Sec 3 G3 Maths Tutor Bukit Timah | Fractional Equations

Secondary 3 G3 Mathematics Tutor Bukit Timah | Quadratic Fractional Equations, Restrictions and Checking — Did you know? A strong mathematical explanation makes the first move clear before showing a final answer. This Secondary 3 G3 Mathematics tutor Bukit Timah guide begins with quick information, then carefully develops one distinct skill with worked examples, ways to check, misconceptions and practical next steps for learners and parents.

The goal is not to finish a stack of questions by recognising repeated page layouts. It is to read an unfamiliar problem, decide which relationship applies, perform a legal operation and verify the result. The examples use approachable numbers so readers can see the method rather than get distracted by heavy arithmetic.

For families comparing Secondary 3 G3 Mathematics tuition in Bukit Timah, the useful question is what the student can do independently after a lesson. The Singapore–Cambridge 2027 SEC G3 Mathematics K310 syllabus gives the curriculum framework, while each school determines the timing of teaching within the full secondary course.

Quick answers: the rules worth remembering

Record excluded values from every ORIGINAL denominator before clearing fractions.
Multiply every term by a valid common denominator, keeping complete brackets.
Solve the resulting quadratic by factorisation or the quadratic formula.
Reject candidates that make an original denominator zero, even if they solve the transformed polynomial.
An identity may leave infinitely many allowed values; a forbidden-only candidate may leave none.

On this page: Understand the key ideas · 50 checked examples · Mistakes to avoid · A six-week plan · Frequently asked questions.

Understand the mathematical relationship

Why an algebraic fraction carries a hidden condition

A fraction whose denominator contains x is undefined at every value that makes that original denominator zero. That restriction applies before and after the equation is rearranged. For 3/(x + 2) = x + 4, x = −2 is forbidden from the beginning, no matter what polynomial equation appears after multiplying.

A student should write the excluded values on the first line, then work with the equation only for permitted x. This saves trouble when multiplication or factorisation accidentally hides the denominator. An excluded value cannot become a legitimate answer because its factor later disappears.

Clearing denominators creates a quadratic, not permission

For 3/(x + 2) = x + 4, multiply both sides by x + 2 only with x ≠ −2. The resulting equation 3 = (x + 2)(x + 4) becomes x² + 6x + 5 = 0 and factorises as (x + 1)(x + 5) = 0. The solutions x = −1 and x = −5 satisfy the starting equation.

The multiplication step converts a rational equation into a polynomial one, but it does not erase the original restriction. The polynomial may allow values that the original fractional expression could not accept. Always compare candidate roots with the excluded list and, if practical, check them by direct substitution into the starting fractions.

The common denominator must contain every factor

When a question has 1/(x − 4) + 1/(x − 1), neither denominator can be zero. A common denominator is (x − 4)(x − 1), provided x ≠ 4 and x ≠ 1. Multiplying the entire equation by that product can remove fractions from each term and produce a quadratic.

Students sometimes multiply one term while leaving another unchanged. Write the common denominator over the entire equation and show what remains after each cancellation. Maintaining brackets is essential: a negative coefficient distributed across x − 4 changes both the x-term and constant.

The apparent solution may be forbidden

Consider (x − 2)/(x − 2) = x − 1. If x ≠ 2, the fraction on the left equals one, so the equation would require 1 = x − 1, giving x = 2. But x = 2 is forbidden. The original equation therefore has no solution.

This small example explains why cross-multiplication is not a license to ignore denominators. Multiplying by x − 2 yields x − 2 = (x − 1)(x − 2), and the transformed equation has a root at x = 2. That candidate belongs to the transformed polynomial but not the original fraction equation. Excluding it is not optional.

Sometimes infinitely many values survive

The identity (x² − 4)/(x − 2) = x + 2 is true for every real x except 2, because x² − 4 factorises into (x − 2)(x + 2). A student’s attempt to solve it may yield 0 = 0 after legal simplification. That does not mean every real number is allowed.

The correct answer is all real numbers except the excluded value x = 2. This is the opposite of the no-solution example, and practising both gives students a stronger understanding of what the algebra actually says. A result that simplifies to an identity still carries the original domain restriction.

A radical root is not automatically an error

Not every fractional equation reduces to a quadratic with convenient integer factors. For 1/(x + 1) + 1/(x − 1) = 1, multiplying through yields 2x = x² − 1, or x² − 2x − 1 = 0. The exact roots are 1 ± √2, neither of which makes an original denominator zero.

Keep exact surd form when appropriate. If the question requires approximate values, calculate carefully using the standard accuracy convention or the stated instruction. A neat decimal copied too early may fail a substitution check because of rounding rather than algebra.

The fastest route still needs visible reasoning

A fluent pupil may mentally spot the factorisation immediately. That is useful, but a written solution should expose the denominator restrictions, the equation obtained after clearing, the candidate roots and the final eligibility check. A correct-looking answer without these steps may conceal a forbidden candidate.

The syllabus includes fractional equations reducible to quadratic equations. The mathematical goal is not to practise algebraic fractions endlessly; it is to see when an equation changes form and what conditions remain. A tutor can reduce worksheet volume by choosing one example that exposes each distinct failure mode.

Check the starting equation, not only the new polynomial

If the transformed polynomial is x² + 6x + 5, verifying a root in that polynomial does not independently establish that the fractional equation was valid at that x. The polynomial was derived through multiplication by a variable expression and may be defined at additional values.

The safest final check is two-part. First, confirm no denominator vanishes. Second, substitute the candidate into the original equation and compare the two sides. For a long fraction sum, this last numerical step may be slower, but it makes the logic explicit and helps students recognise when an algebraic candidate is extraneous.

