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AMC 10 & 12

Rational Expressions

Domain restrictions, simplifying, and catching extraneous solutions — the fine print that separates a correct answer from a nearly-correct one.

A rational expression is just a fraction with polynomials in the numerator and denominator. The algebra is usually easy; the discipline is remembering that division by zero is never allowed, which restricts the domain and can silently invalidate a solution you find later.

Domain first, always

Before doing anything else, find every value that makes any denominator zero — those values are excluded from the domain, permanently, no matter what algebra happens afterward.

  • 1x\dfrac{1}{x} excludes x=0x=0
  • 1x2\dfrac{1}{x-2} excludes x=2x=2
  • x+1x24\dfrac{x+1}{x^2-4} excludes x=±2x = \pm 2 (factor the denominator first)

Solving rational equations

The standard method: factor everything, note the excluded values, clear denominators by multiplying both sides by the least common denominator, solve the resulting polynomial equation, then check every solution against the excluded values. A solution that happens to equal an excluded value is called extraneous — it must be discarded even though it satisfies the cleared-denominator equation.

Worked example

Problem: Solve xx3=3x3+2\dfrac{x}{x-3} = \dfrac{3}{x-3} + 2.

Solution: The domain excludes x=3x = 3. Multiply both sides by(x3)(x-3):

x=3+2(x3)=3+2x6=2x3x = 3 + 2(x-3) = 3 + 2x - 6 = 2x - 3

So x=2x3x=3x = 2x - 3 \Rightarrow x = 3. But x=3x=3 is exactly the excluded value — it's extraneous. This equation has no solution.

Practice problems

1.State the domain restriction(s) for x+5x29\dfrac{x+5}{x^2-9}.

2.Simplify x24x25x+6\dfrac{x^2-4}{x^2-5x+6}.

3.Solve 2x+13=5x\dfrac{2}{x} + \dfrac{1}{3} = \dfrac{5}{x}.

4.Solve xx11x1=x1x1\dfrac{x}{x-1} - \dfrac{1}{x-1} = \dfrac{x-1}{x-1} and note anything unusual.