Multiplying an equation by the LCD turns it into a polynomial equation you can already solve. That step is legal and it is not reversible, so it can hand you numbers that solve the new equation and not the old one.
Core ideas
This is the first lesson where an entirely correct piece of algebra can produce a wrong answer. Understanding why is more useful than remembering to check.
Multiply every term on both sides by the LCD. The denominators cancel and you are left with a linear equation from Linear Equations and Inequalities or a factorable quadratic from Polynomials and Factoring.
If A = B then A·M = B·M, for any M at all. The converse needs M ≠ 0, because A·0 = B·0 holds for every A and B. So solutions of the cleared equation need not be solutions of the original.
It is the predictable residue of a legal step. You cannot avoid it by being careful; you can only detect it, by substituting into the equation you were actually given.
The material
The method, the reason it leaks, and the check that plugs it.
The method is short. Factor every denominator, list the excluded values, multiply both sides by the LCD, solve whatever polynomial equation appears, and then test each candidate in the original.
An extraneous solution is a value that satisfies an equation derived from the original but not the original itself. For rational equations these are precisely the candidates that are excluded values.
The mechanism is worth stating exactly. Multiplying both sides by the LCD is multiplication by an expression whose value depends on x. At an excluded value that expression is 0, and multiplying an equation by 0 produces 0 = 0 — true regardless of what the original said.
A = B implies AM = BM always. AM = BM implies A = B only when M ≠ 0. Clearing denominators uses the first implication, so the solution set can only grow, never shrink. Every candidate must be tested against the original.
x/(x − 3) = 3/(x − 3) + 2. The LCD is x − 3, and x = 3 is excluded. Clearing gives x = 3 + 2(x − 3), so x = 2x − 3 and x = 3. The only candidate is the excluded value, so it is extraneous and the equation has no solution.
Nothing went wrong there. The algebra is correct at every line; the equation simply has no solution, and the method's honest report of that fact takes the form of a candidate you must throw away.
A quadratic can produce two candidates of which only one survives, and that is the case worth practising, because discarding both is easy and discarding neither is easier still.
1/(x − 2) + 1/x = (x² + x − 4)/(x(x − 2)) has LCD x(x − 2) and exclusions x ≠ 0, x ≠ 2. Clearing gives x + (x − 2) = x² + x − 4, that is x² − x − 2 = 0, or (x − 2)(x + 1) = 0. The candidates are 2 and −1; 2 is excluded, −1 checks out, and the solution set is {−1}.
Substitute into the original equation, never into the cleared one. Every candidate satisfies the cleared equation by construction, so checking there confirms nothing at all — it is the one test guaranteed to pass.
One special case deserves a name. When both sides have the same denominator, or the equation is a single fraction equal to a single fraction, clearing is the familiar cross-multiplication: a/b = c/d becomes ad = bc. It is the same step, with the same obligation to check.
Interactive
Clear the denominators, solve, then substitute into the ORIGINAL
Method
Four steps, and the fourth is not optional.
Before any solving. These are the values that will disqualify a candidate later, and they are easiest to find now.
Every term on both sides, including terms that are not fractions. Bracket each numerator as you go, exactly as in “Adding and Subtracting Rational Expressions”.
Use Linear Equations and Inequalities for a linear result and factoring from Polynomials and Factoring for the quadratics chosen in this course. The quadratic formula arrives in Quadratics and Complex Numbers and is not required here.
Any candidate on the exclusion list is extraneous and is discarded. State the surviving solution set, which may be empty.
Two candidates come out of the quadratic. One of them is not a solution of the equation we were given.
The candidate x = 2 is not a slip. It is the value at which the LCD x(x − 2) is zero, so it is exactly the value at which multiplying through stopped being reversible. The method predicts where its own failures will appear, which is why the exclusion list is written first.
Rational and Radical Expressions does this twice: “Solving Radical Equations” squares both sides of a radical equation, which is legal in the same one-way manner and produces extraneous roots for the same structural reason.
For a faded solve, take 1/(x − 1) + 1/(x + 1) = (x² − x + 2)/(x² − 1). The supplied LCD is (x − 1)(x + 1) and the exclusions are x ≠ −1, 1. Clear every term, factor the resulting quadratic and check both candidates in the original. You should reach candidates 1 and 2, discard 1, and keep solution set {2}. If both survived, you checked only the cleared quadratic.
Given a rational equation you should list the exclusions before solving, clear correctly, test each candidate in the original, and say why the discarded one appeared — naming the value at which the LCD was zero. That explanation is what transfers to “Solving Radical Equations”.
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