Sec 4 G3 Maths Bukit Timah | Fractional Equations

Did you know? Multiplying a fractional equation by its denominator can produce a quadratic equation — and that quadratic may have a solution that the original fraction does not allow. It is possible to perform the algebra correctly and still report an invalid root.

Quick answer: list every forbidden denominator value before clearing fractions. Multiply both sides by an appropriate common denominator, solve the resulting quadratic, then substitute the candidate roots into the original equation. Reject any root that makes an original denominator zero or violates the context.

The 2027 SEC G3 Mathematics K310 syllabus explicitly includes solving fractional equations that can be reduced to quadratic equations. This Bukit Timah learning guide focuses on the transition from fraction to quadratic, where the hidden restrictions matter as much as the factorisation.

Quick route: the essentials

Exclude every value that makes an original denominator zero.
Clear the fractions, solve the resulting quadratic, then check every candidate.
Reject a root only for a valid original restriction or contextual condition.

Jump directly to a worked answer

Start with the denominator, not the numerator
Why does clearing a denominator change the form of a problem?
Worked example 1: two valid quadratic roots
Worked example 2: a different linear denominator
The most revealing example: an answer that is not allowed
Another equation whose candidate must be rejected
Worked example 3: quadratic formula when factorisation is awkward
How to choose a common denominator safely
When cross-multiplying is appropriate
A real-world speed problem that becomes quadratic
What is a valid reason for rejecting a root?
Four stages to write clearly in an exam script
A seven-day revision plan for fractional equations
Frequently asked questions about fractional quadratic equations

Start with the denominator, not the numerator

For a fraction such as 6/x, the condition is x ≠ 0. For 12/(x + 1), the condition is x ≠ −1. These are part of the original expression’s domain.

If an equation contains two denominators, identify the zeros of both. For example, 1/(x − 2) + 3/(x + 4) requires x ≠ 2 and x ≠ −4.

Did you know? The denominator restrictions remain in force even when later algebraic cancellation makes a forbidden factor disappear from the expression.

Why does clearing a denominator change the form of a problem?

Consider 6/x = x − 1. Because x is in the denominator, this is a fractional equation. For x ≠ 0, multiply both sides by x: 6 = x(x − 1).

Expand to get x² − x − 6 = 0. The fraction has become a quadratic equation. The step is legitimate only for the allowed values of x, because multiplication by zero would conceal part of the original problem.

The result is not automatically the end of the task. You still need to solve the quadratic and check which roots are permitted.

Worked example 1: two valid quadratic roots

Factorise x² − x − 6 as (x − 3)(x + 2). The candidate solutions are x = 3 and x = −2.

Test in the original equation 6/x = x − 1. If x = 3, the left side is 2 and the right side is 2. If x = −2, the left side is −3 and the right side is −3.

Neither candidate makes x = 0, so both are valid. Negative numbers are not automatically forbidden: only the original denominator and any contextual restrictions decide which roots survive.

Worked example 2: a different linear denominator

Solve 12/(x + 1) = x + 2. First note x ≠ −1. Multiply by x + 1: 12 = (x + 1)(x + 2).

Expand and collect: 12 = x² + 3x + 2, so x² + 3x − 10 = 0. Factorise as (x + 5)(x − 2) = 0.

The candidates are x = −5 and x = 2. For x = −5, the original left side is 12/(−4) = −3 and the right side is −3. For x = 2, both sides give 4. Thus both are permitted.

The most revealing example: an answer that is not allowed

Consider (x² − 9)/(x − 3) = 6. The original denominator requires x ≠ 3.

Factor the numerator: x² − 9 = (x − 3)(x + 3). For allowed x, cancelling (x − 3) gives x + 3 = 6 and therefore x = 3.

But the only candidate is explicitly forbidden by the original denominator. Hence the correct conclusion is there is no solution. Reporting x = 3 because the simplified equation said so would ignore a condition imposed from the beginning.

This example is useful because the algebra is easy. The real skill is noticing that cancellation produces an expression which is only equivalent to the original when x ≠ 3.

Another equation whose candidate must be rejected

Solve x/(x − 1) = 1/(x − 1) + 2, with x ≠ 1. Multiply by x − 1: x = 1 + 2(x − 1).

Expand: x = 1 + 2x − 2 = 2x − 1, so x = 1. However x = 1 makes both fractions’ denominators zero.

The equation therefore has no solution. The result may feel surprising after a neat line of algebra, but the original condition is decisive.

Worked example 3: quadratic formula when factorisation is awkward

Consider 2/(x − 2) = x, where x ≠ 2. Multiplying through gives 2 = x(x − 2), or x² − 2x − 2 = 0.

The quadratic formula gives x = [2 ± √(4 + 8)]/2 = [2 ± √12]/2 = 1 ± √3. Numerically these are approximately 2.732 and −0.732.

Neither equals the forbidden value 2, so both are algebraically valid. This is why a fractional equation does not need to factorise neatly to be solvable.

How to choose a common denominator safely

Suppose an equation contains 2/x and 1/(x − 1). The common denominator can be x(x − 1), provided x ≠ 0 and x ≠ 1.