Why a student can follow the tutor but still struggle alone

A typical lesson demonstrates a familiar calculation and the student understands while watching. An unfamiliar question then removes the clue. The issue may not be a total lack of knowledge. It may be difficulty selecting the idea, identifying a necessary condition, interpreting a diagram or checking whether an algebraic answer is eligible. Those are separate problems needing separate corrections.

Use a three-question diagnosis. First, let the student solve a routine exercise from memory. Second, offer a look-alike for which an important detail changes the correct first move. Third, use a contextual question that does not announce the chapter. Ask the student to name the unknown, the governing relationship and the final checking method. Record where assistance was needed, not merely whether the final number matched.

The official G3 Mathematics syllabus emphasises techniques, problem-solving in varied contexts and mathematical reasoning. For longer questions, essential working matters; calculations should be clear enough for another reader to trace. The 2027 syllabus also gives standard reporting conventions for approximate numbers, subject to the specific question’s instructions. Those facts inform the teaching approach, not a promise of any future examination question.

50 worked examples: attempt, explain and check

The problems below are original teaching exercises created to illustrate valid methods. They are not official examination questions or predictions of future SEC papers. Use them as a bank, not a fifty-question sprint. Work on a few suitable examples independently, compare your reasoning with the method, study the check and then revisit a variation after a short delay.

One denominator and an integer-root quadratic

This cluster focuses on one denominator and an integer-root quadratic. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 1. Question: Let t = x − 0. Solve 3/(t + 2) = t + 4, stating excluded x-values. Method: The denominator requires t ≠ −2, so x ≠ -2. Multiply legally: 3 = (t + 2)(t + 4) = t² + 6t + 8. Thus (t + 1)(t + 5) = 0, giving t = −1 or −5 and x = -1 or -5. Check: Both t candidates differ from −2. At t = −1 the two sides equal 3; at t = −5 the sides equal −1. Watch for: A transformed equation is not allowed to restore x = -2, where the original denominator is zero.

Try next: Try shifting the variable by one more unit and predict how both x-roots change. Read the question once for what is given and again for what is sought. Write down whether the answer is a point, a length, an angle, a value of x or a contextual decision. Sketch the key relationship on a smaller, clearer diagram and label only the necessary lengths, positions or quantities.

Two denominators and a common product

This cluster focuses on two denominators and a common product. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 2. Question: Let t = x − 0. Solve 1/(t − 4) + 1/(t − 1) = 1/2. Method: First exclude t = 4 and 1, or x = 4 and 1. Multiply by 2(t − 4)(t − 1): 2[(t − 1)+(t − 4)] = (t − 4)(t − 1). Simplify to t² − 9t + 14 = 0 = (t − 2)(t − 7), so x = 2 or 7. Check: At t = 2, −1/2 + 1 = 1/2; at t = 7, 1/3 + 1/6 = 1/2. Both are allowed. Watch for: Do not add the denominators directly or forget that two values are excluded.

Try next: Write the full common denominator before distributing the multiplier. Cover the method after reading it, then reproduce the first step without help. A student who only recognises the finished working has not yet demonstrated independent retrieval. Explain the same result using an independent check rather than merely copying the original calculation in reverse.

A numerator expression gives surd roots

This cluster focuses on a numerator expression gives surd roots. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 3. Question: Let t = x − 0. Solve (t + 1)/(t − 2) = t − 1 exactly. Method: Exclude t = 2, or x = 2. Multiply by t − 2: t + 1 = (t − 2)(t − 1) = t² − 3t + 2. Therefore t² − 4t + 1 = 0, giving t = 2 ± √3. Thus x = 2 ± √3. Check: Neither t = 2 + √3 nor t = 2 − √3 equals 2, and both satisfy the original equation. Watch for: An exact surd root is not automatically an error; do not round before deciding eligibility.

Try next: Estimate both roots to three significant figures and compare with the excluded x-value. Explain in one sentence why the chosen operation is allowed. Use the relevant geometry, algebra or units rather than the words ‘because it is the formula’. Record the first possible careless error and the exact wording of a self-check that would have prevented it.

Solve a rational equation with two integer roots

This cluster focuses on solve a rational equation with two integer roots. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 4. Question: Let t = x − 0. Solve 2/(t − 3) = t − 2. Method: First t ≠ 3 (so x ≠ 3). Multiply: 2 = (t − 3)(t − 2) = t² − 5t + 6. Rearranging gives t² − 5t + 4 = (t − 1)(t − 4) = 0, so x = 1 or 4. Check: At t = 1, both sides equal −1; at t = 4, both sides equal 2. Neither makes the denominator vanish. Watch for: Clearing the denominator does not mean t = 3 becomes available.

Try next: Find why the factor pair 1 and 4 matches the transformed coefficients. Return to this example after two days. Keep one dated first attempt and one unprompted second attempt so that progress becomes visible in the actual working. Replace one value and decide whether the first operation remains unchanged. If it changes, say which assumption was responsible.

Fractions with a linear numerator

This cluster focuses on fractions with a linear numerator. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 5. Question: Let t = x − 0. Solve (t + 5)/(t − 1) = t + 1. Method: Exclude t = 1, or x = 1. Multiply: t + 5 = (t − 1)(t + 1) = t² − 1. Hence t² − t − 6 = (t − 3)(t + 2) = 0 and x = 3 or -2. Check: At t = 3, the fraction 8/2 = 4 equals t + 1; at t = −2, 3/(−3) = −1 equals t + 1. Watch for: Both roots must be checked against the original denominator t − 1, not a cancelled expression.

