The image shows the following equation to solve:

$\frac{2x - 3}{x + 4} = 3 - \frac{4x}{x - 2}$

Steps to Solve:

1. Clear the fractions by multiplying both sides of the equation by $(x + 4)(x - 2)$, the least common denominator (LCD) of all fractions involved.

   $(x + 4)(x - 2) \cdot \frac{2x - 3}{x + 4} = (x + 4)(x - 2) \cdot \left(3 - \frac{4x}{x - 2}\right)$

2. Simplify the equation:

   • The left-hand side simplifies to:

     $(x - 2)(2x - 3)$

   • The right-hand side expands to:

     $3(x + 4)(x - 2) - 4x(x + 4)$

3. Expand and simplify both sides:

   • Expanding the left-hand side:

     $2x^2 - 4x - 3x + 6 = 2x^2 - 7x + 6$

   • Expanding the right-hand side:

     $3(x^2 - 2x + 4x - 8) - 4x^2 - 16x$ Simplifies to:

     $3x^2 + 6x - 24 - 4x^2 - 16x = -x^2 - 10x - 24$

4. Combine like terms and set the equation to zero:

   $2x^2 - 7x + 6 = -x^2 - 10x - 24$

   Add $x^2 + 10x + 24$ to both sides:

   $3x^2 + 3x + 30 = 0$

   Simplify by dividing the equation by 3:

   $x^2 + x + 10 = 0$

5. Solve the quadratic equation $x^2 + x + 10 = 0$:

   The discriminant ($b^2 - 4ac$) is:

   $1^2 - 4(1)(10) = 1 - 40 = -39$

   Since the discriminant is negative, there are no real solutions. The solutions are complex:

   $x = \frac{-1 \pm \sqrt{-39}}{2} = \frac{-1 \pm i\sqrt{39}}{2}$

Final Answer:

The solutions are complex numbers: $x = \frac{-1 + i\sqrt{39}}{2}, \quad x = \frac{-1 - i\sqrt{39}}{2}$

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