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The given equation is:

\[
\frac{3}{x+3} = \frac{15}{3-x}
\]

### To solve for \( x \):

1. **Cross-multiply** to eliminate the fractions:

\[
3(3 - x) = 15(x + 3)
\]

2. **Expand both sides**:

\[
9 - 3x = 15x + 45
\]

3. **Move all terms involving \( x \) to one side and constants to the other**:

\[
9 - 45 = 15x + 3x
\]

\[
-36 = 18x
\]

4. **Solve for \( x \)** by dividing both sides by 18:

\[
x = \frac{-36}{18} = -2
\]

### Therefore, the solution is:

\[
x = -2
\]

Let me know if you would like more details or have questions!

Here are 5 related questions to further explore:

1. What happens if you substitute \( x = -2 \) back into the original equation?
2. How would the equation change if the denominator was \( x - 3 \) instead of \( 3 - x \)?
3. Can you solve similar rational equations using a different method, such as factoring?
4. How does cross-multiplication work in general for rational equations?
5. What other types of equations can be solved by eliminating fractions?

**Tip**: Always check your solutions by substituting back into the original equation to avoid extraneous solutions.

Solve the following equation for the unknown x: 3 / (x + 3) = 15 / (3 - x)

Learn how to solve the rational equation 3/(x+3) = 15/(3-x) by cross-multiplying. This step-by-step solution explores key algebraic concepts in rational equations, suitable for students in Grades 8-10.

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