Try next: Sketch why neither result lies at the excluded value x = 1. Identify a plausible wrong approach and name the condition it violates. Recognising the limits of a rule is as valuable as remembering the rule itself. Give the full final answer as a sentence, including relevant units and any exclusions or interpretation.

An identity with one excluded value

This cluster focuses on an identity with one excluded value. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 6. Question: Let t = x − 0. For which real x is (t² − 4)/(t − 2) = t + 2 true? Method: The denominator requires t ≠ 2. Factor the numerator: (t − 2)(t + 2). For all permitted t, the left side equals t + 2, so the equation is true for every real x except x = 2. Check: At any t other than 2, cancellation is lawful. At t = 2, the original expression would divide by zero. Watch for: Do not answer ‘every real x’ without the forbidden value.

Try next: Choose three permitted x-values and check the identity numerically. Check the sign, unit and approximate size of the result. A sensible estimate can expose a wrong interpretation without repeating every calculator step. Create a comparable question with different numbers that a partner can solve without your guidance.

A candidate that must be rejected

This cluster focuses on a candidate that must be rejected. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 7. Question: Let t = x − 0. Solve (t − 2)/(t − 2) = t − 1, including all domain restrictions. Method: The starting denominator requires t ≠ 2, or x ≠ 2. For permitted t the left side is 1, so 1 = t − 1 implies t = 2. Since that is forbidden, the original equation has no solution. Check: There is no permitted t for which t − 1 equals one and the left fraction is defined. Watch for: Multiplying by t − 2 creates an algebraic candidate at the excluded value; it does not make it valid.

Try next: Explain why writing 0 = 0 at an intermediate step would not rescue t = 2. Change one given number, predict the direction of the answer’s change and only then calculate. This tests whether the structure of the problem has been learned. State whether this is exact reasoning, a model-dependent estimate or a numerical approximation.

Another shifted quadratic fraction equation

This cluster focuses on another shifted quadratic fraction equation. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 8. Question: Let t = x − 0. Solve 6/(t + 4) = t + 3. Method: Exclude t = −4, or x = -4. Multiply: 6 = (t + 4)(t + 3) = t² + 7t + 12. Thus t² + 7t + 6 = (t + 1)(t + 6) = 0, giving x = -1 or -6. Check: At t = −1 both sides equal 2; at t = −6 both sides equal −3. The denominators are nonzero. Watch for: Rearrange the quadratic carefully: 12 − 6 = 6, not 18.

Try next: Compare the starting and transformed equations to see why the same roots remain valid. Ask a classmate to interpret the answer in words. A technically correct numerical value may still be incomplete if it does not answer the story or diagram. Ask if every candidate meets the original conditions or whether a domain or physical constraint removes one.

Two reciprocals leading to irrational roots

This cluster focuses on two reciprocals leading to irrational roots. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 9. Question: Let t = x − 0. Solve 1/(t + 1) + 1/(t − 1) = 1 exactly. Method: Exclude t = −1 and t = 1. Multiply by (t + 1)(t − 1): (t − 1)+(t + 1)=t² − 1. Then t² − 2t − 1 = 0, so t = 1 ± √2 and x = 1 ± √2. Check: Neither candidate equals 1 or −1. Substituting back or using the exact quadratic confirms the equality. Watch for: Do not cancel the two denominators against separate numerators before combining the fractions.

Try next: Explain why the surd expressions must be retained exactly if the question asks for exact roots. Trace the final result backwards toward the original condition. If the result fails an original equation, construction or real-world restriction, the calculation is not finished. Link the problem to a neighbouring chapter and explain how the mathematical techniques support each other.

Check discriminant and restrictions together

This cluster focuses on check discriminant and restrictions together. The useful first question is, ‘What does the given information establish?’ Keep the numerical work separate from the explanation for choosing it. Later, mix examples from different clusters so the method must be selected without a worksheet heading giving it away.

Worked example 10. Question: Let t = x − 0. Solve 2/(t + 1) = t + 3. Method: Exclude t = −1. Multiply: 2 = (t + 1)(t + 3) = t² + 4t + 3, so t² + 4t + 1 = 0. The formula gives t = −2 ± √3, or x = -2 ± √3. Check: Neither root equals t = −1, so both are permitted. Their substitution satisfies the cleared quadratic and the original equality. Watch for: Do not round the roots before checking whether they coincide with a forbidden input.

Try next: Find the discriminant 4²−4×1×1 and explain why it gives two distinct real roots. Mix this problem with one from a different cluster so the heading no longer announces the method. Accurate selection is important in unseen assessments. Spend a minute identifying what the question did not specify, so that no extra assumption is slipped into the solution.

One denominator and an integer-root quadratic

Worked example 11. Question: Let t = x − 1. Solve 3/(t + 2) = t + 4, stating excluded x-values. Method: The denominator requires t ≠ −2, so x ≠ -1. Multiply legally: 3 = (t + 2)(t + 4) = t² + 6t + 8. Thus (t + 1)(t + 5) = 0, giving t = −1 or −5 and x = 0 or -4. Check: Both t candidates differ from −2. At t = −1 the two sides equal 3; at t = −5 the sides equal −1. Watch for: A transformed equation is not allowed to restore x = -1, where the original denominator is zero.

Two denominators and a common product

Worked example 12. Question: Let t = x − 1. Solve 1/(t − 4) + 1/(t − 1) = 1/2. Method: First exclude t = 4 and 1, or x = 5 and 2. Multiply by 2(t − 4)(t − 1): 2[(t − 1)+(t − 4)] = (t − 4)(t − 1). Simplify to t² − 9t + 14 = 0 = (t − 2)(t − 7), so x = 3 or 8. Check: At t = 2, −1/2 + 1 = 1/2; at t = 7, 1/3 + 1/6 = 1/2. Both are allowed. Watch for: Do not add the denominators directly or forget that two values are excluded.

