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solve the inequality
[tex] \frac{x+1}{x-8} \leq 0 [/tex]

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Konrad509idn
[tex]D:x\not=8\\ \frac{x+1}{x-8}\leq0\\ (x+1)(x-8)\leq0\\ x\in\langle-1,8\rangle\\\\ x\in\langle-1,8\rangle \wedge x\not=8\\ \boxed{x\in\langle-1,8)} [/tex]
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