### Instruction: Solve
\[\frac{(x + 7)(x - 2)}{x - 1} < 0.\]Enter your answer using interval notation.### Response: To solve this inequality, I need to find the values of x that make the numerator negative and the denominator positive, or vice versa.
I also need to avoid dividing by zero, so x cannot be 1.
To find the sign of the numerator and the denominator, I can use a sign chart.
I'll start by factoring the numerator and denominator completely.
The numerator is already factored, and the denominator is just x - 1.
Now I'll plot the zeros of the numerator and the denominator on a number line.
The zeros are -7, 1, and 2.
These divide the number line into four intervals: (-infinity, -7), (-7, 1), (1, 2), and (2, infinity).
I'll pick a test point in each interval and plug it into the numerator and the denominator to see their signs.
For the first interval, I'll pick -8.
Plugging -8 into the numerator gives (-8 + 7)(-8 - 2) = (-1)(-10) = 10, which is positive.
Plugging -8 into the denominator gives -8 - 1 = -9, which is negative.
So the fraction is positive over negative, which is negative.
I'll write a minus sign below the first interval.
For the second interval, I'll pick 0.
Plugging 0 into the numerator gives (0 + 7)(0 - 2) = (7)(-2) = -14, which is negative.
Plugging 0 into the denominator gives 0 - 1 = -1, which is also negative.
So the fraction is negative over negative, which is positive.
I'll write a plus sign below the second interval.
For the third interval, I'll pick 1.5.
Plugging 1.5 into the numerator gives (1.5 + 7)(1.5 - 2) = (8.5)(-0.5) = -4.25, which is negative.
Plugging 1.5 into the denominator gives 1.5 - 1 = 0.5, which is positive.
So the fraction is negative over positive, which is negative.
I'll write a minus sign below the third interval.
For the fourth interval, I'll pick 3.
Plugging 3 into the numerator gives (3 + 7)(3 - 2) = (10)(1) = 10, which is positive.
Plugging 3 into the denominator gives 3 - 1 = 2, which is also positive.
So the fraction is positive over positive, which is positive.
I'll write a plus sign below the fourth interval.
Now I have a sign chart that shows me the sign of the fraction in each interval.
I want the fraction to be less than zero, so I need to find the intervals where the sign is negative.
Those are the intervals (-infinity, -7) and (1, 2).
However, I also need to exclude the endpoints, because they make the fraction zero or undefined.
So the final answer is (-infinity, -7) union (1, 2).