What if the LCD is just a constant?

If the LCD contains no variables, the equation is linear or polynomial in form with fractional coefficients. There are no domain restrictions to check for extraneous solutions, but multiplying by the LCD is still the most efficient way to solve.

What if all potential solutions turn out to be extraneous?

If every solution you find is restricted by the original domain, the equation has no solution. You write the answer as 'No solution' or $\emptyset$.

Can cross-multiplication be used instead?

Cross-multiplication is only valid when the equation is a simple proportion: one fraction equal to one fraction (e.g., $\frac{a}{b} = \frac{c}{d}$). If there are multiple terms on either side, you must use the LCD method.

How to solve a rational equation and check for extraneous solutions

A rational equation contains at least one fraction whose denominator includes a variable. To solve it, multiply the entire equation by the least common denominator (LCD) of all the fractions to eliminate the denominators entirely.

This method applies to any equation with algebraic rational expressions. Because multiplying by an expression containing a variable can introduce false roots, you must always check all potential solutions against the restricted domain of the original equation.

The setup

Identify all denominators in the equation. Factor each denominator completely to determine the least common denominator (LCD). Note any values of the variable that make any denominator equal to zero; these are the domain restrictions. A solution is extraneous if it equals one of these restricted values.

The steps

Factor all denominators in the equation completely.
Determine the LCD for all rational expressions in the equation.
State the domain restrictions (values where the LCD is zero).
Multiply both sides of the entire equation by the LCD. Distribute carefully to cancel out all denominators.
Solve the resulting polynomial equation.
Compare your solutions to the domain restrictions.

Checking the result

Evaluate each potential solution. If a potential solution causes any original denominator to become zero, it is an extraneous solution and must be discarded. If it does not, it is a valid solution. You can also plug the valid solution back into the original equation to verify the equality holds.

Common errors

Failing to distribute the LCD to every term: Students often forget to multiply the LCD by terms that are not fractions.
Sign errors during distribution: Pay strict attention to negative signs, especially when subtracting rational expressions.
Accepting extraneous solutions: Forgetting to check the final answers against the original domain restrictions is the most frequent error.

Worked example

Solve for $x$ : $\frac{x}{x-2} + \frac{1}{x+2} = \frac{8}{x^2-4}$

Factor the denominators: $x^2-4 = (x-2)(x+2)$

Identify the LCD and domain restrictions: LCD: $(x-2)(x+2)$ Restrictions: $x eq 2$ , $x eq -2$

Multiply the entire equation by the LCD: $(x-2)(x+2) \left[ \frac{x}{x-2} + \frac{1}{x+2} \right] = (x-2)(x+2) \left[ \frac{8}{(x-2)(x+2)} \right]$

Distribute and cancel denominators: $x(x+2) + 1(x-2) = 8$

Expand and collect terms to form a polynomial equation: $x^2 + 2x + x - 2 = 8$ $x^2 + 3x - 10 = 0$

Solve the polynomial by factoring: $(x+5)(x-2) = 0$ $x = -5$ or $x = 2$

Check for extraneous solutions: The domain restrictions are $x eq 2$ and $x eq -2$ . Since $x=2$ is a restricted value, it is an extraneous solution.

Final answer: $x = -5$

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References: OpenStax College Algebra, Chapter 2: Equations and Inequalities · Khan Academy, Algebra 2: Rational equations · Paul's Online Math Notes: Algebra, Solving Equations and Inequalities