - 1 )} \geq 0

\frac{( x + 5 )}{( 20 - 5 x )} \geq 0

What is ((x^2-3x+4))/((x+5))< 1/3 ?
The solution to ((x^2-3x+4))/((x+5))< 1/3 is x<-5\lor 1<x< 7/3