Why x=2 stays excluded after cancellation: check a rational equation in its original form

Cancellation preserves equality on a restricted domain

The OpenStax rational-equations chapter explains excluding denominator zeros and checking candidates in the original equation. The particular exercises here are original editorial examples, not copied textbook questions or a measured learning intervention.

At x=2, (x²−4)/(x−2) becomes 0/0 and is undefined. For any other x, factor the numerator as (x−2)(x+2), then cancel the nonzero factor to obtain x+2. The restriction enables that step; it is not a disposable note after the algebra.

Think of the original expression as allowing one fewer input than x+2. Their values agree where x≠2, but their domains are not identical. A straight-line sketch without its missing point can hide precisely this distinction.

Question one: why does a right-hand side of four give no solution?

Start with (x²−4)/(x−2)=4 and state x≠2. Cancellation gives x+2=4, producing the candidate x=2. It lies outside the domain and must be rejected. With no other candidate, the solution set is empty.

Substitution into the original gives numerator 2²−4=0 and denominator 2−2=0, not a valid value equal to four. Checking only 2+2=4 verifies an equation with an enlarged domain, not the original problem.

Question two: establish that three is valid

Change only the right side. Keep x≠2, solve x+2=5 to obtain x=3, then substitute into the original: (9−4)/(3−2)=5/1=5. The denominator is nonzero and the equality holds, so the solution set is {3}.

Domain checking does not mean rational equations never have solutions. It excludes illegal inputs rather than treating every result as invalid simply because a denominator appeared.

Try two variations before reading their answers

Cover the answers and solve A: (x²−9)/(x−3)=6 and B: (x²−9)/(x−3)=7. For each, write the restriction, candidate and original substitution rather than only a final number.

For A, x≠3. Factoring and cancelling gives x+3=6; candidate three is excluded, so there is no solution. For B, again x≠3, and x+3=7 gives four. The original evaluates to (16−9)/(4−3)=7, so the solution set is {4}.

Now explain the identity (x²−4)/(x−2)=x+2. It holds throughout the allowed domain, so its solutions are all real x except two. It is neither unrestricted because of cancellation nor unsolvable because no single candidate appears. This is a different task from screening one candidate.

Turn the error into an observable checking action

A useful error-log instruction is: “Mark denominator zeros before cancelling; carry the restriction through the reduction; check the original afterward.” “Be more careful” does not identify an action you can inspect next time.

For squaring and roots, use the radical-equation candidate check. For negative multiplication or division of inequalities, use the boundary and sign-reversal guide. These concern different operation conditions; one slogan cannot replace their reasons.