Solve the following inequality. \[ \frac{9}{x+1}<3 \] Write your answer using interval notation

To solve the inequality \(\frac{9}{x+1} < 3\), we start by isolating the fraction: 1. Subtract 3 from both sides to get: \[\frac{9}{x+1} - 3 < 0\] 2. To combine the terms, we need a common denominator: \[\frac{9 - 3(x + 1)}{x + 1} < 0\] \[\frac{9 - 3x - 3}{x + 1} < 0\] \[\frac{6 - 3x}{x + 1} < 0\] 3. Factor out a 3 from the numerator: \[\frac{3(2 - x)}{x + 1} < 0\] 4. The inequality \(\frac{3(2 - x)}{x + 1} < 0\) will change signs at the critical points where the numerator or denominator equals zero: - Numerator \(3(2 - x) = 0 \Rightarrow x = 2\) - Denominator \(x + 1 = 0 \Rightarrow x = -1\) 5. Now, we will use these points to test intervals around them: - Interval 1: \((- \infty, -1)\) - Interval 2: \((-1, 2)\) - Interval 3: \((2, \infty)\) 6. Choose test points from each interval to determine the sign of the expression: - For \(x = -2\): \[\frac{3(2 - (-2))}{-2 + 1} = \frac{3(4)}{-1} = -12\] (negative) - For \(x = 0\): \[\frac{3(2 - 0)}{0 + 1} = \frac{3(2)}{1} = 6\] (positive) - For \(x = 3\): \[\frac{3(2 - 3)}{3 + 1} = \frac{3(-1)}{4} = -\frac{3}{4}\] (negative) 7. Thus, the expression is negative in the intervals \((- \infty, -1)\) and \((2, \infty)\). 8. We also note that at \(x = -1\), the expression is undefined, so we do not include that point. 9. Finally, we have our solution in interval notation: \[ (-\infty, -1) \cup (2, \infty) \]