Rational equations
Equations containing rational expressions can be cleared of fractions by multiplying by the LCD. The extra step is checking for answers that would divide by zero.
Clearing denominators
To solve a rational equation, multiply every term by the LCD to remove the fractions, then solve the resulting polynomial equation. For \(\frac{x}{3} + \frac{x}{6} = 5\), multiply by 6: \(2x + x = 30\), so \(x = 10\).
Variables in denominators
For \(\frac{4}{x} + \frac{1}{2} = \frac{5}{x}\), the LCD is 2x. Multiplying gives \(8 + x = 10\), so \(x = 2\). Check: \(2 + 0.5 = 2.5\) on both sides.
Extraneous solutions
Multiplying by an expression that can be zero may create extraneous solutions, answers to the new equation that fail the original. Solving \(\frac{x}{x - 3} = \frac{3}{x - 3} + 2\) by multiplying by \(x - 3\) gives \(x = 3 + 2x - 6\), so \(x = 3\). But x = 3 makes the denominators zero, so there is no solution.
Always check
List the excluded values before solving, and reject any answer equal to one of them. This single habit prevents the most common mistake with rational equations.
Proportions
A proportion sets two ratios equal: \(\frac{a}{b} = \frac{c}{d}\). Cross-multiplying gives \(ad = bc\). If 3 cm on a map represents 45 km, then 7 cm represents x km where \(\frac{3}{45} = \frac{7}{x}\), so \(3x = 315\) and \(x = 105\) km.
Work problems
If one painter can paint a room in 4 hours and another in 6 hours, working together for t hours gives \(\frac{t}{4} + \frac{t}{6} = 1\). Multiplying by 12: \(3t + 2t = 12\), so \(t = 2.4\) hours, or 2 hours 24 minutes.
Distance-rate-time problems
A cyclist rides 30 km with the wind and 20 km against it in the same time; the wind adds or subtracts 5 km/h. With still-air speed r: \(\frac{30}{r + 5} = \frac{20}{r - 5}\). Cross-multiplying gives \(30r - 150 = 20r + 100\), so \(r = 25\) km/h.
Worked example
Solve \(\frac{2}{x - 1} + \frac{1}{x + 1} = \frac{5}{x^2 - 1}\). The excluded values are 1 and -1, and the LCD is \((x - 1)(x + 1)\). Multiplying: \(2(x + 1) + (x - 1) = 5\), so \(3x + 1 = 5\) and \(x = \tfrac{4}{3}\). It is not excluded, so it is the solution.
Key takeaways
Solve rational equations by multiplying every term by the LCD to clear fractions, then solving the polynomial equation. Multiplying by expressions that can be zero can create extraneous solutions, so list excluded values first and reject any answer matching them. Proportions are solved by cross-multiplying, as in map scales. Work problems add rates: if jobs take a and b hours, t/a + t/b = 1. Distance-rate-time problems often set two travel times equal.
Flashcard
Equation with rational expressions
Check your understanding
Solve \(\frac{x}{x - 3} = \frac{3}{x - 3} + 2\).