Section 9.5: Solving Rational Equations
A rational equation is one that contains fractions with polynomials in the denominator. To solve, multiply through by the least common denominator (LCD) to eliminate fractions. Always check for extraneous solutions (values that make denominators zero).
Example 1
Solve: \( \dfrac{x}{2} = \dfrac{3}{4} \)
Multiply both sides by 4: \( 2x = 3 \).
\( x = \tfrac{3}{2} \).
Example 2
Solve: \( \dfrac{2}{x} = \dfrac{3}{5} \)
Cross multiply: \( 2 \cdot 5 = 3x \).
\( 10 = 3x \) ⟹ \( x = \tfrac{10}{3} \).
Check: denominator not zero, valid.
Example 3
Solve: \( \dfrac{x+1}{x} = 2 \)
Multiply both sides by \( x \): \( x+1 = 2x \).
\( 1 = x \).
Check: \( x \neq 0 \), so \( x=1 \) is valid.
Example 4
Solve: \( \dfrac{1}{x-2} + \dfrac{1}{x+2} = \dfrac{1}{2} \)
LCD = \( 2(x-2)(x+2) \).
Multiply through: \( 2(x+2) + 2(x-2) = (x-2)(x+2) \).
Left: \( 2x+4 + 2x-4 = 4x \).
Right: \( x^2 - 4 \).
Equation: \( 4x = x^2 - 4 \).
\( x^2 - 4x - 4 = 0 \).
Quadratic formula: \( x = \dfrac{4 \pm \sqrt{16+16}}{2} = 2 \pm 2\sqrt{2} \).
Restrictions: \( x \neq \pm 2 \). Both solutions valid.
Practice Problems
- \( \dfrac{x}{3} = \dfrac{2}{5} \)
- \( \dfrac{4}{y} = \dfrac{1}{2} \)
- \( \dfrac{x+2}{x} = 3 \)
- \( \dfrac{1}{z-1} + \dfrac{1}{z+1} = 1 \)
- \( \dfrac{2}{x} - \dfrac{3}{x+1} = 1 \)