A numerator expression gives surd roots

Worked example 13. Question: Let t = x − 1. Solve (t + 1)/(t − 2) = t − 1 exactly. Method: Exclude t = 2, or x = 3. Multiply by t − 2: t + 1 = (t − 2)(t − 1) = t² − 3t + 2. Therefore t² − 4t + 1 = 0, giving t = 2 ± √3. Thus x = 3 ± √3. Check: Neither t = 2 + √3 nor t = 2 − √3 equals 2, and both satisfy the original equation. Watch for: An exact surd root is not automatically an error; do not round before deciding eligibility.

Solve a rational equation with two integer roots

Worked example 14. Question: Let t = x − 1. Solve 2/(t − 3) = t − 2. Method: First t ≠ 3 (so x ≠ 4). Multiply: 2 = (t − 3)(t − 2) = t² − 5t + 6. Rearranging gives t² − 5t + 4 = (t − 1)(t − 4) = 0, so x = 2 or 5. Check: At t = 1, both sides equal −1; at t = 4, both sides equal 2. Neither makes the denominator vanish. Watch for: Clearing the denominator does not mean t = 3 becomes available.

Fractions with a linear numerator

Worked example 15. Question: Let t = x − 1. Solve (t + 5)/(t − 1) = t + 1. Method: Exclude t = 1, or x = 2. Multiply: t + 5 = (t − 1)(t + 1) = t² − 1. Hence t² − t − 6 = (t − 3)(t + 2) = 0 and x = 4 or -1. Check: At t = 3, the fraction 8/2 = 4 equals t + 1; at t = −2, 3/(−3) = −1 equals t + 1. Watch for: Both roots must be checked against the original denominator t − 1, not a cancelled expression.

Try next: Sketch why neither result lies at the excluded value x = 2. Identify a plausible wrong approach and name the condition it violates. Recognising the limits of a rule is as valuable as remembering the rule itself. Give the full final answer as a sentence, including relevant units and any exclusions or interpretation.

An identity with one excluded value

Worked example 16. Question: Let t = x − 1. For which real x is (t² − 4)/(t − 2) = t + 2 true? Method: The denominator requires t ≠ 2. Factor the numerator: (t − 2)(t + 2). For all permitted t, the left side equals t + 2, so the equation is true for every real x except x = 3. Check: At any t other than 2, cancellation is lawful. At t = 2, the original expression would divide by zero. Watch for: Do not answer ‘every real x’ without the forbidden value.

A candidate that must be rejected

Worked example 17. Question: Let t = x − 1. Solve (t − 2)/(t − 2) = t − 1, including all domain restrictions. Method: The starting denominator requires t ≠ 2, or x ≠ 3. For permitted t the left side is 1, so 1 = t − 1 implies t = 2. Since that is forbidden, the original equation has no solution. Check: There is no permitted t for which t − 1 equals one and the left fraction is defined. Watch for: Multiplying by t − 2 creates an algebraic candidate at the excluded value; it does not make it valid.

Another shifted quadratic fraction equation

Worked example 18. Question: Let t = x − 1. Solve 6/(t + 4) = t + 3. Method: Exclude t = −4, or x = -3. Multiply: 6 = (t + 4)(t + 3) = t² + 7t + 12. Thus t² + 7t + 6 = (t + 1)(t + 6) = 0, giving x = 0 or -5. Check: At t = −1 both sides equal 2; at t = −6 both sides equal −3. The denominators are nonzero. Watch for: Rearrange the quadratic carefully: 12 − 6 = 6, not 18.

Two reciprocals leading to irrational roots

Worked example 19. Question: Let t = x − 1. Solve 1/(t + 1) + 1/(t − 1) = 1 exactly. Method: Exclude t = −1 and t = 1. Multiply by (t + 1)(t − 1): (t − 1)+(t + 1)=t² − 1. Then t² − 2t − 1 = 0, so t = 1 ± √2 and x = 2 ± √2. Check: Neither candidate equals 1 or −1. Substituting back or using the exact quadratic confirms the equality. Watch for: Do not cancel the two denominators against separate numerators before combining the fractions.

Check discriminant and restrictions together

Worked example 20. Question: Let t = x − 1. Solve 2/(t + 1) = t + 3. Method: Exclude t = −1. Multiply: 2 = (t + 1)(t + 3) = t² + 4t + 3, so t² + 4t + 1 = 0. The formula gives t = −2 ± √3, or x = -1 ± √3. Check: Neither root equals t = −1, so both are permitted. Their substitution satisfies the cleared quadratic and the original equality. Watch for: Do not round the roots before checking whether they coincide with a forbidden input.

One denominator and an integer-root quadratic

Worked example 21. Question: Let t = x − 2. Solve 3/(t + 2) = t + 4, stating excluded x-values. Method: The denominator requires t ≠ −2, so x ≠ 0. Multiply legally: 3 = (t + 2)(t + 4) = t² + 6t + 8. Thus (t + 1)(t + 5) = 0, giving t = −1 or −5 and x = 1 or -3. Check: Both t candidates differ from −2. At t = −1 the two sides equal 3; at t = −5 the sides equal −1. Watch for: A transformed equation is not allowed to restore x = 0, where the original denominator is zero.

