A rational equation is an equation that involves at least one rational expression, which is a fraction where both the numerator and the denominator are polynomials. The general form of a rational equation looks like:

$\frac{P(x)}{Q(x)} = \frac{R(x)}{S(x)}$ where $P(x), Q(x), R(x),$ and $S(x)$ are polynomials, and $Q(x)$ and $S(x)$ are not equal to zero.

Steps to solve a rational equation:

1. Find the Least Common Denominator (LCD): Determine the LCD of all the rational expressions involved. This helps eliminate the fractions.

2. Multiply through by the LCD: To get rid of the denominators, multiply both sides of the equation by the LCD.

3. Simplify the equation: After clearing the fractions, simplify the resulting equation.

4. Solve for the variable: Solve the simplified equation, which could now be a linear, quadratic, or other form of polynomial equation.

5. Check for extraneous solutions: Plug the solutions back into the original equation to ensure they do not make any denominators zero. If a solution causes a zero denominator, it is not valid and must be excluded.

Example:

Solve the equation: $\frac{2x}{x-3} = \frac{x+1}{x+2}$

Step 1: Find the LCD

The LCD is $(x-3)(x+2)$.

Step 2: Multiply both sides by the LCD

$(x-3)(x+2) \cdot \frac{2x}{x-3} = (x-3)(x+2) \cdot \frac{x+1}{x+2}$ This simplifies to: $2x(x+2) = (x+1)(x-3)$

Step 3: Simplify the equation

$2x^2 + 4x = x^2 - 2x - 3$ Subtract all terms from one side: $2x^2 + 4x - x^2 + 2x + 3 = 0$ $x^2 + 6x + 3 = 0$

Step 4: Solve the quadratic equation

Using the quadratic formula: $x = \frac{-6 \pm \sqrt{6^2 - 4(1)(3)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 12}}{2} = \frac{-6 \pm \sqrt{24}}{2} = \frac{-6 \pm 2\sqrt{6}}{2} = -3 \pm \sqrt{6}$ So the solutions are: $x = -3 + \sqrt{6}, \quad x = -3 - \sqrt{6}$

Step 5: Check for extraneous solutions

Neither solution causes the denominators $x-3$ or $x+2$ to become zero, so both solutions are valid.

---

Would you like further details or explanations on any step?

Follow-up questions:

1. Can you solve rational equations with more than two rational expressions?
2. How do you handle cases where rational equations involve complex fractions?
3. What happens if the equation involves higher-degree polynomials in the denominator?
4. How would you solve a rational equation where the variables are in both the numerator and denominator of multiple terms?
5. Could you provide an example of a rational equation that has extraneous solutions?

Tip:

Always check for extraneous solutions, as they frequently occur when solving rational equations due to restrictions on the denominators (division by zero is undefined).