We are asked to solve the inequality:

$\frac{3x + 7}{x - 1} < 2$

Step 1: Rewrite the inequality

First, let's move everything to one side so we can analyze it properly:

$\frac{3x + 7}{x - 1} - 2 < 0$

We can rewrite the expression on the left side by putting everything over a common denominator:

$\frac{3x + 7 - 2(x - 1)}{x - 1} < 0$

Simplify the numerator:

$3x + 7 - 2x + 2 = x + 9$

So the inequality becomes:

$\frac{x + 9}{x - 1} < 0$

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Step 2: Find critical points

We need to find the points where the expression is zero or undefined.

1. Numerator = 0:
   $x + 9 = 0 \quad \Rightarrow \quad x = -9$

2. Denominator = 0:
   $x - 1 = 0 \quad \Rightarrow \quad x = 1$

So, the critical points are $x = -9$ and $x = 1$.

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Step 3: Analyze the sign on each interval

We now analyze the sign of the expression $\frac{x + 9}{x - 1}$ on the intervals determined by the critical points. These intervals are:

1. $(-\infty, -9)$
2. $(-9, 1)$
3. $(1, \infty)$

We will pick a test point from each interval and determine the sign of the expression $\frac{x + 9}{x - 1}$.

• For $x = -10$ (in $(-\infty, -9)$):
  $\frac{-10 + 9}{-10 - 1} = \frac{-1}{-11} = \frac{1}{11}$ (positive)

• For $x = 0$ (in $(-9, 1)$):
  $\frac{0 + 9}{0 - 1} = \frac{9}{-1} = -9$ (negative)

• For $x = 2$ (in $(1, \infty)$):
  $\frac{2 + 9}{2 - 1} = \frac{11}{1} = 11$ (positive)

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Step 4: Determine where the expression is negative

We are looking for where $\frac{x + 9}{x - 1} < 0$. From our sign analysis:

• The expression is negative on the interval $(-9, 1)$.

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Step 5: Handle critical points

• At $x = -9$, the expression equals 0 (so it is not included in the solution).
• At $x = 1$, the expression is undefined (so it is also not included in the solution).

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Step 6: Write the solution in interval notation

The solution to the inequality is:

$(-9, 1)$

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Would you like further explanation or have any questions?

Related Questions:

1. How do you solve inequalities with rational functions in general?
2. What is the process for graphing rational inequalities?
3. How do vertical asymptotes affect inequalities?
4. What happens when there are multiple undefined points?
5. Can inequalities like this appear in real-world applications?

Tip: Always pay attention to where the denominator becomes zero, as it divides the number line into key regions for testing signs.