Two denominators and a common product

Worked example 22. Question: Let t = x − 2. Solve 1/(t − 4) + 1/(t − 1) = 1/2. Method: First exclude t = 4 and 1, or x = 6 and 3. Multiply by 2(t − 4)(t − 1): 2[(t − 1)+(t − 4)] = (t − 4)(t − 1). Simplify to t² − 9t + 14 = 0 = (t − 2)(t − 7), so x = 4 or 9. Check: At t = 2, −1/2 + 1 = 1/2; at t = 7, 1/3 + 1/6 = 1/2. Both are allowed. Watch for: Do not add the denominators directly or forget that two values are excluded.

A numerator expression gives surd roots

Worked example 23. Question: Let t = x − 2. Solve (t + 1)/(t − 2) = t − 1 exactly. Method: Exclude t = 2, or x = 4. Multiply by t − 2: t + 1 = (t − 2)(t − 1) = t² − 3t + 2. Therefore t² − 4t + 1 = 0, giving t = 2 ± √3. Thus x = 4 ± √3. Check: Neither t = 2 + √3 nor t = 2 − √3 equals 2, and both satisfy the original equation. Watch for: An exact surd root is not automatically an error; do not round before deciding eligibility.

Solve a rational equation with two integer roots

Worked example 24. Question: Let t = x − 2. Solve 2/(t − 3) = t − 2. Method: First t ≠ 3 (so x ≠ 5). Multiply: 2 = (t − 3)(t − 2) = t² − 5t + 6. Rearranging gives t² − 5t + 4 = (t − 1)(t − 4) = 0, so x = 3 or 6. Check: At t = 1, both sides equal −1; at t = 4, both sides equal 2. Neither makes the denominator vanish. Watch for: Clearing the denominator does not mean t = 3 becomes available.

Fractions with a linear numerator

Worked example 25. Question: Let t = x − 2. Solve (t + 5)/(t − 1) = t + 1. Method: Exclude t = 1, or x = 3. Multiply: t + 5 = (t − 1)(t + 1) = t² − 1. Hence t² − t − 6 = (t − 3)(t + 2) = 0 and x = 5 or 0. Check: At t = 3, the fraction 8/2 = 4 equals t + 1; at t = −2, 3/(−3) = −1 equals t + 1. Watch for: Both roots must be checked against the original denominator t − 1, not a cancelled expression.

Try next: Sketch why neither result lies at the excluded value x = 3. Identify a plausible wrong approach and name the condition it violates. Recognising the limits of a rule is as valuable as remembering the rule itself. Give the full final answer as a sentence, including relevant units and any exclusions or interpretation.

An identity with one excluded value

Worked example 26. Question: Let t = x − 2. For which real x is (t² − 4)/(t − 2) = t + 2 true? Method: The denominator requires t ≠ 2. Factor the numerator: (t − 2)(t + 2). For all permitted t, the left side equals t + 2, so the equation is true for every real x except x = 4. Check: At any t other than 2, cancellation is lawful. At t = 2, the original expression would divide by zero. Watch for: Do not answer ‘every real x’ without the forbidden value.

A candidate that must be rejected

Worked example 27. Question: Let t = x − 2. Solve (t − 2)/(t − 2) = t − 1, including all domain restrictions. Method: The starting denominator requires t ≠ 2, or x ≠ 4. For permitted t the left side is 1, so 1 = t − 1 implies t = 2. Since that is forbidden, the original equation has no solution. Check: There is no permitted t for which t − 1 equals one and the left fraction is defined. Watch for: Multiplying by t − 2 creates an algebraic candidate at the excluded value; it does not make it valid.

Another shifted quadratic fraction equation

Worked example 28. Question: Let t = x − 2. Solve 6/(t + 4) = t + 3. Method: Exclude t = −4, or x = -2. Multiply: 6 = (t + 4)(t + 3) = t² + 7t + 12. Thus t² + 7t + 6 = (t + 1)(t + 6) = 0, giving x = 1 or -4. Check: At t = −1 both sides equal 2; at t = −6 both sides equal −3. The denominators are nonzero. Watch for: Rearrange the quadratic carefully: 12 − 6 = 6, not 18.

Two reciprocals leading to irrational roots

Worked example 29. Question: Let t = x − 2. Solve 1/(t + 1) + 1/(t − 1) = 1 exactly. Method: Exclude t = −1 and t = 1. Multiply by (t + 1)(t − 1): (t − 1)+(t + 1)=t² − 1. Then t² − 2t − 1 = 0, so t = 1 ± √2 and x = 3 ± √2. Check: Neither candidate equals 1 or −1. Substituting back or using the exact quadratic confirms the equality. Watch for: Do not cancel the two denominators against separate numerators before combining the fractions.

Check discriminant and restrictions together

Worked example 30. Question: Let t = x − 2. Solve 2/(t + 1) = t + 3. Method: Exclude t = −1. Multiply: 2 = (t + 1)(t + 3) = t² + 4t + 3, so t² + 4t + 1 = 0. The formula gives t = −2 ± √3, or x = 0 ± √3. Check: Neither root equals t = −1, so both are permitted. Their substitution satisfies the cleared quadratic and the original equality. Watch for: Do not round the roots before checking whether they coincide with a forbidden input.

One denominator and an integer-root quadratic

Worked example 31. Question: Let t = x − 3. Solve 3/(t + 2) = t + 4, stating excluded x-values. Method: The denominator requires t ≠ −2, so x ≠ 1. Multiply legally: 3 = (t + 2)(t + 4) = t² + 6t + 8. Thus (t + 1)(t + 5) = 0, giving t = −1 or −5 and x = 2 or -2. Check: Both t candidates differ from −2. At t = −1 the two sides equal 3; at t = −5 the sides equal −1. Watch for: A transformed equation is not allowed to restore x = 1, where the original denominator is zero.

