Rational Expressions and Equations
Students simplify rational expressions and solve rational equations while identifying domain restrictions and rejecting extraneous solutions.
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Domain Restrictions
A rational expression is defined only when its denominator is not zero. To identify domain restrictions, set each original denominator equal to zero and exclude the resulting values. For example, consider (x + 3)/(x² − 9). Because x² − 9 factors as (x − 3)(x + 3), the denominator is zero when x = 3 or x = −3. Therefore, the restrictions are x ≠ 3 and x ≠ −3. These restrictions come from the original expression and remain in effect even if a factor later cancels. At allowed values, the expression behaves like an ordinary quotient. At excluded values, division by zero would occur, so the expression has no value. Recording restrictions first helps prevent an undefined value from being accepted as a solution later.
Simplifying Rational Expressions
To simplify a rational expression, factor the numerator and denominator completely, then cancel common factors. Factors may be canceled because they represent multiplication; individual terms joined by addition or subtraction cannot be canceled. Consider (x² − 9)/(x² + x − 6). Factoring gives [(x − 3)(x + 3)]/[(x + 3)(x − 2)]. The common factor x + 3 cancels, leaving (x − 3)/(x − 2). However, the original denominator was zero at x = −3 and x = 2, so both restrictions must be retained. Thus, the simplified expression is (x − 3)/(x − 2), with x ≠ −3 and x ≠ 2. Simplifying changes the form of the expression, but it does not restore values excluded from its original domain.
Finding Common Denominators
Rational expressions must have a common denominator before they can be added or subtracted. Factor each denominator, then form the least common denominator using every distinct factor at its greatest needed power. For example, to add 2/(x − 1) + 3/(x + 2), use the least common denominator (x − 1)(x + 2). Multiply the first fraction by (x + 2)/(x + 2) and the second by (x − 1)/(x − 1). The sum becomes [2(x + 2) + 3(x − 1)]/[(x − 1)(x + 2)]. Combining like terms in the numerator gives (5x + 1)/[(x − 1)(x + 2)]. The restrictions are x ≠ 1 and x ≠ −2 because those values make an original denominator zero.
Solving Rational Equations
To solve a rational equation, first state all domain restrictions. Then identify the least common denominator and multiply every term on both sides by it. This clears the fractions and produces an equation that is usually easier to solve. For example, solve 2/(x − 1) + 1 = 5/(x − 1). The restriction is x ≠ 1. The least common denominator is x − 1. Multiplying every term by x − 1 gives 2 + (x − 1) = 5. Simplifying produces x + 1 = 5, so x = 4. Because 4 is allowed by the restriction, substitute it into the original equation: 2/3 + 1 = 5/3. Both sides equal 5/3, so x = 4 is a valid solution.
Extraneous Solutions
An extraneous solution is a value produced during algebraic work that does not satisfy the original equation. In rational equations, this often happens when clearing denominators leads to a value that makes an original denominator zero. Consider (x + 1)/(x − 1) = 2/(x − 1). The domain restriction is x ≠ 1. Multiplying both sides by x − 1 gives x + 1 = 2, which leads to x = 1. However, substituting 1 into the original equation makes both denominators zero. The original equation is undefined at that value, so x = 1 must be rejected and the equation has no solution. Always compare every candidate with the restrictions and, when possible, substitute it into the original equation. Checking is essential because transformed equations can include values that were never permitted.