Secondary Mathematics · Worked Repair Guide 13
Algebraic fractions and formulae are difficult for the same reason many later Mathematics topics are difficult: several meanings must stay active at once. A symbol may stand for a number, a denominator may impose a restriction, an expression may need to be factorised before it can simplify, and a formula may preserve a relationship even when the subject changes.
This guide develops one central habit: protect the structure before manipulating the symbols. That means identifying restrictions, preserving equality, substituting with brackets, simplifying only legitimate factors and checking whether the transformed expression still represents the original relationship.
All examples and practice questions are original teaching material. Extension work should be matched to the learner’s actual Secondary Mathematics route and school programme.
1. A formula is a relationship, not a sentence to memorise
The formula A=lw states that the area A of a rectangle is related to length l and width w by multiplication. If l=8 and w=5, substitution gives A=8×5=40.
The letters are not decorations. They represent quantities whose values can change while the relationship remains the same.
Entry check: if P=2l+2w, l=7 and w=3, then P=2(7)+2(3)=20. Brackets are useful because they preserve the substituted value as one object.
When the substituted number is negative, brackets become essential: if x=−4, then x² means (−4)²=16.
2. Substitute first, then respect order of operations
Suppose y=3x²−2x+5 and x=−2. Write y=3(−2)²−2(−2)+5. Then y=12+4+5=21.
A common error is to write −2² instead of (−2)², changing the meaning of the square.
Another common error is to perform the multiplication before the exponent has been evaluated. The original order of operations still applies after substitution.
A good routine is: copy the formula, substitute with brackets, simplify powers, multiply or divide, then add or subtract.
3. A denominator creates a restriction
The algebraic fraction 5/(x−3) is undefined when x=3 because the denominator becomes zero.
Therefore the expression carries the restriction x≠3.
Restrictions are part of the mathematical object. They should be noticed before simplification because later algebra can hide the value that originally caused division by zero.
Worked example: (x²−9)/(x−3) factorises to (x−3)(x+3)/(x−3)=x+3, but only for x≠3. The simplified expression x+3 is defined at x=3; the original fraction is not. The restriction must survive.
4. Cancel factors, not terms
In 6x/9, the common factor 3 can be cancelled, giving 2x/3.
In (x+3)/x, the x in the denominator cannot cancel the x inside the sum x+3. Cancellation works across multiplication, not across addition.
Factorisation can expose legitimate factors. For example (x²+5x)/(x)=x(x+5)/x=x+5 for x≠0.
Repair rule: if you cannot point to a multiplication sign joining the factor to the rest of the numerator or denominator, cancellation is probably not legal yet.
5. Multiplying algebraic fractions uses factor structure
To simplify (3x/4)×(8/x²), factor and cancel common multiplicative factors:
(3x/4)×(8/x²)=24x/(4x²)=6/x, for x≠0.
You can also cancel before multiplying: 8/4 becomes 2 and x/x² becomes 1/x.
The restriction x≠0 comes from the original denominator x².
Multiplication is usually easier than addition because no common denominator is required beyond the existing product structure.
6. Dividing by an algebraic fraction means multiplying by its reciprocal
For ordinary fractions, a/b ÷ c/d = a/b × d/c when the divisions are defined. Algebraic fractions follow the same structure.
Worked example: (2x/3) ÷ (4/x) = (2x/3)×(x/4)=x²/6.
Restrictions come from both the original denominators and from the divisor being non-zero. Here x≠0 because 4/x must exist and because 4/x cannot equal zero anyway.
Do not invert the first fraction. Invert the divisor.
7. Adding algebraic fractions needs a common denominator
For 2/x + 3/y, the common denominator is xy, giving 2y/(xy)+3x/(xy)=(2y+3x)/(xy), with x≠0 and y≠0.
You cannot add denominators: 2/x+3/y is not 5/(x+y).
Worked example: 1/x + 2/(3x) = 3/(3x)+2/(3x)=5/(3x), x≠0.
Choose the least useful common denominator that contains all required factors; then multiply each numerator by exactly what its denominator is missing.
8. Subtracting fractions requires brackets around the second numerator
Consider 3/x − (x+1)/x. Since the denominators match, subtract the numerators:
[3−(x+1)]/x=(2−x)/x.
If the brackets are omitted, the negative sign may fail to distribute through every term of the second numerator.
This is the same sign discipline used in ordinary algebraic subtraction.
When denominators differ, build the common denominator first and keep each converted numerator grouped.
9. Factorising denominators can reveal the true common denominator
Suppose we add 1/(x−2) and 3/(x²−4). Since x²−4=(x−2)(x+2), the second denominator already contains the first factor.
The common denominator is (x−2)(x+2). Rewrite the first fraction as (x+2)/[(x−2)(x+2)]. Then:
1/(x−2)+3/(x²−4)=(x+5)/[(x−2)(x+2)].
The restrictions are x≠2 and x≠−2.
10. Solving equations with fractions begins by clearing denominators carefully
Consider 2/x=3/5 with x≠0. Multiply both sides by 5x to clear the denominators: 10=3x, so x=10/3.
The shortcut called cross-multiplication is justified because both sides are multiplied by the product of the denominators.
Worked example: 1/(x−1)=2/3. Restriction x≠1. Multiply by 3(x−1): 3=2(x−1), so 3=2x−2, hence x=5/2. The solution does not violate the restriction.
Always check candidate solutions against the original denominator restrictions.
11. Changing the subject preserves equality
From A=lw, making w the subject means isolating w while preserving equality. Divide both sides by l, assuming l≠0: w=A/l.