Multiplying every term of the equation by this product removes each fraction. Be careful: a term without a denominator must also be multiplied by the full product, not left unchanged.

Writing all four stages — restrictions, common denominator, expanded equation and checked roots — makes the reasoning easier to follow and reduces the temptation to cross-multiply where that shortcut does not fit.

When cross-multiplying is appropriate

If a valid equality has exactly one fraction on each side, such as a/b = c/d with b and d nonzero, then cross-multiplication produces ad = bc.

However, an equation containing 1/x + 2/(x − 1) = 3 is a sum of fractions, not one fraction set equal to another. Write an appropriate common denominator or multiply the entire equation through before simplifying.

The warning is not that cross-multiplication is forbidden. It is that a shortcut must match the structure of the equation and keep all denominator restrictions.

A real-world speed problem that becomes quadratic

Fictional trip. A 120 km journey takes one hour less when a vehicle’s constant speed increases by 20 km/h. Let x be the original speed in km/h, with x greater than zero.

Original journey time is 120/x hours; the faster journey takes 120/(x + 20) hours. Their difference is 1 hour, so 120/x − 120/(x + 20) = 1.

Multiply by x(x + 20), which is nonzero for the stated positive x: 120(x + 20) − 120x = x(x + 20). Simplify to 2400 = x² + 20x, or x² + 20x − 2400 = 0.

Factorise: (x + 60)(x − 40) = 0. The candidates are x = −60 and x = 40. A negative speed is invalid for this stated problem, so the original speed is 40 km/h.

Check the times: 120/40 = 3 hours; 120/60 = 2 hours. The difference is exactly one hour. Both algebra and the story agree.

What is a valid reason for rejecting a root?

There are two main reasons. First, the root might make an original denominator zero, as in the x = 3 cancellation example. Second, a real-world context might require a positive length, an integer count or another specified condition.

A negative algebraic solution is not automatically invalid if the unknown represents something that can be negative, such as a signed coordinate. In the 120 km trip example, negative speed is invalid because x was defined as a positive physical speed.

State the reason for rejection in words or a short mathematical condition. A student who silently removes a root can appear not to understand what the equation allows.

Four stages to write clearly in an exam script

Stage one: restrictions. Write values the original denominators exclude. Stage two: clearing fractions. Multiply every term by a common denominator and show the resulting equation.

Stage three: solve. Expand, rearrange and factorise, or use the quadratic formula. Stage four: verify. Substitute candidates into the original and apply the problem’s context.

An elegant final answer with no essential working may not earn the available method credit under the actual marking scheme. The 2027 K310 examination notes explicitly warn that omitting essential working can lose marks.

A seven-day revision plan for fractional equations

Day 1: list denominator restrictions without solving. Day 2: clear a single fraction to make a quadratic. Day 3: factorise the resulting quadratic and check every root.

Day 4: practise an equation with an excluded candidate. Day 5: use the quadratic formula when factors are not simple. Day 6: translate a journey-time story into fractions. Day 7: solve an unseen mixed question with no worked answer in sight.

An error ledger might classify mistakes as forgotten restriction, missed term when clearing fractions, incorrect factorisation, unchecked candidate or wrong contextual decision. Retesting after a gap shows which principle has actually been learned.

Frequently asked questions about fractional quadratic equations

Why do I list excluded values before clearing denominators? Because multiplication and cancellation can hide restrictions that were present in the original expression.

Can a quadratic equation have two valid roots? Yes, as with 6/x = x − 1; both 3 and −2 satisfy that equation.

Can it have no valid roots? Yes. A derived root may make an original denominator zero.

Should I reject every negative root? No. Reject a root only when it is forbidden by the original equation or by the question’s defined context.

Does the quadratic always factorise into integers? No. Use the quadratic formula when factorisation is not convenient.

Is this part of G3 Mathematics? Yes. The K310 syllabus explicitly includes fractional equations reducible to quadratics and applying equations to problems.

How Bukit Timah three-pupil Mathematics tuition supports this skill

The original eduKateSG Mathematics small-group reference describes premium classes with three pupils for 1.5 hours weekly, carefully sequenced examples, guided correction and home practice. Its stated teaching location is 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT. This guide does not assert another outlet or a vacancy.

A practical lesson starts with the student’s unprompted attempt, finds the first incorrect line, repairs the idea and retests using a changed question without hints. Different pupils can need different corrections even when their paper scores are similar. Progress is independent success on unfamiliar work, not the number of completed worksheets.

Discuss a parent–student Mathematics consultation with eduKate Singapore. A recent independent script, the relevant exam year and specific difficult questions help focus the discussion.

Check the correct examination year

The first Singapore–Cambridge SEC is in 2027, with G3 Mathematics K310. The official SEAB subject listing shows the earlier GCE O-Level Mathematics reference as 4052. Secondary 4 students sitting the 2026 O-Level must use the applicable 2026 syllabus and instructions, not assume the 2027 K310 document describes their own paper.

Secondary 4 G3 Mathematics Tutor Bukit Timah | Fractional Equations: Quadratic Solutions and Restrictions