The problem is to solve the following equation for $x$: $\frac{2x+3}{2} = \frac{2x-3}{2x} + x$

1. Problem Description

The problem is to solve the following equation for $x$:

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

2. Solution Steps

First, we multiply both sides of the equation by $2x$ to eliminate the fractions.

$2x * \frac{2x+3}{2} = 2x * (\frac{2x-3}{2x} + x)$

$x(2x+3) = (2x-3) + 2x^2$

$2x^2+3x = 2x-3+2x^2$

Subtract $2x^2$ from both sides:

$3x = 2x - 3$

Subtract $2x$ from both sides:

$3x - 2x = -3$

$x = -3$

Now, let's check the solution by substituting $x = -3$ back into the original equation:

$\frac{2(-3)+3}{2} = \frac{2(-3)-3}{2(-3)} + (-3)$

$\frac{-6+3}{2} = \frac{-6-3}{-6} - 3$

$\frac{-3}{2} = \frac{-9}{-6} - 3$

$-\frac{3}{2} = \frac{3}{2} - 3$

$-\frac{3}{2} = \frac{3}{2} - \frac{6}{2}$

$-\frac{3}{2} = -\frac{3}{2}$

The solution is valid.

3. Final Answer

x = -3