A **rational equation** is an equation that contains one or more rational expressions, which are fractions where the numerator and denominator are both polynomials. Solving a rational equation often involves finding the values of the variable that make the equation true, while being cautious of any restrictions (such as values that make the denominator zero).

### Steps to Solve a Rational Equation:
1. **Find a common denominator** (if necessary).
2. **Multiply both sides of the equation** by the least common denominator (LCD) to eliminate the fractions.
3. **Simplify the resulting equation** (which is now polynomial).
4. **Solve the simplified equation**.
5. **Check for extraneous solutions**, which are solutions that may result in division by zero in the original equation.

### Example:
Solve the rational equation:

\[
\frac{2x}{x - 3} = 3 + \frac{5}{x - 3}
\]

**Step 1**: Multiply both sides by the common denominator \((x - 3)\):

\[
2x = (3(x - 3)) + 5
\]

**Step 2**: Distribute and simplify:

\[
2x = 3x - 9 + 5
\]

\[
2x = 3x - 4
\]

**Step 3**: Solve for \(x\):

\[
2x - 3x = -4
\]

\[
-x = -4 \quad \Rightarrow \quad x = 4
\]

**Step 4**: Check for any restrictions. Since \(x = 3\) would make the original denominator zero, we need to ensure \(x = 4\) is valid. The solution \(x = 4\) is acceptable.

Would you like more details on rational equations or have any specific examples you'd like to solve?

### 5 Related Questions:
1. What happens if a solution makes the denominator zero?
2. How do you solve a rational equation with two different denominators?
3. Can rational equations have more than one solution?
4. How are extraneous solutions identified in rational equations?
5. What role does the least common denominator (LCD) play in solving rational equations?

**Tip**: Always simplify the equation after multiplying by the LCD to avoid mistakes when solving.

This guide explains how to solve rational equations using a step-by-step approach. Learn about key concepts like finding the least common denominator (LCD), eliminating fractions, and identifying extraneous solutions with a detailed example: 2x/(x - 3) = 3 + 5/(x - 3). Suitable for students in Grades 8-10.

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