What is (x^2-3x-7)/(x-1)<=-1 ?

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

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(x^2-3x-7)/(x-1)<=-1

\frac{x ^{2} - 3 x - 7}{x - 1} \leq - 1

x \leq - 2 or 1 < x \leq 4

Interval Notation

(- \infty, - 2] \cup (1, 4]

Solution steps

\frac{x ^{2} - 3 x - 7}{x - 1} \leq - 1

Rewrite in standard form

\frac{x ^{2} - 2 x - 8}{x - 1} \leq 0

Factor

\frac{x ^{2} - 2 x - 8}{x - 1} : \frac{( x + 2 ) ( x - 4 )}{x - 1}

\frac{( x + 2 ) ( x - 4 )}{x - 1} \leq 0

Identify the intervals

x \leq - 2 or 1 < x \leq 4

\frac{15}{5 20}

\frac{3 a - 12}{3}

3 - 3.7 b = 10.3 b + 10

2 - x \leq \frac{5 x + 6}{x + 3}

- 7 x + 7 (- 2 x + 25) = - 56

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