We are given the equation:

\[
\frac{2x + 1}{x - 3} = - \frac{2x - 1}{x + 3}
\]

To solve for \(x\), we will follow these steps:

### Step 1: Eliminate the denominators
We can eliminate the denominators by cross-multiplying:

\[
(2x + 1)(x + 3) = - (2x - 1)(x - 3)
\]

### Step 2: Expand both sides
Expand both sides of the equation:

\[
(2x + 1)(x + 3) = 2x(x + 3) + 1(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3
\]

\[
-(2x - 1)(x - 3) = -(2x(x - 3) - 1(x - 3)) = -(2x^2 - 6x - x + 3) = -2x^2 + 7x - 3
\]

### Step 3: Set up the new equation
Now we have:

\[
2x^2 + 7x + 3 = -2x^2 + 7x - 3
\]

### Step 4: Move all terms to one side
Move all terms to the left side of the equation:

\[
2x^2 + 7x + 3 + 2x^2 - 7x + 3 = 0
\]

Simplify:

\[
4x^2 + 6 = 0
\]

### Step 5: Solve for \(x\)
Now, solve for \(x\):

\[
4x^2 = -6
\]

\[
x^2 = -\frac{6}{4} = -\frac{3}{2}
\]

However, this equation has no real solutions because the square of a real number cannot be negative. Therefore, the equation has **no real solutions**.

Would you like more details or have any questions?

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Here are 5 related questions to explore further:
1. How would the solution differ if the equation had real solutions?
2. What is the importance of checking for extraneous solutions in rational equations?
3. How do you solve quadratic equations when the discriminant is negative?
4. Can you describe how to solve a rational equation step by step?
5. How do imaginary solutions arise in quadratic equations?

**Tip:** Always check for restrictions in the domain of rational equations to avoid dividing by zero.

Learn how to solve the rational equation (2x + 1)/(x-3) = -(2x-1)/(x + 3) through cross-multiplication and analyzing quadratic equations. This problem demonstrates key algebraic techniques, including the identification of imaginary solutions. Suitable for students in Grades 10-12.

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