Let's work through the math problems in the image you provided step-by-step, focusing on Activity 1 first.

Activity 1, Problem 1:

We are given the rational equation:

$\frac{x - 2}{x^2 - 4} + \frac{1}{x + 2} = \frac{1}{x - 2}$

Step 1: Find the Least Common Denominator (LCD)

The denominators are $x^2 - 4$, $x + 2$, and $x - 2$.

• Factor $x^2 - 4$ as $(x - 2)(x + 2)$.
• The LCD will be $(x - 2)(x + 2)$.

Step 2: Multiply both sides by the LCD

To eliminate the denominators, multiply each term by $(x - 2)(x + 2)$.

$(x - 2)(x + 2) \times \left(\frac{x - 2}{(x - 2)(x + 2)} + \frac{1}{x + 2}\right) = (x - 2)(x + 2) \times \frac{1}{x - 2}$

Simplifying each term:

• The first term becomes $x - 2$.
• The second term becomes $(x - 2)$.
• The right-hand side becomes $x + 2$.

Now, the equation becomes: $x - 2 + (x - 2) = x + 2$

Step 3: Apply the distributive property and simplify

Combine like terms: $2x - 4 = x + 2$

Step 4: Solve for $x$

Subtract $x$ from both sides: $x - 4 = 2$

Add 4 to both sides: $x = 6$

Step 5: Check for extraneous roots

Substitute $x = 6$ into the original equation. It does not cause any denominator to be zero, so $x = 6$ is a valid solution.

Activity 1, Problem 2:

We are given the rational equation:

$\frac{x - 6}{x^2 - 4x - 12} + \frac{2}{x + 2} = \frac{1}{x - 6}$

Step 1: Find the Least Common Denominator (LCD)

• Factor $x^2 - 4x - 12$ as $(x - 6)(x + 2)$.
• The LCD is $(x - 6)(x + 2)$.

Step 2: Multiply both sides by the LCD

$(x - 6)(x + 2) \times \left( \frac{x - 6}{(x - 6)(x + 2)} + \frac{2}{x + 2} \right) = (x - 6)(x + 2) \times \frac{1}{x - 6}$

Simplifying:

• The first term becomes $x - 6$.
• The second term becomes $2(x - 6)$.
• The right-hand side becomes $x + 2$.

Now, the equation is: $x - 6 + 2(x - 6) = x + 2$

Step 3: Apply the distributive property and simplify

Expand and combine like terms: $x - 6 + 2x - 12 = x + 2$ $3x - 18 = x + 2$

Step 4: Solve for $x$

Subtract $x$ from both sides: $2x - 18 = 2$ Add 18 to both sides: $2x = 20$ Divide by 2: $x = 10$

Step 5: Check for extraneous roots

Substitute $x = 10$ into the original equation. It does not cause any denominator to be zero, so $x = 10$ is a valid solution.

Activity 2:

Activity 2 involves solving several equations and inequalities and matching them to the correct answer from the list.

Would you like me to proceed with the detailed solution for Activity 2, or do you have any specific questions?

Here are some follow-up questions to extend the learning:

1. How can you identify extraneous roots in rational equations?
2. Why is factoring essential when solving rational equations?
3. What strategies can help simplify complex rational expressions?
4. How would you solve similar rational equations with different types of numerators?
5. What is the significance of checking the solution back in the original equation?

Tip:

Always check for restrictions on the variable by identifying values that make any denominator zero. These values will be excluded from the solution set.