The image you provided contains a set of rational equations for solving. Let's work through the first few problems to illustrate the steps required.

Problem 1:

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

Solution:

1. Cross multiply to eliminate the fractions: $1 \cdot x = 2 \cdot (3x - 1)$
2. Expand and simplify: $x = 6x - 2$
3. Move all terms involving $x$ to one side: $x - 6x = -2$ $-5x = -2$
4. Solve for $x$: $x = \frac{2}{5}$

Problem 2:

$\frac{x}{2x+4} = \frac{3}{5}$

Solution:

1. Cross multiply: $x \cdot 5 = 3 \cdot (2x + 4)$
2. Expand and simplify: $5x = 6x + 12$
3. Move all terms involving $x$ to one side: $5x - 6x = 12$ $-x = 12$
4. Solve for $x$: $x = -12$

Problem 3:

$\frac{2x + 4}{3} = 5$

Solution:

1. Multiply both sides by 3 to clear the fraction: $2x + 4 = 15$
2. Subtract 4 from both sides: $2x = 11$
3. Divide by 2: $x = \frac{11}{2}$

These steps can be applied similarly to solve the remaining equations on the page. Would you like detailed solutions for other specific problems or further explanations on any of these steps?

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Follow-up Questions:

1. What happens if the solution to a rational equation makes a denominator zero?
2. How do you solve a rational equation with more than two fractions?
3. What methods can you use to check your solutions in rational equations?
4. How does cross-multiplication work when dealing with rational equations?
5. Can a rational equation have no solution? Why or why not?

Tip:

Always check your solutions by substituting them back into the original equation to ensure they do not make any denominator zero or cause any inconsistencies.