A student solves x/(x−2) = 2/(x−2) by multiplying both sides by (x−2), arriving at x = 2. What is the correct solution set?
A{2} — x = 2 satisfies the simplified equation, so it is valid
BAll real numbers — the equation simplifies to a tautology
CNo solution — x = 2 makes both denominators zero and is excluded from the domain
DThe equation cannot be solved by this method
Multiplying both sides by (x−2) is only valid when x ≠ 2. The 'solution' x = 2 makes both original denominators zero — it is an extraneous solution produced by multiplying by zero. Since x = 2 is the only candidate and it is excluded from the domain, the equation has no solution. This is the most important check in solving rational equations.
Question 2 Multiple Choice
Solving 1/(x+3) + 1/(x−3) = 2/(x²−9) yields x = ±3 after clearing fractions. A student reports both as solutions. What error did the student make?
AThe student used the wrong LCD; the correct LCD produces different solutions
BNo error — both ±3 satisfy the polynomial equation obtained after clearing fractions
CBoth x = 3 and x = −3 are extraneous; they make the original denominators zero
DOnly x = 3 is extraneous; x = −3 is a valid solution
x = 3 makes (x−3) = 0, and x = −3 makes (x+3) = 0; both also make x²−9 = 0. These are extraneous solutions — they appear because multiplying both sides by (x²−9) is invalid when x²−9 = 0. The polynomial equation may accept these values, but the original rational equation is undefined there. The equation has no solution.
Question 3 True / False
An extraneous solution to a rational equation may satisfy the polynomial equation obtained after clearing fractions, yet be rejected as a solution to the original equation.
TTrue
FFalse
Answer: True
Clearing fractions by multiplying by the LCD is only reversible when the LCD is nonzero. If a candidate solution makes the LCD equal zero, the multiplication step was invalid at that point — it was equivalent to multiplying by zero, which can create false solutions. The polynomial equation may produce that value as a root, but the original rational equation is undefined there, so it must be rejected.
Question 4 True / False
Extraneous solutions to rational equations can generally be identified because they are negative numbers or zero.
TTrue
FFalse
Answer: False
Extraneous solutions are not identified by their sign or magnitude. A solution is extraneous if and only if it makes at least one denominator in the original equation equal to zero. An extraneous solution can be any real number — positive, negative, or zero. The only reliable identification method is to substitute every candidate solution back into the original (unmodified) equation and check that it is defined and balanced.
Question 5 Short Answer
Why can clearing fractions in a rational equation produce extraneous solutions, and what is the only reliable method to identify them?
Think about your answer, then reveal below.
Model answer: Clearing fractions requires multiplying both sides by the LCD. If the LCD contains the variable, this step is only valid when the LCD is nonzero. A candidate solution that makes the LCD zero corresponds to a step where both sides were multiplied by zero — an irreversible operation that can introduce false solutions. The only reliable method is to substitute every candidate solution back into the original equation. Any value that makes a denominator zero, or that fails to balance the equation, must be rejected as extraneous.
The parallel with radical equations is useful: squaring both sides to clear a square root is also irreversible and also produces extraneous solutions. In both cases, the algebra can yield results that satisfy the transformed equation but not the original. Substitution back into the original is the universal check.