Solve the inequality $\frac{t-1}{4t+5} < \frac{t-3}{4t-3}$.

1. Problem Description

Solve the inequality $\frac{t-1}{4t+5} < \frac{t-3}{4t-3}$.

2. Solution Steps

First, subtract $\frac{t-3}{4t-3}$ from both sides to get:

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

Next, find a common denominator and combine the fractions:

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

Expand the numerator:

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

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

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 positive (18 > 0), the inequality is satisfied when the denominator is negative:

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

Find the critical points by setting each factor to zero:

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

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

Now, test the intervals determined by these critical points:

Interval 1: $t < -\frac{5}{4}$

Let $t = -2$. Then $(4(-2)+5)(4(-2)-3) = (-8+5)(-8-3) = (-3)(-11) = 33 > 0$.

Interval 2: $-\frac{5}{4} < t < \frac{3}{4}$

Let $t = 0$. Then $(4(0)+5)(4(0)-3) = (5)(-3) = -15 < 0$.

Interval 3: $t > \frac{3}{4}$

Let $t = 1$. Then $(4(1)+5)(4(1)-3) = (4+5)(4-3) = (9)(1) = 9 > 0$.

The inequality is satisfied when $-\frac{5}{4} < t < \frac{3}{4}$.

3. Final Answer

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