2х+3/х+2=3х+2/х

To solve the equation \( \frac{2x + 3}{x + 2} = \frac{3x + 2}{x} \), we can start by cross-multiplying to eliminate the fractions. The equation becomes:

\[ (2x + 3) \cdot x = (3x + 2) \cdot (x + 2) \]

Expanding both sides:

Left side:

\[ 2x^2 + 3x \]

Right side:

\[ (3x + 2)(x + 2) = 3x^2 + 6x + 2x + 4 = 3x^2 + 8x + 4 \]

Now we have:

\[ 2x^2 + 3x = 3x^2 + 8x + 4 \]

Next, we will move all terms to one side of the equation:

\[ 2x^2 + 3x - 3x^2 - 8x - 4 = 0 \]

This simplifies to:

\[ -x^2 - 5x - 4 = 0 \]

To make the leading coefficient positive, multiply the entire equation by -1:

\[ x^2 + 5x + 4 = 0 \]

Now, we can factor this quadratic equation:

\[ (x + 1)(x + 4) = 0 \]

Setting each factor to zero gives the solutions:

\[ x + 1 = 0 \implies x = -1 \] \[ x + 4 = 0 \implies x = -4 \]

Thus, the solutions to the equation are:

\[ x = -1 \quad \text{and} \quad x = -4 \]

Finally, we need to check for any restrictions from the original equation. The denominators \(x + 2\) and \(x\) must not be zero. Therefore, \(x \neq -2\) and \(x \neq 0\).

Since both \(x = -1\) and \(x = -4\) do not violate these restrictions, they are valid solutions.

The final solutions are:

\[ \boxed{-1 \text{ and } -4} \]