How do you solve $(x+3)/(x-8)<0$ ?

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Answer 1 · 2017-07-22T11:22:48

Solution: $-3 < x <8$ , in interval notation: $(-3,8)$

$(x+3)/(x-8) < 0 ; x !=8$ critical points are $x=8 , x = -3$

Sign Change:

For $x < -3$ , sign of $(x+3)/(x-8)$ is $(-)/(-) = + i.e >0$

For $-3 < x <8$ , sign of $(x+3)/(x-8)$ is $(+)/(-) = - i.e <0$

For $x > 8$ , sign of $(x+3)/(x-8)$ is $(+)/(+) = + i.e >0$

So, Solution: $-3 < x <8$ , in interval notation: $(-3,8)$

The graph also confirms above findings.

graph{(x+3)/(x-8) [-11.25, 11.25, -5.625, 5.625]} [Ans]

How do you solve (x+3)/(x-8)<0(x+3)/(x-8)<0?