The uploaded image contains two inequalities:

6) $\frac{1}{x+1} < \frac{2}{3x-1}$

7) $\frac{x+2}{x+4} < \frac{x-1}{x-2}$

Let's solve both inequalities one by one.

Problem 6:

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

To solve this, follow these steps:

1. Cross-multiply the terms, but be careful with signs of $x+1$ and $3x-1$. This is valid as long as both sides have the same sign, which we must account for later: $1(3x-1) < 2(x+1)$ Expanding both sides: $3x - 1 < 2x + 2$

2. Simplify the inequality: $3x - 2x < 2 + 1$ $x < 3$

3. Now consider the restrictions on the domain of the inequality. Both $x+1$ and $3x-1$ must be nonzero (otherwise, division by zero occurs):

   • $x + 1 \neq 0 \Rightarrow x \neq -1$
   • $3x - 1 \neq 0 \Rightarrow x \neq \frac{1}{3}$

Thus, the solution is: $x < 3 \quad \text{and} \quad x \neq -1, \frac{1}{3}$

Problem 7:

$\frac{x+2}{x+4} < \frac{x-1}{x-2}$

Again, let's solve this step by step:

1. Cross-multiply the terms (considering the signs of $x+4$ and $x-2$): $(x+2)(x-2) < (x-1)(x+4)$ Use the distributive property to expand both sides: $(x^2 - 4) < (x^2 + 3x - 4)$

2. Simplify the inequality: $x^2 - 4 < x^2 + 3x - 4$ Cancel out the $x^2$ terms: $-4 < 3x - 4$ Add 4 to both sides: $0 < 3x$ Divide by 3: $0 < x \quad \text{or} \quad x > 0$

3. Consider the restrictions on the domain:

   • $x+4 \neq 0 \Rightarrow x \neq -4$
   • $x-2 \neq 0 \Rightarrow x \neq 2$

Thus, the solution is: $x > 0 \quad \text{and} \quad x \neq -4, 2$

Do you want further details or have any questions?

Here are 5 related questions you might want to explore:

1. How do we deal with inequalities when cross-multiplying fractions?
2. What are the steps to solve rational inequalities involving more complex expressions?
3. How do you determine the domain restrictions when solving rational inequalities?
4. Can rational inequalities have multiple valid solution intervals?
5. How would you graph the solution sets for these inequalities?

Tip: Always check for restrictions (like division by zero) when solving rational inequalities to avoid invalid solutions.