To solve the inequality $\frac{2x - 7}{x - 5} \geq 3$, we can proceed as follows:

Step 1: Rewrite the inequality

Rewrite the inequality so that it has a single fraction on one side:

$\frac{2x - 7}{x - 5} - 3 \geq 0$

Step 2: Combine into a single fraction

To subtract 3, express it with the same denominator:

$\frac{2x - 7 - 3(x - 5)}{x - 5} \geq 0$

Simplify the numerator:

$2x - 7 - 3x + 15 = -x + 8$

So, the inequality becomes:

$\frac{-x + 8}{x - 5} \geq 0$

Step 3: Find the critical points

The critical points are where the numerator and denominator are zero:

1. Numerator: $-x + 8 = 0 \Rightarrow x = 8$
2. Denominator: $x - 5 = 0 \Rightarrow x = 5$

Step 4: Test the intervals

These critical points divide the number line into three intervals:

1. $(-\infty, 5)$
2. $(5, 8)$
3. $(8, \infty)$

We will test the sign of the expression $\frac{-x + 8}{x - 5}$ in each interval.

• For $x < 5$ (e.g., $x = 0$): $\frac{-0 + 8}{0 - 5} = \frac{8}{-5} = -\text{negative}$

• For $5 < x < 8$ (e.g., $x = 6$): $\frac{-6 + 8}{6 - 5} = \frac{2}{1} = \text{positive}$

• For $x > 8$ (e.g., $x = 9$): $\frac{-9 + 8}{9 - 5} = \frac{-1}{4} = -\text{negative}$

Step 5: Determine the solution

We need the fraction to be non-negative ($\geq 0$), so the solution occurs in the interval where the fraction is positive or zero.

• The interval $(5, 8]$ satisfies the inequality. Note that $x = 8$ makes the numerator zero, so it is included.
• $x = 5$ is excluded because it makes the denominator zero.

Final Answer:

The interval notation for the solution is:

$(5, 8]$

Would you like further details or explanations? Here are some related questions:

1. How do we handle inequalities involving quadratic expressions?
2. What happens when the inequality is strict (i.e., $>$ instead of $\geq$)?
3. Can you solve similar inequalities with different numbers?
4. What changes if we move the term to the other side before simplifying?
5. How do these intervals relate to the graph of the function?

Tip: Always check the sign of both the numerator and denominator separately when dealing with inequalities involving fractions.