Solving rational equations | PPT - Free Printable

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Definition of Rational Equations

A rational equation is an equation that contains one or more rational expressions. A rational expression is a fraction in which the numerator and/or the denominator are polynomials.

Examples Provided

1. $$ \frac{1}{x} = \frac{2}{x+3} $$
2. $$ \frac{x^2 - 3}{x + 1} = \frac{x - 1}{x^3 + 2x} $$

General Steps to Solve Rational Equations

To solve rational equations, follow these steps:

1. Identify the Least Common Denominator (LCD): Find the LCD of all the denominators in the equation.
2. Eliminate the Denominators: Multiply both sides of the equation by the LCD to clear the fractions.
3. Solve the Resulting Equation: Simplify and solve the resulting equation, which should no longer contain fractions.
4. Check for Extraneous Solutions: Verify that the solutions do not make any denominator in the original equation equal to zero. If they do, discard those solutions as they are extraneous.

Example 1: Solve \( \frac{1}{x} = \frac{2}{x+3} \)

#### Step 1: Identify the LCD
The denominators are \( x \) and \( x + 3 \). The LCD is \( x(x + 3) \).

#### Step 2: Eliminate the Denominators
Multiply both sides of the equation by \( x(x + 3) \):
\[
x(x + 3) \cdot \frac{1}{x} = x(x + 3) \cdot \frac{2}{x + 3}
\]

This simplifies to:
\[
(x + 3) \cdot 1 = x \cdot 2
\]
\[
x + 3 = 2x
\]

#### Step 3: Solve the Resulting Equation
Rearrange the equation to isolate \( x \):
\[
x + 3 = 2x
\]
\[
3 = 2x - x
\]
\[
3 = x
\]

#### Step 4: Check for Extraneous Solutions
Substitute \( x = 3 \) back into the original denominators \( x \) and \( x + 3 \):
- For \( x \): \( x = 3 \neq 0 \)
- For \( x + 3 \): \( x + 3 = 3 + 3 = 6 \neq 0 \)

Since neither denominator is zero, \( x = 3 \) is a valid solution.

Final Answer for Example 1:

\[
\boxed{x = 3}
\]

Example 2: Solve \( \frac{x^2 - 3}{x + 1} = \frac{x - 1}{x^3 + 2x} \)

#### Step 1: Identify the LCD
The denominators are \( x + 1 \) and \( x^3 + 2x \). Factor \( x^3 + 2x \):
\[
x^3 + 2x = x(x^2 + 2)
\]
So, the LCD is \( (x + 1)x(x^2 + 2) \).

#### Step 2: Eliminate the Denominators
Multiply both sides of the equation by \( (x + 1)x(x^2 + 2) \):
\[
(x + 1)x(x^2 + 2) \cdot \frac{x^2 - 3}{x + 1} = (x + 1)x(x^2 + 2) \cdot \frac{x - 1}{x(x^2 + 2)}
\]

This simplifies to:
\[
x(x^2 + 2)(x^2 - 3) = (x + 1)(x - 1)
\]

#### Step 3: Solve the Resulting Equation
Expand both sides:
- Left side: \( x(x^2 + 2)(x^2 - 3) \)
\[
x(x^4 - 3x^2 + 2x^2 - 6) = x(x^4 - x^2 - 6) = x^5 - x^3 - 6x
\]

- Right side: \( (x + 1)(x - 1) \)
\[
x^2 - 1
\]

So the equation becomes:
\[
x^5 - x^3 - 6x = x^2 - 1
\]

Rearrange all terms to one side:
\[
x^5 - x^3 - x^2 - 6x + 1 = 0
\]

This is a polynomial equation of degree 5, which may not have simple solutions. Solving this requires numerical methods or factoring techniques beyond basic algebra.

#### Step 4: Check for Extraneous Solutions
Any solution must not make the original denominators \( x + 1 \), \( x \), or \( x^2 + 2 \) equal to zero. Specifically:
- \( x + 1 \neq 0 \) implies \( x \neq -1 \)
- \( x \neq 0 \)
- \( x^2 + 2 \neq 0 \) (always true since \( x^2 + 2 > 0 \) for all real \( x \))

Final Answer for Example 2:

The equation \( x^5 - x^3 - x^2 - 6x + 1 = 0 \) needs further solving, but we know that any solution must satisfy \( x \neq -1 \) and \( x \neq 0 \).

If you need further assistance with solving the polynomial equation, let me know!

Summary of Solutions:

1. For \( \frac{1}{x} = \frac{2}{x+3} \): \( \boxed{x = 3} \)
2. For \( \frac{x^2 - 3}{x + 1} = \frac{x - 1}{x^3 + 2x} \): Further solving required, but \( x \neq -1 \) and \( x \neq 0 \).