The image contains three inequality problems. Let's solve each of them step-by-step:

1. Problem 1

$\frac{3x - 15}{x} \leq 5 - x$

Solution:

1. Clear the fraction by multiplying both sides by $x$ (assuming $x \neq 0$): $3x - 15 \leq x(5 - x)$

2. Distribute and simplify: $3x - 15 \leq 5x - x^2$

3. Rearrange the inequality to form a quadratic equation: $x^2 - 2x - 15 \geq 0$

4. Factor the quadratic: $(x - 5)(x + 3) \geq 0$

5. Determine the critical points: $x = 5$ and $x = -3$. Analyze the intervals: $-\infty, -3$, $(-3, 5)$, and $(5, \infty)$.

6. Test values from each interval to find where the inequality holds: $\text{The solution is: } x \in (-\infty, -3] \cup [5, \infty).$

2. Problem 2

$\frac{4t + 5}{t} > 3$

Solution:

1. Clear the fraction by multiplying both sides by $t$ (assuming $t \neq 0$): $4t + 5 > 3t$

2. Rearrange the inequality: $t > -5$

3. Determine the valid range of $t$: $\text{The solution is: } t > 0 \text{ and } t > -5 \implies t > 0.$

3. Problem 3

$\frac{2x}{x+1} \geq 1$

Solution:

1. Move 1 to the left side: $\frac{2x}{x+1} - 1 \geq 0$

2. Combine fractions: $\frac{2x - (x + 1)}{x+1} \geq 0 \implies \frac{x - 1}{x+1} \geq 0$

3. Find critical points: $x = -1, 1$.

4. Determine where the inequality holds: [ \text{Solution: } (-\1]