Secondary Mathematics · Worked Repair Guide 43
A fractional equation can look difficult because the unknown appears inside denominators as well as numerators. The important structure is simpler: first identify values that are forbidden, then multiply every term by a common denominator to obtain an equivalent equation on the allowed domain.
This guide develops one central habit: write the restrictions before clearing denominators. Multiplying by an expression that can be zero is only safely reversible on the part of the domain where that expression is non-zero. Once a forbidden value is written down, later cancellation or quadratic reduction cannot accidentally restore it.
All equations below are original teaching constructions. This page extends the equation-solving layer of Algebraic Fractions, Formulae and Substitution and Quadratic Equations, Factorisation and Roots.
1. Denominators define forbidden values before solving begins
For 3/(x−2)=5, the original denominator requires x≠2.
This restriction exists before any algebra is performed.
Entry check: for 1/x+2/(x−3)=4, the restrictions are x≠0 and x≠3.
A candidate equal to a forbidden value must be rejected even if it appears after later simplification.
2. A single fractional term can be cleared directly
Solve 3/(x−2)=5, with x≠2.
Multiply both sides by x−2:
3=5(x−2)=5x−10.
Thus 5x=13 and x=13/5, which is allowed.
3. Cross-multiplication is shorthand for multiplying both sides
For a/b=c/d with b,d non-zero, multiplying both sides by bd gives ad=bc.
Example: 2/(x+1)=3/5, x≠−1.
10=3(x+1), so 3x=7 and x=7/3.
“Cross-multiply” is therefore a compressed description of a legal common-denominator multiplication, not a separate rule.
4. Multiple terms require the common denominator to multiply every term
Solve 1/x+1/2=3/4, x≠0.
A common denominator is4x. Multiply the entire equation by4x:
4+2x=3x.
Hence x=4.
Multiplying only the fractional terms and not every term changes the equation.
5. Factor denominators before choosing the least common denominator
Consider 1/(x−2)+3/(x²−4)=2, with x≠2,−2.
Since x²−4=(x−2)(x+2), the least common denominator is(x−2)(x+2).
Multiplying gives (x+2)+3=2(x²−4).
So x+5=2x²−8, hence 2x²−x−13=0. This quadratic does not factor neatly over the integers, so another quadratic method may be appropriate.
6. Clearing denominators can produce a quadratic
Solve 1/x+1/(x−2)=1, with x≠0,2.
Multiply by x(x−2):
(x−2)+x=x(x−2).
2x−2=x²−2x, so x²−4x+2=0.
Thus x=[4±√(16−8)]/2=2±√2. Both are allowed because neither equals0 or2.
7. Some fractional quadratics factor neatly
Solve 6/x=x+1, x≠0.
Multiply by x:6=x²+x.
x²+x−6=0=(x+3)(x−2).
Therefore x=−3 or2. Both satisfy x≠0.
8. A forbidden value can appear as an algebraic root and must still be rejected
Solve (x+1)/(x−2)=3x/(x−2), with x≠2.
Multiply by x−2 on the allowed domain:
x+1=3x.
So x=1/2, which is allowed.
If later algebra had produced x=2, it would have to be rejected because the original equation is undefined there.
9. Cancellation does not erase restrictions
Consider (x²−4)/(x−2)=x+2.
The left side simplifies to x+2 only for x≠2.
Thus the equation is true for every real x except2.
If the restriction is forgotten after cancellation, one may incorrectly claim x=2 is included even though the original left side is undefined.
10. A fraction equals zero when its numerator is zero and denominator is non-zero
Solve (x²−9)/(x−3)=0.
Restriction: x≠3.
Numerator zero gives x²−9=0, so x=±3.
Reject x=3. The only solution is x=−3.
11. Reciprocal equations often become quadratics
Solve x+6/x=5, x≠0.
Multiply by x:
x²+6=5x.
x²−5x+6=0=(x−2)(x−3).
Thus x=2 or3.
12. Reciprocal symmetry can provide a check
In x+6/x=5, the two roots are2 and3.
Their product is6, matching the constant associated with the reciprocal term after multiplying through.
Substitution confirms:2+3=5 and3+2=5.
This symmetry is a useful check, but the formal solution still comes from the cleared quadratic.
13. Equations with two shifted denominators need careful expansion
Solve 1/(x−1)+1/(x+1)=1, with x≠±1.
Multiply by(x−1)(x+1):
(x+1)+(x−1)=x²−1.
2x=x²−1, so x²−2x−1=0.
Thus x=1±√2. Neither candidate equals±1, so both are allowed.
