Sec 2 G2 Maths Bukit Timah | Fraction Equations

Did you know that a fractional equation becomes much easier when you multiply every term on both sides by the same suitable number? For x/2+1/3=x/3+1/2, the least common multiple of the denominators is six. That gives 3x+2=2x+3, so x=1. Nothing mysterious happened: the equality stayed balanced and the fractions became whole coefficients.

The difficulty for many Secondary 2 G2 Mathematics students is not fractions themselves. It is forgetting that a multiplier affects every term, dropping a sign from a bracket, or deciding that a fraction answer cannot be correct. For families looking for a Secondary 2 G2 Mathematics tutor in Bukit Timah, the best diagnostic is often the first line after the denominators are cleared.

This guide develops numeric-denominator equations with fractions on both sides, exact checks, contradictions and identities, with short rational-equation extensions where appropriate. Each worked example is original educational material and should be matched to the student’s present school assignments rather than treated as an examination prediction.

Quick answer: the four-step method

Step	What to do
1. Read the fractions	Identify every complete numerator and numeric denominator.
2. Choose a multiplier	Find a nonzero common multiple, preferably the least common multiple.
3. Apply it everywhere	Multiply every additive term on both sides, keeping brackets.
4. Solve and verify	Collect x, solve or classify the equality, then check the original equation.
If x terms cancel and constants disagree	There is no solution.
If both sides become identical	Every permitted x satisfies the equation.
If x appears in the denominator	Record excluded values before clearing fractions.

How fractional equations fit Secondary 2 G2 Mathematics

G2 Mathematics develops algebraic manipulation, equations, fractions and problem solving. Multi-step fractional equations build upon the same rules used to solve whole-number linear equations. Variable-denominator rational equations and more complicated identity classification can be extension work according to the school. Full Subject-Based Banding treats G2 as a subject level; the student’s current teacher instructions and marked work determine the immediate learning goals.

Curriculum background: MOE’s published G2 Mathematics syllabus and a 2026 school Secondary 2 G2 content outline.

Twenty central ideas behind fractional-equation accuracy

1. A fraction is a number, even when it multiplies a letter

In x/2, the fraction one-half multiplies x. In (x+1)/3, the whole bracket x+1 is divided by three. The difference matters, because a student who reads x+1/3 instead of (x+1)/3 has changed the original equation before any algebra begins. The fraction bar is therefore a grouping instruction as well as a division sign.

A useful demonstration: Have the learner substitute x=5 into the two expressions and compare their numerical values.

Common error: Do not read a numerator containing addition as though only the last term is divided.

2. Both sides of an equation must stay balanced

An equation states that its two sides have the same numerical value. Adding, subtracting, multiplying or dividing both complete sides by the same permissible number preserves the equality. In a fractional equation, the goal is to transform the statement without losing any terms. Rather than speaking vaguely of ‘moving a denominator’, name the operation used on every side.

A useful demonstration: Draw a balance with two equal collections of objects and apply the same multiplier to each collection.

Common error: Multiplying the left by six and leaving the right unchanged creates a different equation.

3. Clear numeric denominators with a common positive multiple

For x/2+1/3=x/3+1/2, the numerical denominators two and three have least common multiple six. Multiplying every additive term on both sides by six gives 3x+2=2x+3. Fractions disappear because each denominator divides the common multiplier exactly. This is an efficient way to reveal ordinary linear algebra.

A useful demonstration: List all denominators before choosing the multiplier, then tick off every term after multiplication.

Common error: Do not multiply just the terms with x while forgetting the fractional or whole-number constants.

4. The least common multiple avoids unnecessary arithmetic

If denominators are four, six and twelve, the least common multiple is twelve, not necessarily their product 288. Using twelve keeps coefficients smaller and reduces arithmetic slips. A larger common multiple is mathematically valid if applied to every term, but it may make the work harder. The choice should be based on exact divisibility and the learner’s ability to track every additive piece.

A useful demonstration: Find multiples of four and six, then check twelve divides all of them.

Common error: The smallest number visible in the problem is not automatically a valid common denominator.

5. The multiplier applies to every term, including integers

In x/2+3=7, multiplying both sides by two yields x+6=14, not x+3=14. The whole number three is equivalent to 3/1, so it also gets multiplied. This common error is often disguised because the student has correctly cleared x/2 but silently changed the remaining constant. A quick original-equation check reveals the difference.

A useful demonstration: Write 2(x/2+3)=2×7 as one line before distributing.

Common error: An integer has denominator one, not an exemption from balance operations.

6. Brackets protect a multi-term numerator

For (x+1)/3=(x−1)/2, multiplying through by six produces 2(x+1)=3(x−1). Only then expand to 2x+2=3x−3, which yields x=5. The brackets tell us to multiply both parts of the numerator, including its sign. Without them, a student might write 2x+1 and lose the original relationship.

A useful demonstration: Place a visible bracket around each numerator before clearing the denominator.

Common error: Writing 2x+1 when distributing two over x+1 changes the equation.

7. Negative signs outside a bracket affect the entire numerator

In 2−(x+1)/3=(x−2)/2, multiplying by six gives 12−2(x+1)=3(x−2). When expanding, 12−2x−2=3x−6, so x=16/5. The minus in front of the fraction belongs to the complete quotient, not only the x part. This is a signed-number and distribution issue, not a fundamentally different fraction rule.

A useful demonstration: Keep the term −2(x+1) in brackets for an extra line before expanding.

Common error: Dropping the negative sign on the constant one can reverse the final answer.

