Sec 4 G3 Maths Bukit Timah | Linear Fractions

Did you know? A cancelled denominator can still rule out an answer

A Secondary 4 G3 Mathematics learner might solve a fractional equation correctly, cancel a factor and then report a value that was never allowed in the first place. The original denominator is part of the question. Simplifying an expression does not make a division by zero legal.

The quick answer is to write every prohibited denominator value first. Then use a common denominator, multiply each term, solve the resulting linear equation, and check the candidate in the original expression.

The 2027 SEC G3 Mathematics K310 syllabus expressly includes simple fractional equations reducible to linear equations. This guide concentrates on those questions rather than the separate topic of fractional equations that become quadratics.

Start with what must not equal zero

For 5/(x − 2), the restriction is x ≠ 2. For a fraction with denominator x + 4, the restriction is x ≠ −4. If an equation contains both, keep both restrictions.

These exclusions do not automatically mean the equation has no solution. They simply identify values for which the original statement is undefined. I find it helps students to put ‘Restrictions:’ above their first algebra line.

A denominator may disappear later, but its original restriction remains relevant when reporting the solution set.

Worked example 1: one fraction equals a constant

Solve (2x + 1)/(x − 1) = 3. First note that x ≠ 1. Multiply both sides by x − 1 to get 2x + 1 = 3(x − 1).

Expand: 2x + 1 = 3x − 3. Rearranging gives x = 4. Checking the original fraction gives (2(4) + 1)/(4 − 1) = 9/3 = 3, so the candidate is valid.

Look for the first wrong line if a student reports a different answer. Was the bracket 3(x − 1) expanded incorrectly, or was the fraction incorrectly multiplied?

Worked example 2: two different denominators

Now solve 3/(x − 2) = 6/(x + 4). Here x ≠ 2 and x ≠ −4. With those restrictions, multiply by the common denominator (x − 2)(x + 4).

The resulting equation is 3(x + 4) = 6(x − 2). Expand: 3x + 12 = 6x − 12. So 24 = 3x and x = 8.

Check both sides of the original: 3/(8 − 2) = 1/2, while 6/(8 + 4) = 1/2. That direct verification is often faster and safer than replaying every algebra line.

Worked example 3: an integer term must also be multiplied

Consider 1/x + 1/2 = 3/x, where x ≠ 0. Multiply the entire equation by 2x, a common denominator for the fractions.

The first term becomes 2, the middle term becomes x and the right side becomes 6. Thus 2 + x = 6, giving x = 4.

Verify: 1/4 + 1/2 = 3/4, equal to 3/4 on the right. A common mistake is to clear only the visible 1/x fractions and leave the 1/2 term unchanged.

A cancellation that leads to no valid solution

Take (x² − 16)/(x − 4) = 8. The original denominator forbids x = 4. Factorising gives (x − 4)(x + 4)/(x − 4).

For allowed x the fraction simplifies to x + 4, so the derived equation x + 4 = 8 produces candidate x = 4.

But x = 4 makes the original expression undefined. Since no other candidate remains, the correct answer is no solution. The working has proved that conclusion; it is not a sign the student failed to solve the equation.

A no-solution trap without quadratic factorisation

Solve 2x/(x − 2) = 4 + 4/(x − 2), starting with x ≠ 2. Multiply all terms by x − 2.

This gives 2x = 4(x − 2) + 4, or 2x = 4x − 4. Rearranging gives only x = 2, which is forbidden.

Therefore the original equation has no solution. Do not award yourself a valid answer just because the resulting linear algebra produces a neat integer.

When the equation is true for every permitted value

Compare 2/(x − 3) = 4/(2x − 6). Both denominators require x ≠ 3. Since 2x − 6 = 2(x − 3), the right side simplifies to 2/(x − 3).

So the two expressions are identical for every real x except 3. The solution is a set of many values, not one number.

This is a useful contrast with the no-solution example. If clearing the denominator produces an identity such as 2 = 2, always return to the original domain before stating the final set.

Can a valid answer be negative?

Yes. Take 2/(x + 1) = −1, where x ≠ −1. Multiplying both sides by x + 1 gives 2 = −x − 1.

So x = −3. Checking the original, 2/(−2) = −1. Nothing in the problem forbids negative x, so −3 is valid.

