https://file.aiquickdraw.com/mathbot/file/172590661737456auzdu5.jpg

The equation in the image is:

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

To solve this equation, we will cross-multiply to eliminate the denominators. Cross-multiplying gives:

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

Now, expand both sides:

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

Next, bring all terms to one side of the equation:

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

Simplify the terms:

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

This is a quadratic equation. We can factor it:

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

So, the solutions are:

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

However, we need to check for any restrictions based on the original equation's denominators. The denominators are \( x - 1 \) and \( x + 1 \), so \( x \neq 1 \) and \( x \neq -1 \) to avoid division by zero. Neither 2 nor 3 violates these restrictions.

Thus, the solutions are:

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

Would you like more details on the solution, or do you have any other questions?

Here are 5 related questions to deepen your understanding:
1. How do you solve quadratic equations using the quadratic formula?
2. What happens when an equation has no real solutions?
3. How do you check for extraneous solutions in rational equations?
4. Can you solve equations with more complex denominators?
5. How would you solve this problem graphically?

**Tip:** Always check for restrictions in rational equations to ensure solutions don't make any denominator zero!

Solve the equation: \(\frac{x}{x - 1} = \frac{6}{x + 1}\)

Learn how to solve the rational equation \(\frac{x}{x - 1} = \frac{6}{x + 1}\) using cross-multiplication and factoring. This problem introduces quadratic equations and restrictions on denominators. Suitable for students in Grades 8-10.

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