Two denominators and a common product

Worked example 32. Question: Let t = x − 3. Solve 1/(t − 4) + 1/(t − 1) = 1/2. Method: First exclude t = 4 and 1, or x = 7 and 4. Multiply by 2(t − 4)(t − 1): 2[(t − 1)+(t − 4)] = (t − 4)(t − 1). Simplify to t² − 9t + 14 = 0 = (t − 2)(t − 7), so x = 5 or 10. Check: At t = 2, −1/2 + 1 = 1/2; at t = 7, 1/3 + 1/6 = 1/2. Both are allowed. Watch for: Do not add the denominators directly or forget that two values are excluded.

A numerator expression gives surd roots

Worked example 33. Question: Let t = x − 3. Solve (t + 1)/(t − 2) = t − 1 exactly. Method: Exclude t = 2, or x = 5. Multiply by t − 2: t + 1 = (t − 2)(t − 1) = t² − 3t + 2. Therefore t² − 4t + 1 = 0, giving t = 2 ± √3. Thus x = 5 ± √3. Check: Neither t = 2 + √3 nor t = 2 − √3 equals 2, and both satisfy the original equation. Watch for: An exact surd root is not automatically an error; do not round before deciding eligibility.

Solve a rational equation with two integer roots

Worked example 34. Question: Let t = x − 3. Solve 2/(t − 3) = t − 2. Method: First t ≠ 3 (so x ≠ 6). Multiply: 2 = (t − 3)(t − 2) = t² − 5t + 6. Rearranging gives t² − 5t + 4 = (t − 1)(t − 4) = 0, so x = 4 or 7. Check: At t = 1, both sides equal −1; at t = 4, both sides equal 2. Neither makes the denominator vanish. Watch for: Clearing the denominator does not mean t = 3 becomes available.

Fractions with a linear numerator

Worked example 35. Question: Let t = x − 3. Solve (t + 5)/(t − 1) = t + 1. Method: Exclude t = 1, or x = 4. Multiply: t + 5 = (t − 1)(t + 1) = t² − 1. Hence t² − t − 6 = (t − 3)(t + 2) = 0 and x = 6 or 1. Check: At t = 3, the fraction 8/2 = 4 equals t + 1; at t = −2, 3/(−3) = −1 equals t + 1. Watch for: Both roots must be checked against the original denominator t − 1, not a cancelled expression.

Try next: Sketch why neither result lies at the excluded value x = 4. Identify a plausible wrong approach and name the condition it violates. Recognising the limits of a rule is as valuable as remembering the rule itself. Give the full final answer as a sentence, including relevant units and any exclusions or interpretation.

An identity with one excluded value

Worked example 36. Question: Let t = x − 3. For which real x is (t² − 4)/(t − 2) = t + 2 true? Method: The denominator requires t ≠ 2. Factor the numerator: (t − 2)(t + 2). For all permitted t, the left side equals t + 2, so the equation is true for every real x except x = 5. Check: At any t other than 2, cancellation is lawful. At t = 2, the original expression would divide by zero. Watch for: Do not answer ‘every real x’ without the forbidden value.

A candidate that must be rejected

Worked example 37. Question: Let t = x − 3. Solve (t − 2)/(t − 2) = t − 1, including all domain restrictions. Method: The starting denominator requires t ≠ 2, or x ≠ 5. For permitted t the left side is 1, so 1 = t − 1 implies t = 2. Since that is forbidden, the original equation has no solution. Check: There is no permitted t for which t − 1 equals one and the left fraction is defined. Watch for: Multiplying by t − 2 creates an algebraic candidate at the excluded value; it does not make it valid.

Another shifted quadratic fraction equation

Worked example 38. Question: Let t = x − 3. Solve 6/(t + 4) = t + 3. Method: Exclude t = −4, or x = -1. Multiply: 6 = (t + 4)(t + 3) = t² + 7t + 12. Thus t² + 7t + 6 = (t + 1)(t + 6) = 0, giving x = 2 or -3. Check: At t = −1 both sides equal 2; at t = −6 both sides equal −3. The denominators are nonzero. Watch for: Rearrange the quadratic carefully: 12 − 6 = 6, not 18.

Two reciprocals leading to irrational roots

Worked example 39. Question: Let t = x − 3. Solve 1/(t + 1) + 1/(t − 1) = 1 exactly. Method: Exclude t = −1 and t = 1. Multiply by (t + 1)(t − 1): (t − 1)+(t + 1)=t² − 1. Then t² − 2t − 1 = 0, so t = 1 ± √2 and x = 4 ± √2. Check: Neither candidate equals 1 or −1. Substituting back or using the exact quadratic confirms the equality. Watch for: Do not cancel the two denominators against separate numerators before combining the fractions.

Check discriminant and restrictions together

Worked example 40. Question: Let t = x − 3. Solve 2/(t + 1) = t + 3. Method: Exclude t = −1. Multiply: 2 = (t + 1)(t + 3) = t² + 4t + 3, so t² + 4t + 1 = 0. The formula gives t = −2 ± √3, or x = 1 ± √3. Check: Neither root equals t = −1, so both are permitted. Their substitution satisfies the cleared quadratic and the original equality. Watch for: Do not round the roots before checking whether they coincide with a forbidden input.

One denominator and an integer-root quadratic

Worked example 41. Question: Let t = x − 4. Solve 3/(t + 2) = t + 4, stating excluded x-values. Method: The denominator requires t ≠ −2, so x ≠ 2. Multiply legally: 3 = (t + 2)(t + 4) = t² + 6t + 8. Thus (t + 1)(t + 5) = 0, giving t = −1 or −5 and x = 3 or -1. Check: Both t candidates differ from −2. At t = −1 the two sides equal 3; at t = −5 the sides equal −1. Watch for: A transformed equation is not allowed to restore x = 2, where the original denominator is zero.

