To solve the equation \( \frac{2x - 3x}{x + 3} = \frac{x}{x + 3} \), we first simplify the left-hand side.
The left side simplifies as follows:
\[ \frac{2x - 3x}{x + 3} = \frac{-x}{x + 3} \]
So the equation now looks like:
\[ \frac{-x}{x + 3} = \frac{x}{x + 3} \]
Next, since both sides have the same denominator, we can multiply both sides by \( x + 3 \) (assuming \( x + 3 \neq 0 \)):
\[ -x = x \]
To solve for \( x \), add \( x \) to both sides:
\[ -x + x = x + x \implies 0 = 2x \]
Thus,
\[ 2x = 0 \implies x = 0 \]
Now, we must also check if substituting \( x = 0 \) results in any restrictions on the equation regarding the denominator:
For \( x = 0 \):
\[ x + 3 = 0 + 3 = 3 \quad (\text{This does not cause division by zero.}) \]
Thus, \( x = 0 \) is a valid solution.
Now let's summarize the solutions found based on the response options given:
- \(-1\)
- \(\frac{3}{2}\)
- \(1\)
- \(-\frac{3}{2}\)
- \(0\)
The solution we derived is:
\[ \boxed{0} \]
If 0 was not listed in the options, it appears there may be a miscommunication or an oversight in the question's provided options. In terms of your selection, none of the given non-zero responses match the found solution.