Solve for x: x-1x+3>0 inquality with rational expression and interval, test points graphics 

Let's solve the inequality \( \frac{x - 1}{x + 3} > 0 \).

### Step-by-Step Solution:

1. **Find Critical Points**:
   First, we need to find the points where the expression is either zero or undefined.

   - The numerator \( x - 1 = 0 \) when \( x = 1 \).
   - The denominator \( x + 3 = 0 \) when \( x = -3 \) (this makes the expression undefined).

   So, the critical points are \( x = 1 \) (where the expression is zero) and \( x = -3 \) (where the expression is undefined).

2. **Divide the Number Line into Intervals**:
   Based on the critical points \( x = 1 \) and \( x = -3 \), divide the number line into the following intervals:
   - \( (-\infty, -3) \)
   - \( (-3, 1) \)
   - \( (1, \infty) \)

3. **Test Each Interval**:
   Choose a test point in each interval and check the sign of the rational expression \( \frac{x - 1}{x + 3} \) in that interval.

   - For \( (-\infty, -3) \), choose \( x = -4 \):
     \[
     \frac{-4 - 1}{-4 + 3} = \frac{-5}{-1} = 5 \quad (\text{positive})
     \]

   - For \( (-3, 1) \), choose \( x = 0 \):
     \[
     \frac{0 - 1}{0 + 3} = \frac{-1}{3} \quad (\text{negative})
     \]

   - For \( (1, \infty) \), choose \( x = 2 \):
     \[
     \frac{2 - 1}{2 + 3} = \frac{1}{5} \quad (\text{positive})
     \]

4. **Determine the Solution**:
   The expression \( \frac{x - 1}{x + 3} \) is positive in the intervals \( (-\infty, -3) \) and \( (1, \infty) \).

   - At \( x = -3 \), the expression is undefined, so we exclude \( x = -3 \).
   - At \( x = 1 \), the expression equals zero, but we want \( \frac{x - 1}{x + 3} > 0 \), so exclude \( x = 1 \).

   Therefore, the solution to the inequality is:
   \[
   x \in (-\infty, -3) \cup (1, \infty)
   \]

### Graphing the Solution:
To graph the solution:
- Draw a number line with the critical points \( x = -3 \) and \( x = 1 \) marked.
- Use open circles at \( x = -3 \) and \( x = 1 \) to show these points are excluded from the solution.
- Shade the intervals \( (-\infty, -3) \) and \( (1, \infty) \).

### Conclusion:
The solution to \( \frac{x - 1}{x + 3} > 0 \) is \( x \in (-\infty, -3) \cup (1, \infty) \).

Would you like the graph, or do you have questions?

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### 5 Related Questions:
1. How do we solve inequalities involving polynomials?
2. What happens if the inequality were \( \frac{x - 1}{x + 3} \geq 0 \)?
3. Can you explain how to solve absolute value inequalities?
4. How can we solve inequalities with quadratic expressions?
5. What are the key differences between solving linear equations and inequalities?

### Tip:
When solving inequalities with rational expressions, always remember to check where the denominator is zero and test the intervals accordingly.

Solve for x: x-1/(x+3)>0 inequality with rational expression and interval, test points, graphics.

Learn how to solve the inequality x-1/(x+3)>0 with a detailed, step-by-step approach. This problem explains critical points, dividing the number line into intervals, and testing points in each interval. The solution covers essential algebraic concepts related to inequalities and rational expressions, suitable for students in Grades 9-11.

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