Two denominators and a common product

Worked example 42. Question: Let t = x − 4. Solve 1/(t − 4) + 1/(t − 1) = 1/2. Method: First exclude t = 4 and 1, or x = 8 and 5. Multiply by 2(t − 4)(t − 1): 2[(t − 1)+(t − 4)] = (t − 4)(t − 1). Simplify to t² − 9t + 14 = 0 = (t − 2)(t − 7), so x = 6 or 11. Check: At t = 2, −1/2 + 1 = 1/2; at t = 7, 1/3 + 1/6 = 1/2. Both are allowed. Watch for: Do not add the denominators directly or forget that two values are excluded.

A numerator expression gives surd roots

Worked example 43. Question: Let t = x − 4. Solve (t + 1)/(t − 2) = t − 1 exactly. Method: Exclude t = 2, or x = 6. Multiply by t − 2: t + 1 = (t − 2)(t − 1) = t² − 3t + 2. Therefore t² − 4t + 1 = 0, giving t = 2 ± √3. Thus x = 6 ± √3. Check: Neither t = 2 + √3 nor t = 2 − √3 equals 2, and both satisfy the original equation. Watch for: An exact surd root is not automatically an error; do not round before deciding eligibility.

Solve a rational equation with two integer roots

Worked example 44. Question: Let t = x − 4. Solve 2/(t − 3) = t − 2. Method: First t ≠ 3 (so x ≠ 7). Multiply: 2 = (t − 3)(t − 2) = t² − 5t + 6. Rearranging gives t² − 5t + 4 = (t − 1)(t − 4) = 0, so x = 5 or 8. Check: At t = 1, both sides equal −1; at t = 4, both sides equal 2. Neither makes the denominator vanish. Watch for: Clearing the denominator does not mean t = 3 becomes available.

Fractions with a linear numerator

Worked example 45. Question: Let t = x − 4. Solve (t + 5)/(t − 1) = t + 1. Method: Exclude t = 1, or x = 5. Multiply: t + 5 = (t − 1)(t + 1) = t² − 1. Hence t² − t − 6 = (t − 3)(t + 2) = 0 and x = 7 or 2. Check: At t = 3, the fraction 8/2 = 4 equals t + 1; at t = −2, 3/(−3) = −1 equals t + 1. Watch for: Both roots must be checked against the original denominator t − 1, not a cancelled expression.

Try next: Sketch why neither result lies at the excluded value x = 5. Identify a plausible wrong approach and name the condition it violates. Recognising the limits of a rule is as valuable as remembering the rule itself. Give the full final answer as a sentence, including relevant units and any exclusions or interpretation.

An identity with one excluded value

Worked example 46. Question: Let t = x − 4. For which real x is (t² − 4)/(t − 2) = t + 2 true? Method: The denominator requires t ≠ 2. Factor the numerator: (t − 2)(t + 2). For all permitted t, the left side equals t + 2, so the equation is true for every real x except x = 6. Check: At any t other than 2, cancellation is lawful. At t = 2, the original expression would divide by zero. Watch for: Do not answer ‘every real x’ without the forbidden value.

A candidate that must be rejected

Worked example 47. Question: Let t = x − 4. Solve (t − 2)/(t − 2) = t − 1, including all domain restrictions. Method: The starting denominator requires t ≠ 2, or x ≠ 6. For permitted t the left side is 1, so 1 = t − 1 implies t = 2. Since that is forbidden, the original equation has no solution. Check: There is no permitted t for which t − 1 equals one and the left fraction is defined. Watch for: Multiplying by t − 2 creates an algebraic candidate at the excluded value; it does not make it valid.

Another shifted quadratic fraction equation

Worked example 48. Question: Let t = x − 4. Solve 6/(t + 4) = t + 3. Method: Exclude t = −4, or x = 0. Multiply: 6 = (t + 4)(t + 3) = t² + 7t + 12. Thus t² + 7t + 6 = (t + 1)(t + 6) = 0, giving x = 3 or -2. Check: At t = −1 both sides equal 2; at t = −6 both sides equal −3. The denominators are nonzero. Watch for: Rearrange the quadratic carefully: 12 − 6 = 6, not 18.

Two reciprocals leading to irrational roots

Worked example 49. Question: Let t = x − 4. Solve 1/(t + 1) + 1/(t − 1) = 1 exactly. Method: Exclude t = −1 and t = 1. Multiply by (t + 1)(t − 1): (t − 1)+(t + 1)=t² − 1. Then t² − 2t − 1 = 0, so t = 1 ± √2 and x = 5 ± √2. Check: Neither candidate equals 1 or −1. Substituting back or using the exact quadratic confirms the equality. Watch for: Do not cancel the two denominators against separate numerators before combining the fractions.

Check discriminant and restrictions together

Worked example 50. Question: Let t = x − 4. Solve 2/(t + 1) = t + 3. Method: Exclude t = −1. Multiply: 2 = (t + 1)(t + 3) = t² + 4t + 3, so t² + 4t + 1 = 0. The formula gives t = −2 ± √3, or x = 2 ± √3. Check: Neither root equals t = −1, so both are permitted. Their substitution satisfies the cleared quadratic and the original equality. Watch for: Do not round the roots before checking whether they coincide with a forbidden input.