Changing the subject is equation solving with several variables rather than one numerical unknown.
Worked example: v=u+at. Make a the subject. Subtract u: v−u=at. Divide by t, t≠0: a=(v−u)/t.
The final form should be checked by substituting it back into the original relationship.
12. When the subject appears twice, factor it
Suppose P=2l+2w and we want l as the subject. Subtract 2w: P−2w=2l. Divide by 2: l=(P−2w)/2.
Now consider y=ax+bx. If x is the subject, first factor: y=x(a+b). Then x=y/(a+b), provided a+b≠0.
Collecting the subject into one multiplicative factor is often the key move.
If the subject appears inside several terms, resist dividing term-by-term before factoring unless the algebra clearly supports it.
13. Subjects inside powers require inverse operations
From A=πr², making r the subject gives r²=A/π, then r=√(A/π) when r represents a non-negative radius.
Algebraically the equation r²=A/π can have two real roots where A/π≥0, but the geometric context restricts radius to non-negative values.
Worked example: E=mv²/2. Make v the subject: 2E=mv², so v²=2E/m and v=±√(2E/m) algebraically when defined. A physical model may impose an additional interpretation on the sign.
Context can therefore narrow the algebraic solution set.
14. Formulae can be checked dimensionally
If distance d, speed v and time t satisfy d=vt, the units agree: (distance/time)×time=distance.
If someone proposes d=v+t, the units do not match, so the formula cannot generally describe the intended relationship.
Dimensional checking does not prove a formula is correct, but a unit mismatch is enough to reveal that it cannot be right as written.
Use the Units, Scale and Measurement guide for more on dimensional reasoning.
15. Substitution can verify a rearranged formula
Suppose C=5(F−32)/9 and a learner rearranges to F=9C/5+32. Test C=100. The rearranged formula gives F=180+32=212.
Substitute F=212 into the original: C=5(212−32)/9=5×180/9=100. The pair is consistent.
One numerical test does not constitute a formal proof of equivalence, but it is an excellent error detector.
A stronger symbolic check substitutes the new expression for F into the original and simplifies back to C.
16. Capstone: restrictions, simplification and verification
Simplify (x²−4)/(x²−x−2) and state all restrictions.
Factorise: x²−4=(x−2)(x+2), while x²−x−2=(x−2)(x+1).
Therefore the fraction simplifies to (x+2)/(x+1), but the original denominator requires x≠2 and x≠−1.
The cancelled factor x−2 does not erase the original restriction x≠2.
Check with x=3. Original: (9−4)/(9−3−2)=5/4. Simplified: 5/4. The forms agree at an allowed value.
17. Independent practice
- Evaluate y=2x²−3x+4 when x=−2.
- State the restriction on 7/(x+5).
- Simplify 12x/18.
- Simplify (x²−16)/(x−4) and state the restriction.
- Simplify (3x/5)(10/x²).
- Simplify (4x/7)÷(2/x).
- Simplify 1/x+1/(2x).
- Simplify 5/y−2/y.
- Simplify 1/(x+1)+2/[(x+1)(x−3)].
- Solve 3/x=2/7.
- Solve 1/(x+2)=3/5.
- Make w the subject of A=lw.
- Make t the subject of d=vt.
- Make x the subject of y=ax+b.
- Make x the subject of y=px+qx.
- Make r the subject of A=πr² for a non-negative radius.
- Explain why x cannot cancel in (x+4)/x.
- State the restrictions on (x+1)/[(x−2)(x+3)].
- Check whether x=2 is allowed in (x²−4)/(x−2).
- Explain why a restriction can survive after a factor is cancelled.
18. Worked answers
1. 18. Use brackets: 2(4)+6+4.
2. x≠−5.
3. 2x/3.
4. x+4, with x≠4.
5. 6/x, x≠0.
6. 2x²/7.
7. 3/(2x), x≠0.
8. 3/y, y≠0.
9. (x−1)/[(x+1)(x−3)]. Restrictions x≠−1,3.
10. x=21/2.
11. x=−1/3. From 5=3(x+2).
12. w=A/l. Require l≠0 where division is used.
13. t=d/v. Require v≠0.
14. x=(y−b)/a. Require a≠0.
15. x=y/(p+q). Require p+q≠0.
16. r=√(A/π).
17. Because x is a term inside a sum, not a common multiplicative factor of the whole numerator.
18. x≠2 and x≠−3.
19. No. The original denominator is zero at x=2.
20. Simplification changes form but does not change where the original expression was undefined.
19. Diagnose the first structural error
Common failures include cancelling terms across addition, losing denominator restrictions, substituting negative values without brackets, changing the subject by moving terms without preserving equality, or clearing denominators without multiplying every term.
A useful correction note names the structure: “factor before cancelling”, “restriction survives cancellation”, “subject appears twice—factor it”, or “substitute with brackets”.
Then vary the surface so the learner must rebuild the reasoning instead of copying a sequence of moves.
20. Continue through the BTT Mathematics library
Return to the BTT Mathematics Hub. Use Equations, Balance and Checking for equality-preserving transformations, Signed Numbers, Brackets and Algebraic Structure for factorisation and sign control, and Indices, Roots and Standard Form for powers inside formulae.
The BTT Mathematical Lab is the diagnostic route when the same symbolic-structure error keeps returning across different chapters.
21. Sources and scope
The examples, formulae contexts and practice questions in this guide are original teaching material. They are designed to expose algebraic structure rather than reproduce examination questions.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match rational-expression and subject-changing extensions to the learner’s actual subject level and school programme.