Reject negative candidates only when there is a reason, such as a word problem defining x as a nonnegative count or a positive physical measurement.

A word problem about sharing a fixed cost

Fictional group-budget exercise: A shared outing costs S$48 before a subsidy and S$36 after. If each member pays equally and saves S$3, how many people are in the group?

Let n be the positive whole number of people. Per-person saving is 48/n − 36/n = 3. Combining the fractions gives 12/n = 3, so multiplying by nonzero n gives 12 = 3n.

Therefore n = 4. Check: the original per-person payment was S$12 and the reduced payment S$9, saving S$3. The answer makes sense as a whole-number count.

Why cross-multiplication is sometimes a poor shortcut

When one fraction equals another, a/b = c/d with b and d nonzero, the equality can be rearranged as ad = bc. But not every equation containing fractions has that single-fraction-on-each-side structure.

For 1/x + 1/2 = 3/x, multiplying all three terms by a common denominator is safer than treating the plus sign as if it did not exist.

A tutor can ask, ‘Where is the addition happening?’ before any manipulation. That question often reveals why the pupil cross-multiplied terms that did not actually form one fraction.

How to verify a candidate in under a minute

First, compare it against the exclusions written before the algebra. Second, substitute into the original fractions, not only the cleared equation. Third, calculate both sides independently and ask whether they agree.

If a candidate is forbidden, reject it even if the derived linear equation says it works. If a permitted candidate fails the original test, inspect the first denominator-clearing or expansion error.

Did you know? This checking routine is useful across other G3 Mathematics chapters too, including formula rearrangement, simultaneous equations and quadratic equations.

Do not mix simple linear and quadratic fractional cases

Some fractional equations become linear after denominators are cleared, but others become quadratic. For example, 6/x = x − 1, with x ≠ 0, yields x² − x − 6 = 0 after multiplication by x.

That equation needs a quadratic-solving method, not only linear rearrangement. The checking process is the same: enforce the original domain and substitute candidate roots into the given statement.

The 2027 K310 syllabus names both types. Learn to recognise what kind of equation appears after the fractions are cleared rather than assuming all questions use the same last step.

A one-week repair plan

Day 1: identify forbidden denominator values. Day 2: solve a fraction equal to an integer. Day 3: solve equations with two different denominators. Day 4: multiply correctly through a three-term equation.

Day 5: analyse a no-solution case. Day 6: recognise an identity true across a permitted domain. Day 7: translate one short budgeting story and retest two previous mistakes without model answers.

Keep an error log labelled forgotten restriction, missing common denominator, invalid cross-product, sign error, unchecked candidate or misread context. The diagnosis should decide the next practice question.

Frequently asked questions

Why write excluded values before solving? They cannot be restored by later cancellation.

Can a fractional equation have no solution? Yes, when every candidate is forbidden or the valid algebra is contradictory.

Can there be infinitely many solutions? Yes, if both sides are identical for every permissible x.

Are negative solutions allowed? Unless the domain or story excludes them.

When is cross-multiplication valid? For an equality of two single fractions with nonzero denominators.

Is this in SEC G3 Mathematics? Yes. K310 explicitly includes simple fractional equations reducible to linear equations.

Small-group Mathematics tutorials in Bukit Timah

The unchanged eduKateSG Mathematics tutorial reference describes premium tutorials for groups of up to three learners, with weekly 1.5-hour lessons, close individual feedback and corrections. The stated teaching venue is 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT. This page does not imply any new branch or confirm an available seat.

For algebraic fractions, one student may fail at factorisation, another may miss a denominator restriction, and a third may choose the wrong common denominator. A tutor can diagnose the first wrong step for each pupil and test an altered unseen equation after reteaching.

Enquire about an eduKate Singapore parent–student Mathematics consultation with one independent school script and the learner’s actual national examination year.

Examination-year guidance and official references

The first Singapore–Cambridge SEC is in 2027, with G3 Mathematics code K310 on the official SEAB list. A student taking the 2026 GCE O-Level Mathematics examination should follow their applicable subject 4052 syllabus and rules instead of treating the 2027 papers as identical.

All worked examples are original learning exercises, not official or predicted national examination questions. Outcomes depend on independent understanding, not promises of a particular grade.