What is (x^2+3x-18)/(x-7)>= 0 ?

The solution to (x^2+3x-18)/(x-7)>= 0 is -6 7

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(x^2+3x-18)/(x-7)>= 0

\frac{x ^{2} + 3 x - 18}{x - 7} \geq 0

- 6 \leq x \leq 3 or x > 7

Interval Notation

[- 6, 3] \cup (7, \infty)

Solution steps

\frac{x ^{2} + 3 x - 18}{x - 7} \geq 0

Factor

\frac{x ^{2} + 3 x - 18}{x - 7} : \frac{( x - 3 ) ( x + 6 )}{x - 7}

\frac{( x - 3 ) ( x + 6 )}{x - 7} \geq 0

Identify the intervals

- 6 \leq x \leq 3 or x > 7

\frac{6}{75}

x^{2} + 15 x + 44

- 3.6 - (1.5 x + 1) = - 4 x - 0.8 - (0.4 x - 2)

\frac{1}{5} (x - 8) < x + 16

7 y = 3 y + 32

What is (x^2+3x-18)/(x-7)>= 0 ?
The solution to (x^2+3x-18)/(x-7)>= 0 is -6<= x<= 3\lor x>7