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

\frac{6 ^{3} \cdot 2 0 ^{2} \cdot 125}{2 5 ^{2} \cdot 2 ^{3} \cdot 1 2 ^{2}}

- 104 = 8 x

5 x = 25

(7 + 5 i) (- 6 + 9 i)

(\frac{6}{17})^{9 x}

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