Solving Rational Equations
Overview
Rational equations are equations that involve rational expressions (fractions). Solving rational equations requires eliminating the fractions and solving the resulting equation. The process involves multiplying both sides of the equation by the least common denominator (LCD) of all the fractions, simplifying, and then solving for the unknown variable.
Steps to Solve Rational Equations
- Identify all the rational expressions in the equation.
- Find the least common denominator (LCD) of all the fractions involved.
- Multiply both sides of the equation by the LCD to eliminate the fractions.
- Simplify the equation and solve for the variable.
- Check for extraneous solutions by substituting the solution back into the original equation.
Example 1: Solving a Simple Rational Equation
Consider the following equation:
\( \frac{1}{x} + 2 = \frac{3}{x} \)
Step 1: Identify the rational expressions. In this case, \( \frac{1}{x} \) and \( \frac{3}{x} \) are the rational expressions.
Step 2: The LCD of the fractions is \( x \).
Step 3: Multiply both sides of the equation by \( x \) to eliminate the fractions:
\( x \cdot \left( \frac{1}{x} + 2 \right) = x \cdot \left( \frac{3}{x} \right) \)
Step 4: Simplify both sides:
\( 1 + 2x = 3 \)
Step 5: Solve for \( x \):
\( 2x = 2 \) → \( x = 1 \)
Step 6: Check for extraneous solutions. Substitute \( x = 1 \) back into the original equation:
\( \frac{1}{1} + 2 = \frac{3}{1} \) → \( 1 + 2 = 3 \)
The solution \( x = 1 \) is valid.
Example 2: Solving a More Complex Rational Equation
Consider the following equation:
\( \frac{2}{x} = \frac{5}{x + 1} - \frac{3}{x + 1} \)
Step 1: Identify the rational expressions. In this case, \( \frac{2}{x} \), \( \frac{5}{x + 1} \), and \( \frac{3}{x + 1} \) are the rational expressions.
Step 2: The LCD of the fractions is \( x(x + 1) \).
Step 3: Multiply both sides of the equation by \( x(x + 1) \):
\( x(x + 1) \cdot \frac{2}{x} = x(x + 1) \cdot \left( \frac{5}{x + 1} - \frac{3}{x + 1} \right) \)
Step 4: Simplify both sides:
\( 2(x + 1) = 5x - 3x \)
Step 5: Simplify further:
\( 2x + 2 = 2x \)
Step 6: Solve for \( x \):
\( 2 = 0 \)
This is a contradiction, so there is no solution. The equation has no solution, and there are no extraneous solutions.
Practice Questions
- Question 1: Solve the equation:
\( \frac{3}{x} + 5 = \frac{4}{x} \)
Solution
Step 1: The rational expressions are \( \frac{3}{x} \) and \( \frac{4}{x} \).
Step 2: The LCD is \( x \).
Step 3: Multiply both sides by \( x \) to eliminate the fractions:
\( x \cdot \left( \frac{3}{x} + 5 \right) = x \cdot \left( \frac{4}{x} \right) \)
Step 4: Simplify:
\( 3 + 5x = 4 \)
Step 5: Solve for \( x \):
\( 5x = 1 \) → \( x = \frac{1}{5} \)
Check: \( x = \frac{1}{5} \) is a valid solution.
- Question 2: Solve the equation:
\( \frac{4}{x - 2} = \frac{6}{x + 3} - \frac{2}{x - 2} \)
Solution
Step 1: The rational expressions are \( \frac{4}{x - 2} \), \( \frac{6}{x + 3} \), and \( \frac{2}{x - 2} \).
Step 2: The LCD is \( (x - 2)(x + 3) \).
Step 3: Multiply both sides by \( (x - 2)(x + 3) \):
\( (x - 2)(x + 3) \cdot \frac{4}{x - 2} = (x - 2)(x + 3) \cdot \left( \frac{6}{x + 3} - \frac{2}{x - 2} \right) \)
Step 4: Simplify:
\( 4(x + 3) = 6x - 12 - 2x - 6 \)
Step 5: Simplify further:
\( 4x + 12 = 4x - 18 \)
Step 6: Solve for \( x \):
\( 12 = -18 \) (No solution)
There is no solution, and no extraneous solution exists.
- Question 3: Solve the equation:
\( \frac{5}{x + 1} = \frac{7}{x - 2} + \frac{4}{x + 1} \)
Solution
Step 1: The rational expressions are \( \frac{5}{x + 1} \), \( \frac{7}{x - 2} \), and \( \frac{4}{x + 1} \).
Step 2: The LCD is \( (x + 1)(x - 2) \).
Step 3: Multiply both sides by \( (x + 1)(x - 2) \):
\( (x + 1)(x - 2) \cdot \frac{5}{x + 1} = (x + 1)(x - 2) \cdot \left( \frac{7}{x - 2} + \frac{4}{x + 1} \right) \)
Step 4: Simplify:
\( 5(x - 2) = 7(x + 1) + 4(x - 2) \)
Step 5: Solve:
\( 5x - 10 = 7x + 7 + 4x - 8 \)
\( 5x - 10 = 11x - 1 \)
\( -10 + 1 = 11x - 5x \)
\( -9 = 6x \) → \( x = -\frac{3}{2} \)
Check: \( x = -\frac{3}{2} \) is a valid solution.