Let's solve the two problems from the image:

A. Solve the Rational Equation:

$\frac{x-3}{x^2 - 25} + \frac{1}{x+5} = \frac{1}{x-5}$

Step 1: Factor the denominator

The first term has a denominator $x^2 - 25$, which is a difference of squares: $x^2 - 25 = (x-5)(x+5)$

So the equation becomes: $\frac{x-3}{(x-5)(x+5)} + \frac{1}{x+5} = \frac{1}{x-5}$

Step 2: Combine the fractions

To combine the left-hand side, we'll first write each term with a common denominator. The second term, $\frac{1}{x+5}$, can be written as: $\frac{x-3}{(x-5)(x+5)} + \frac{1(x-5)}{(x-5)(x+5)} = \frac{1}{x-5}$

Now the left-hand side has a common denominator: $\frac{(x-3) + (x-5)}{(x-5)(x+5)} = \frac{1}{x-5}$

Simplifying the numerator: $(x-3) + (x-5) = 2x - 8$

So the equation becomes: $\frac{2x - 8}{(x-5)(x+5)} = \frac{1}{x-5}$

Step 3: Eliminate the denominators

Multiply both sides by $(x-5)$ to cancel the denominator: $\frac{2x - 8}{x+5} = 1$

Step 4: Solve for $x$

Multiply both sides by $x+5$ to eliminate the remaining fraction: $2x - 8 = x + 5$

Now, solve for $x$: $2x - x = 5 + 8$ $x = 13$

However, check the excluded values. Since the original equation has denominators $x-5$ and $x+5$, $x$ cannot be $5$ or $-5$. Since $x = 13$ does not violate this, the solution is valid.

Solution for A: $x = 13$

B. Solve the Rational Inequality:

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

Step 1: Solve the equation first

We will start by solving the equation: $\frac{3}{x-2} = -1$

Multiply both sides by $x-2$: $3 = -(x-2)$ $3 = -x + 2$ $x = -1$

Step 2: Analyze the inequality

Next, we analyze the inequality: $\frac{3}{x-2} \leq -1$

The critical points are where the denominator is zero and where the expression equals $-1$, which we already found to be $x = -1$.

• Excluded point: $x = 2$ (since division by zero occurs here).
• Test intervals: Use test points in the intervals $(-\infty, -1)$, $(-1, 2)$, and $(2, \infty)$.

1. For $x = -2$ (in $(-\infty, -1)$): $\frac{3}{-2-2} = \frac{3}{-4} = -\frac{3}{4} \quad \text{which is greater than} \ -1.$ This does not satisfy the inequality.

2. For $x = 0$ (in $(-1, 2)$): $\frac{3}{0-2} = \frac{3}{-2} = -1.5 \quad \text{which satisfies} \ \leq -1.$

3. For $x = 3$ (in $(2, \infty)$): $\frac{3}{3-2} = 3 \quad \text{which is greater than} \ -1.$ This does not satisfy the inequality.

Step 3: Conclusion

The solution to the inequality is $x \in (-1, 2)$, since it satisfies the inequality condition.

Solution for B: $x \in (-1, 2)$

Would you like further details or explanations?

Here are five related questions for further understanding:

1. How do you solve rational equations involving multiple fractions?
2. Why do we exclude certain values when solving rational equations?
3. How can you solve more complex rational inequalities?
4. What is the significance of factoring when solving rational equations?
5. How do you interpret the solution intervals in inequalities?

Tip: Always check for extraneous solutions when solving rational equations, especially when the original equation has variables in the denominator.