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

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

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

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

x < - 1 or 1 \leq x \leq 2

Interval Notation

(- \infty, - 1) \cup [1, 2]

Solution steps

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

Rewrite in standard form

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

Factor

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

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

Identify the intervals

x < - 1 or 1 \leq x \leq 2

x^{2} - 4 x - 7 < 0

(x^{2} - 4) (x - 1)^{6} < 0

\frac{( 6 + 3 j ) ^{2}}{( 2 - 4 j ) + 8} < 30

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

(x^{2} + 8) (x - 4) (x + 2) < 0

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