https://file.aiquickdraw.com/mathbot/file/1725900764071drryv51f.jpg

The equation given in the image is a rational equation:

\[
\frac{x}{x-1} = \frac{6}{x+1}
\]

To solve this equation, we can cross multiply to eliminate the fractions:

\[
x(x + 1) = 6(x - 1)
\]

Expanding both sides:

\[
x^2 + x = 6x - 6
\]

Next, we'll move all terms to one side of the equation:

\[
x^2 + x - 6x + 6 = 0
\]

Simplify the equation:

\[
x^2 - 5x + 6 = 0
\]

Now, we factor the quadratic equation:

\[
(x - 3)(x - 2) = 0
\]

Setting each factor equal to zero gives us the solutions:

\[
x - 3 = 0 \quad \text{or} \quad x - 2 = 0
\]

So, \( x = 3 \) or \( x = 2 \).

### Checking the solutions:
We must check if these solutions are valid by substituting them back into the original equation to ensure they don't cause division by zero.

- For \( x = 3 \):
  \[
  \frac{3}{3-1} = \frac{6}{3+1} \implies \frac{3}{2} = \frac{6}{4} = \frac{3}{2} \quad \text{(Valid solution)}
  \]

- For \( x = 2 \):
  \[
  \frac{2}{2-1} = \frac{6}{2+1} \implies \frac{2}{1} = \frac{6}{3} = 2 \quad \text{(Valid solution)}
  \]

Both solutions are valid.

### Final Answer:
The solutions to the equation are \( x = 3 \) and \( x = 2 \).

Would you like more details or have any questions?

Here are five related questions you might find interesting:
1. How do you solve a rational equation with multiple fractions?
2. What are the conditions that cause extraneous solutions in rational equations?
3. How can we solve rational inequalities?
4. Can a rational equation have no solution? If so, under what conditions?
5. How would the equation change if the fractions were added instead of equated?

**Tip:** Always check your solutions in rational equations to avoid including values that make the denominator zero.

Solve the rational equation: x / (x-1) = 6 / (x+1).

Learn how to solve the rational equation x / (x-1) = 6 / (x+1) using cross multiplication and factoring techniques. This solution involves simplifying the equation, factoring, and checking for extraneous solutions. Suitable for students in Grades 8-10.

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