Solve the inequality and graph the solution set on a real number line. Express the solution set in interval notation. \( \frac{x^{2}-x-12}{x^{2}-8 x+7}>0 \) The solution set is

To solve the inequality \( \frac{x^{2}-x-12}{x^{2}-8x+7} > 0 \), we start by factoring the numerator and the denominator: 1. **Numerator:** \( x^2 - x - 12 = (x - 4)(x + 3) \) 2. **Denominator:** \( x^2 - 8x + 7 = (x - 1)(x - 7) \) Now, the inequality can be rewritten as: \[ \frac{(x - 4)(x + 3)}{(x - 1)(x - 7)} > 0 \] Next, we identify the critical points where the expression is zero or undefined, which are \( x = -3, 1, 4, 7 \). These points will divide the number line into intervals that we need to test. The intervals are: 1. \( (-\infty, -3) \) 2. \( (-3, 1) \) 3. \( (1, 4) \) 4. \( (4, 7) \) 5. \( (7, \infty) \) Now we test each interval to see where the inequality holds true. - **Interval \( (-\infty, -3) \):** Choose \( x = -4 \): \(\frac{(-4 - 4)(-4 + 3)}{(-4 - 1)(-4 - 7)} = \frac{(-8)(-1)}{(-5)(-11)} > 0\) (True) - **Interval \( (-3, 1) \):** Choose \( x = 0 \): \(\frac{(0 - 4)(0 + 3)}{(0 - 1)(0 - 7)} = \frac{(-4)(3)}{(-1)(-7)} < 0\) (False) - **Interval \( (1, 4) \):** Choose \( x = 2 \): \(\frac{(2 - 4)(2 + 3)}{(2 - 1)(2 - 7)} = \frac{(-2)(5)}{(1)(-5)} > 0\) (True) - **Interval \( (4, 7) \):** Choose \( x = 5 \): \(\frac{(5 - 4)(5 + 3)}{(5 - 1)(5 - 7)} = \frac{(1)(8)}{(4)(-2)} < 0\) (False) - **Interval \( (7, \infty) \):** Choose \( x = 8 \): \(\frac{(8 - 4)(8 + 3)}{(8 - 1)(8 - 7)} = \frac{(4)(11)}{(7)(1)} > 0\) (True) From the tests, the intervals that satisfy the inequality are \( (-\infty, -3) \), \( (1, 4) \), and \( (7, \infty) \). Remember, at \( x = -3, 1, 4, \) and \( 7 \), we have points where the expression equals zero or is undefined, thus they are not included in the solution. In interval notation, the solution set is: \[ (-\infty, -3) \cup (1, 4) \cup (7, \infty) \] And if we want to graph this on a number line, we show open circles at -3, 1, 4, and 7 to indicate that these points are not included in the solution set, and shade the regions corresponding to the intervals.