Fractional equations and checking forbidden roots
Clear algebraic denominators only after identifying where the original equation is defined.
Extended only.
Before you begin
- Add algebraic fractions with a common denominator.
- Solve linear and quadratic equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which values are forbidden in 1/[x(x − 3)]?
What you will learn
- Solve equations containing linear algebraic denominators.
- Recognise a quadratic formed by clearing denominators.
- Reject candidates that fail an original restriction.
Record exclusions before changing the equation
In 3/(x − 2) = 1, the expression is defined only for x ≠ 2. Multiplication by x − 2 gives 3 = x − 2, so x = 5. The multiplication is valid on the original domain. Check 3/(5 − 2) = 1 rather than stopping at the cleared equation.
For x/(x − 1) = 2/(x − 1), clearing gives x = 2, which is allowed. A candidate at x = 1 would have to be rejected regardless of any simplified equality because the original equation would divide by zero. Restrictions belong to the original problem, not just its final algebraic form.
Multiply every term by a common product
For 2/x + 1/(x + 1) = 1, exclude x = 0 and −1, then multiply all three terms by x(x + 1). This gives 2(x + 1) + x = x(x + 1). Simplification produces x² − 2x − 2 = 0. Its roots 1 ± √3 are both allowed and can be checked in the original fractions.
For x/(x − 2) = 3/(x + 2), cross multiplication gives x(x + 2) = 3(x − 2) with x ≠ ±2. Expanding can yield no real solutions; cross multiplication is a transformation, not a promise that there is a root. Do not cancel an unknown denominator unless its non-zero condition is preserved.
Check the candidate and the question's context
An equation (x² − 4)/(x − 2) = 4 excludes two. Cancelling on the allowed domain gives x + 2 = 4 and the sole candidate x = 2. Since it was excluded at the start, the original equation has no solution. Simplification can expose a forbidden candidate without making it valid.
A rate problem often produces fractions because time equals amount divided by rate. If a rate is x, the physical model requires x > 0, as well as any algebraic denominator conditions. Write units, solve the equation and compare every candidate with both the original equation and physical restrictions.
Pause and explain
If clearing denominators produces a forbidden candidate, what should you do?
Worked example
Solve 2/x + 3/(x + 1) = 2.
Show the worked solution
- Exclude x = 0 and −1; multiply by x(x + 1).
- 2(x + 1) + 3x = 2x(x + 1), giving 2x² − 3x − 2 = 0.
- Factorise: (2x + 1)(x − 2) = 0. Both x = −1/2 and x = 2 are allowed and satisfy the original equation.
Answer x = −1/2 or 2
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve 5/(x − 1) = 2.
Give me a hint
First state the excluded value, then clear the denominator.
Compare my reasoning
- x ≠ 1; multiply to get 5 = 2(x − 1).
- Then 2x = 7, so x = 7/2.
- The candidate is allowed and 5/(7/2 − 1) = 2.
x = 7/2
Look for these in your work
- Stated the original restriction.
- Checked the solution in the fraction.
Solve (x² − 9)/(x − 3) = 6.
Give me a hint
Factorise, but keep x ≠ 3 throughout.
Compare my reasoning
- The numerator is (x − 3)(x + 3); the denominator excludes three.
- On the allowed domain the equation becomes x + 3 = 6, giving x = 3.
- This candidate is excluded, so the original equation has no solution.
No solution
Look for these in your work
- Preserved the cancelled factor's restriction.
- Distinguished a candidate from a valid root.
A machine completes 12 items at a rate of r items per minute and then 12 items at r + 2 items per minute. Total time is 5 minutes. Find r, with r > 0.
Give me a hint
Add the two times, not the two rates.
Compare my reasoning
- 12/r + 12/(r + 2) = 5, with r > 0.
- Multiply by r(r + 2): 24r + 24 = 5r² + 10r, giving 5r² − 14r − 24 = 0.
- (5r + 6)(r − 4) = 0, so r = 4; −6/5 is not a positive rate. Times are 3 and 2 minutes.
4 items per minute
Look for these in your work
- Modelled each stage's time as amount divided by rate.
- Rejected the root outside the physical domain.
Common mistakes
- Cross multiplying without recording denominator exclusions.
- Multiplying only some terms by the common denominator.
- Keeping a root that satisfies only the cleared equation.
What can you explain now?
Solve x/(x − 2) = 3, stating the restriction.
Compare with the explanation
x = 3, with original restriction x ≠ 2
x = 3(x − 2) gives 2x = 6. The candidate three is allowed and gives 3/1 = 3.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain a rational equation whose cleared form has a root but whose original form has no solution. Identify the exact forbidden operation.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources