16. \( \frac{2 t}{2 t^{2}-t-3}-\frac{3}{2 t^{2}-t-3} \)

To simplify the expression \( \frac{2 t}{2 t^{2}-t-3} - \frac{3}{2 t^{2}-t-3} \), we first note that the denominators are the same. This allows us to combine the numerators directly: \[ \frac{2 t - 3}{2 t^{2} - t - 3} \] Now, we can simplify the numerator \( 2t - 3 \) and see if the denominator can be factored. The quadratic \( 2t^{2} - t - 3 \) can be factored: 1. Find two numbers that multiply to \( 2 \times (-3) = -6 \) and add to \( -1 \). The numbers \( 2 \) and \( -3 \) work here. 2. Rewrite the middle term: \( 2t^{2} + 2t - 3t - 3 \). 3. Group and factor: \[ 2t(t + 1) - 3(t + 1) = (2t - 3)(t + 1). \] Thus, the denominator is factored as \( (2t - 3)(t + 1) \). Substituting back into our expression gives us: \[ \frac{2t - 3}{(2t - 3)(t + 1)}. \] We can cancel \( 2t - 3 \) from the numerator and the denominator (as long as \( t \neq \frac{3}{2} \)): \[ \frac{1}{t + 1} \quad \text{for } t \neq \frac{3}{2}. \] Thus, the simplified expression is: \[ \frac{1}{t + 1} \quad \text{for } t \neq \frac{3}{2}. \]