Let's solve the inequalities provided in the image.

Problem c:

$\frac{2x - 1}{x - 3} \geq 1$

Step 1: Move everything to one side of the inequality:

$\frac{2x - 1}{x - 3} - 1 \geq 0$

Step 2: Combine the terms on the left-hand side:

$\frac{2x - 1 - (x - 3)}{x - 3} \geq 0$

Simplify the numerator: $\frac{2x - 1 - x + 3}{x - 3} = \frac{x + 2}{x - 3} \geq 0$

Step 3: Solve the inequality

Now we need to determine where the fraction $\frac{x + 2}{x - 3}$ is non-negative (i.e., $\geq 0$).

• Numerator: $x + 2 \geq 0$ gives $x \geq -2$.
• Denominator: $x - 3 > 0$ gives $x > 3$ (since division by zero is undefined).

Analyze the sign of the fraction in the intervals determined by $x = -2$ and $x = 3$:

• Interval $x < -2$: The fraction is negative.
• Interval $-2 \leq x < 3$: The fraction is non-negative (but zero at $x = -2$).
• Interval $x > 3$: The fraction is positive.

Solution:

The inequality $\frac{x + 2}{x - 3} \geq 0$ holds when: $x \in [-2, 3) \cup (3, \infty)$ (Note that $x = 3$ is excluded because it makes the denominator zero.)

Problem d:

$\sqrt{3 - x} \leq \sqrt{2x + 1}$

Step 1: Square both sides to eliminate the square roots:

$3 - x \leq 2x + 1$

Step 2: Solve the resulting linear inequality:

$3 - 1 \leq 2x + x$ $2 \leq 3x$ $\frac{2}{3} \leq x$

Step 3: Consider the domains of the original square root functions:

• $3 - x \geq 0$ implies $x \leq 3$.
• $2x + 1 \geq 0$ implies $x \geq -\frac{1}{2}$.

Solution:

Combine the constraints with the inequality result: $\frac{2}{3} \leq x \leq 3$

Would you like any more details or have any questions about these solutions? Here are some related questions:

1. What happens if the inequality sign in problem c were reversed?
2. How would you solve the inequality if both sides of problem d were cubic roots instead of square roots?
3. What is the impact of squaring both sides of an inequality in problem d?
4. How does one handle inequalities involving absolute values, compared to square roots?
5. Can you describe the solution process for a rational inequality like problem c?

Tip: Always be cautious when squaring both sides of an inequality. Ensure that the solutions fall within the domain of the original functions.