Learning goal and prerequisites
Distinguish an algebraic candidate from a solution in the original domain. Before starting, review asymptotes and holes.
Method and conditions
Write excluded inputs before clearing denominators. Multiply every term by the common denominator, solve the simpler equation, and substitute candidates into the original rule. For inequalities, retain signs of denominator factors.
Worked example: show the decisions
Solve 3/(x+1)=4/(x+2). Inputs −1 and −2 are excluded. Cross multiplication gives 3x+6=4x+4, so x=2. Substitution gives 1 on each side. The equality holds on the original domain.
Contrast: why a plausible shortcut fails
For (x²−9)/(x−3)=6, cancellation gives x+3=6, hence candidate x=3. The original denominator rejects it, so there is no solution. Canceling did not add x=3 back to the domain.
Transfer: try before revealing the answer
Solve 2/(x+1)=3/(x+2), then explain why blindly cross multiplying an inequality would require extra care.
Transfer solution and check
The equation gives 2x+4=3x+3, so x=1, which is allowed. In an inequality the denominator product may be negative, reversing the inequality direction.
How many integers x in [0,6] satisfy 1/(x−3)>0?
- The denominator must be positive; count integers above 3 through 6, including the upper endpoint.
- Using these quantities: 6-3 = 3.
- Check the original domain, requested quantity and units; do not substitute a related parameter.
Compare the other answer choices
Exclude the allowed upper endpoint 6.
Include the undefined input x=3.
Count nonzero integers without checking the denominator sign.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
AP Precalculus Course at a Glance