We need to solve the inequality $\frac{t-1}{4t+5} < \frac{t-3}{4t-3}$.

1. Problem Description

We need to solve the inequality $\frac{t-1}{4t+5} < \frac{t-3}{4t-3}$.

2. Solution Steps

First, move all terms to one side of the inequality:

$\frac{t-1}{4t+5} - \frac{t-3}{4t-3} < 0$.

Next, find a common denominator:

$\frac{(t-1)(4t-3) - (t-3)(4t+5)}{(4t+5)(4t-3)} < 0$.

Expand the numerator:

$\frac{4t^2 - 3t - 4t + 3 - (4t^2 + 5t - 12t - 15)}{(4t+5)(4t-3)} < 0$.

Simplify the numerator:

$\frac{4t^2 - 7t + 3 - 4t^2 + 7t + 15}{(4t+5)(4t-3)} < 0$.

$\frac{18}{(4t+5)(4t-3)} < 0$.

Since the numerator is a positive constant, the fraction will be negative when the denominator is negative.

$(4t+5)(4t-3) < 0$.

To solve this inequality, we need to find the roots of the denominator:

$4t+5 = 0 \Rightarrow t = -\frac{5}{4}$

$4t-3 = 0 \Rightarrow t = \frac{3}{4}$

The expression $(4t+5)(4t-3)$ is a parabola opening upwards. It is negative between the roots. Therefore:

$-\frac{5}{4} < t < \frac{3}{4}$.

3. Final Answer

$-\frac{5}{4} < t < \frac{3}{4}$