Five errors that deserve a targeted correction

1. Dropping the original denominator restrictions after cancellation.

Find the earliest line where the learner’s working stopped matching the definition, diagram or original condition. Repair that line, not merely the final answer. The student should then describe why the earlier step failed and solve one new variation unaided. A short error log with cause and remedy creates a check the student can reuse on a mixed paper.

2. Multiplying only selected terms when clearing fractions.

3. Expanding a bracket incorrectly and obtaining a different quadratic.

4. Treating a transformed-polynomial root as automatically valid.

5. Interpreting 0=0 as permission to include forbidden denominator values.

How a small-group Bukit Timah Maths tutorial can use this guide

A useful tutor begins with a recent marked school exercise, watches an independent attempt and diagnoses the precise decision that failed. The tutor can model one worked problem, ask the student to explain the underlying relationship, then withdraw hints on the next problem. This cycle makes clear whether the student has merely copied a method or can retrieve it.

In a small group, every student still needs individual written attempts and correction. Class size does not automatically guarantee progress. Families should check that the chapter matches the student’s actual G3 level and school schedule, and ask how feedback, independent practice and retrieval after a delay are handled. The quality of thinking matters more than the number of pages finished.

See the year-specific Secondary 3 G3 Mathematics tuition Bukit Timah overview for broader local context. The unchanged Secondary 1 Mathematics Tutor Clementi small-group resource illustrates foundational teaching at a different year level; it is not itself the Secondary 3 syllabus. Neither page has been altered to produce this guide.

A practical six-week repair-and-build plan

Week 1 — Diagnose, do not guess

Gather two marked school questions and one fresh attempt. Identify whether the problem lies in definitions, interpretation, method selection, algebra, arithmetic, units or final checking. Choose one small starting skill with an observable outcome. The focal skill here is quadratic fractional equations, restrictions and checking; one dated first attempt and one later unseen question provide more useful evidence than simply reporting time spent studying.

Week 2 — Rebuild the prerequisite

Explain a governing fact in ordinary words, then connect it to a worked example. The learner should solve one variant after the model is hidden. If a basic definition cannot be explained, repair it before increasing the difficulty. The focal skill here is quadratic fractional equations, restrictions and checking; one dated first attempt and one later unseen question provide more useful evidence than simply reporting time spent studying.

Week 3 — Contrast closely related examples

Put two exercises side by side. Ask why one method is valid for the first but not the other. Require the learner to name the differing condition before any number is entered into a calculator. The focal skill here is quadratic fractional equations, restrictions and checking; one dated first attempt and one later unseen question provide more useful evidence than simply reporting time spent studying.

Week 4 — Train an independent check

Use an equality substitution, reversed calculation, geometric reason, unit check or context constraint. The checking route should be distinct from the main calculation and make a specific error easier to catch. The focal skill here is quadratic fractional equations, restrictions and checking; one dated first attempt and one later unseen question provide more useful evidence than simply reporting time spent studying.

Week 5 — Retrieve after a delay

Mix the topic with a different chapter in a short independent set. Keep the first step unprompted. Record which errors return after several days, and rebuild those decisions without repeating every worksheet from the beginning. The focal skill here is quadratic fractional equations, restrictions and checking; one dated first attempt and one later unseen question provide more useful evidence than simply reporting time spent studying.

Week 6 — Retest transfer

Give a new set with the same mathematical demand but different numbers, story or diagram orientation. Compare correct method choice, written reasoning and self-checking with Week 1, then select the next two learning goals. The focal skill here is quadratic fractional equations, restrictions and checking; one dated first attempt and one later unseen question provide more useful evidence than simply reporting time spent studying.

What parents can notice without solving the homework

Ask the student to tell you what the question wants, which fact justifies the opening step and how the answer can be checked. A calm, intelligible explanation is a useful sign of learning even before a school assessment mark changes. If the pupil can only answer when the chapter heading or a worked model is visible, the next target is independent method selection.

Parents do not need to introduce extra advanced techniques to see progress. Compare the child’s working across two different days. Watch for more accurate labels, fewer unexplained leaps, sensible units and the ability to recover after spotting a mistake. When evaluating tutoring, ask for the student’s own corrected examples and how the tutor plans the next lesson around them.

Frequently asked questions

Why write excluded values first?

Multiplying away denominators can hide values where the original equation was undefined. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

Can a quadratic candidate be wrong after correct factorisation?

Yes, if the candidate zeros an original denominator. That root is extraneous for the starting fraction equation. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

Is cross-multiplying always safe?

Only when denominator restrictions are preserved and the multiplication is justified for the permitted domain. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

What does an answer of no solution look like?

It can occur when all algebraic candidates are excluded from the original equation. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

Can an equation have infinitely many solutions except one value?

Yes. An identity such as (x²−4)/(x−2)=x+2 holds for every real x except 2. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

Are surd roots permitted?

An exact radical root can be valid if it satisfies the equation and the domain restrictions. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

What is a good final check?

Inspect the original denominator, then substitute each allowed candidate into the original equality. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

How is this different from just simplifying an algebraic fraction?

You are solving an equation whose transformed polynomial may have candidates requiring separate eligibility checks. One simple next step is to try a fresh question, explain the choice of method and compare the result with the original conditions.

Official references: SEAB G3 syllabus listing and 2027 SEC G3 Mathematics syllabus K310. Current school chapter sequences may differ; check the student’s actual class programme and assessment instructions before selecting revision tasks.

The takeaway

Did you know? The most useful progress is often visible in the opening decision rather than the final answer. A student starts to identify the mathematical structure, select a valid operation, keep the working readable and check the result without being led through each line. Build that habit one misconception at a time, and quadratic fractional equations, restrictions and checking becomes a clearer part of the wider Secondary 3 G3 Mathematics journey.