To solve the equation
[
\frac{x - 1}{2x + 3} - \frac{2x - 1}{3 - 2x} = 0,
]
we first need to rearrange the equation:
[
\frac{x - 1}{2x + 3} = \frac{2x - 1}{3 - 2x}.
]
Next, we can cross-multiply to eliminate the fractions:
[
(x - 1)(3 - 2x) = (2x - 1)(2x + 3).
]
Expanding both sides:
- Left side:
[
(x - 1)(3 - 2x) = 3x - 2x^2 - 3 + 2x = -2x^2 + 5x - 3.
]
- Right side:
[
(2x - 1)(2x + 3) = 4x^2 + 6x - 2x - 3 = 4x^2 + 4x - 3.
]
Now we set both expanded sides equal to each other:
[
-2x^2 + 5x - 3 = 4x^2 + 4x - 3.
]
Next, we move all terms to one side:
[
-2x^2 + 5x - 3 - 4x^2 - 4x + 3 = 0,
]
which simplifies to:
[
-6x^2 + x = 0.
]
Factoring out (x):
[
x(-6x + 1) = 0.
]
Setting each factor to zero gives us:
- (x = 0)
- (-6x + 1 = 0) or (x = \frac{1}{6}).
So the solutions are:
[
x = 0 \quad \text{and} \quad x = \frac{1}{6}.
]
To ensure these solutions do not make the original denominators equal to zero, we check:
For (x = 0):
- (2(0) + 3 = 3) (not zero)
- (3 - 2(0) = 3) (not zero)
For (x = \frac{1}{6}):
- (2\left(\frac{1}{6}\right) + 3 = \frac{1}{3} + 3 = \frac{10}{3}) (not zero)
- (3 - 2\left(\frac{1}{6}\right) = 3 - \frac{1}{3} = \frac{8}{3}) (not zero)
Thus, both solutions are valid. Therefore, the final solutions are:
[
\boxed{0 \quad \text{and} \quad \frac{1}{6}}.
]