14. A denominator can introduce a hidden condition even when the final equation is linear
Solve (2x+3)/(x−4)=5, with x≠4.
2x+3=5x−20.
23=3x, so x=23/3.
The final equation is linear, but the original domain restriction remains part of the solution process.
15. Equivalent transformations should preserve the allowed solution set
For allowed x, multiplying both sides of an equation by the same non-zero expression preserves equality in both directions.
The complication is that an algebraic expression such as x−2 may be zero at some inputs.
Writing x≠2 first restricts attention to the domain where multiplication by x−2 is reversible.
This is why domain control and equation manipulation belong to the same argument.
16. Context can reject a negative fractional-equation root
Suppose a positive length x is modelled by x+12/x=7.
Multiplying by x gives x²−7x+12=0, so x=3 or4.
Both happen to satisfy x>0.
If the equation instead produced a negative algebraic root, that root could be mathematically valid for the equation but inadmissible for a positive-length model.
17. Verification should use the original fractional equation
After clearing denominators and solving a quadratic, substitute candidates into the original equation rather than only the cleared equation.
The original check confirms both numerical equality and denominator validity.
For x=2 in 6/x=x+1, the original gives3=3.
For x=−3, the original gives−2=−2.
18. Capstone: restrictions, common denominator and quadratic roots
Solve 2/(x−1)+1/(x+1)=2, with x≠±1.
Multiply by(x−1)(x+1):
2(x+1)+(x−1)=2(x²−1).
3x+1=2x²−2.
2x²−3x−3=0.
Thus x=[3±√(9+24)]/4=(3±√33)/4. Neither equals±1, so both are valid.
19. Independent practice
- State the restriction for3/(x−2)=5.
- Solve3/(x−2)=5.
- Solve2/(x+1)=3/5.
- Solve1/x+1/2=3/4.
- State restrictions for1/x+1/(x−2)=1.
- Solve1/x+1/(x−2)=1.
- Solve6/x=x+1.
- Solve(x+1)/(x−2)=3x/(x−2).
- State the solution set of(x²−4)/(x−2)=x+2.
- Solve(x²−9)/(x−3)=0.
- Solve x+6/x=5.
- Solve1/(x−1)+1/(x+1)=1.
- Solve(2x+3)/(x−4)=5.
- State restrictions for2/(x−1)+1/(x+1)=2.
- Solve2/(x−1)+1/(x+1)=2.
- Solve x+12/x=7.
- Explain why cancellation does not erase an original restriction.
- Explain why every term must be multiplied when clearing denominators.
- Why should roots be checked in the original equation?
- Solve5/x=x−4, x≠0.
20. Worked answers
1. x≠2.
2. x=13/5.
3. x=7/3.
4. x=4.
5. x≠0,2.
6. x=2±√2.
7. x=−3 or2.
8. x=1/2.
9. All real x except2.
10. x=−3.
11. x=2 or3.
12. x=1±√2.
13. x=23/3.
14. x≠−1,1.
15. x=(3±√33)/4.
16. x=3 or4.
17. Simplification changes the expression’s form but not where the original denominator was zero.
18. Multiplying only selected terms produces a different equation; equality requires the same multiplier to act on the entire side.
19. The original equation checks both equality and denominator restrictions.
20. x=5 or−1. Multiply by x:5=x²−4x, so x²−4x−5=(x−5)(x+1)=0.
21. Diagnose fractional-equation errors at the restriction line
Common failures include clearing only some denominators, multiplying through before recording forbidden values, cancelling terms across addition, accepting a root that makes an original denominator zero, or solving the cleared quadratic correctly but never checking the original equation.
A useful correction note says “restrictions first”, “factor denominators”, “multiply every term”, or “check roots in the original”.
Then alter one denominator so the least common denominator changes. A secure learner should rebuild the multiplication rather than imitate the previous line.
22. Continue through the BTT learning routes
Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Algebraic Fractions, Formulae and Substitution, Equations, Balance and Checking and Quadratic Equations, Factorisation and Roots for prerequisites.
Within Batch11, continue to Congruence Tests, Triangle Correspondence and Geometric Proof, Algebraic Identities, Expansion and Factorisation, or Pyramids, Prisms, Nets, Slant Heights and Mensuration.
23. Sources and scope
The equations and practice questions are original teaching material. The transformations use standard real algebra with explicit denominator-domain restrictions.
For current Singapore Secondary curriculum and assessment scope, consult the relevant MOE/SEAB syllabus for the learner’s subject level and year.