8. Collecting x on both sides is ordinary linear algebra

After clearing fractions, a question might read 3x−8=2x+4. Subtracting 2x from both sides gives x−8=4, then x=12. The fractions were only a first stage; the rest is the same equation-solving method taught with integers. Students benefit when they see one familiar skill reused rather than treat every denominator arrangement as a new chapter.

A useful demonstration: Ask which single operation removes the x term from the right without changing equality.

Common error: Do not ‘move’ x and reverse its sign without an accompanying balanced operation.

9. A fraction answer can be perfectly exact

Some equations produce x=13/7 or x=25/9 rather than an integer. That is not automatically a mistake. An exact fraction may be the simplest correct answer, and converting it to an early rounded decimal can make substitution appear imperfect. Keep an improper fraction when appropriate, then supply a decimal only if the question requests a particular precision.

A useful demonstration: Evaluate both original sides with the exact fraction before rounding.

Common error: A non-integer solution should not be rounded to a whole number unless the context explicitly requires it.

10. Some equations have no solution

If (2x−3)/4=(x+1)/2, multiplying by four produces 2x−3=2x+2. Subtract 2x and the statement becomes −3=2, a contradiction. There is no value of x for which it holds. A student who expects every worksheet question to produce a number may try to invent one; the contradiction is itself the mathematically correct conclusion.

A useful demonstration: Explain why two parallel algebraic expressions differ by a fixed constant.

Common error: Do not divide by zero or manufacture a root after all variable terms cancel.

11. Some equations are true for every real x

The equation (x+4)/2=(3x+12)/6 clears to 3x+12=3x+12. Both sides were equivalent expressions, so every real x satisfies the equation. This identity is different from a contradiction and from a unique numerical solution. Recognising the three possible outcomes—one solution, none, or infinitely many—builds meaningful equation reasoning.

A useful demonstration: Substitute two different x-values to show both original sides agree.

Common error: A final statement 0=0 does not mean x=0; it means the equality is always true over its permitted domain.

12. Domain restrictions remain if x appears in a denominator

This chapter focuses on numeric denominators, so there are no x-values prohibited by those denominators alone. As an extension, (x+1)/x=2 requires x≠0 before multiplication by x. Solving gives x=1, which is allowed. Equations with variable denominators require special care because clearing a denominator may involve multiplying by an expression that can equal zero.

A useful demonstration: Compare numeric denominator three with variable denominator x and state the difference.

Common error: Do not introduce a forbidden solution by multiplying through by x without listing x≠0.

13. Cross-multiplication is a special balanced multiplication

For a/b=c/d with nonzero denominators b,d, multiplying through by bd yields ad=bc. This is the familiar cross-product method. It is valid because the full products are being multiplied by the same denominator factors. For a sum of fractions, however, you cannot simply cross-multiply one pair and ignore the other terms; use a common multiplier across the entire equation.

A useful demonstration: Derive cross-products from multiplying both sides by bd on a simple two-fraction equality.

Common error: Cross-multiplication is not a general rule for equations with three or four additive fractional terms.

14. An original-equation check is different from a transformed-equation check

Suppose x=5 is obtained from 2(x+1)=3(x−1). It is helpful to verify 2×6=3×4, but the strongest check substitutes into the original (x+1)/3=(x−1)/2, giving 6/3=4/2=2. This catches accidental changes during denominator clearing. Each stage should represent the same original relationship.

A useful demonstration: Have the learner write the original equation again before substituting the final value.

Common error: Checking only the final transformed line can miss an earlier incorrectly distributed bracket.

15. Signs can be predicted before exact calculation

If one side of an equation contains x/3 while the other contains x/2, the larger fractional coefficient is one-half. The difference is 1/6 x, which helps predict the direction of the solution when constants are known. Estimation can catch a negative sign before extensive arithmetic. It does not replace exact balanced operations, but provides a reasonableness check.

A useful demonstration: Compare the coefficients one-third and one-half on a number line.

Common error: An answer with the opposite sign may expose a reversed subtraction or missing negative.

16. Separate algebra from physical-count interpretation

A mathematical equation may give x=13/7, which is a valid real-number solution. If x represents complete tickets, books or buses, the problem’s wording might require a whole-number domain and may have no feasible solution, or need an inequality instead. Do not round simply to create a count while claiming the original exact equation still holds. Context determines permitted values after the mathematics is solved.

A useful demonstration: Ask whether x is an abstract length, a count or another measured quantity before accepting a fractional result.

Common error: Rounding a fractional root for a whole-count story can violate the given equality.

17. Fractions on both sides do not require decimal conversion

In (x−1)/4+(x+3)/6=5/6, multiplying by twelve gives integer coefficients immediately, allowing exact manipulation. Turning one-third into 0.333 prematurely risks rounding differences. Decimal arithmetic is often acceptable for terminating fractions, but it is not always the clearest route. A reliable tutor chooses the exact representation that keeps the structure visible.

A useful demonstration: Compare a common-multiple solution with a calculator-decimal route for the same problem.

Common error: Do not introduce rounding errors into an equation whose data are exact fractions.

18. The expression’s equal sign determines the task

To simplify x/2+x/3, combine like terms and obtain 5x/6; no x-value is requested. To solve x/2+x/3=5, use that simplification to get 5x/6=5 and x=6. The instruction ‘simplify’ asks for equivalent form, whereas ‘solve’ asks for values satisfying a condition. Clear language prevents a student from adding ‘=0’ or inventing roots where there is no equation.

A useful demonstration: Give paired simplify/solve tasks using the same left-hand expression.

Common error: Do not report an x-value when the problem only asked for a simplified expression.

19. A compact working format makes errors visible

A sound solution can use four short lines: write the original equation; multiply by the least common denominator with brackets; collect like terms; substitute the root into the original. Every term remains traceable. Too many unexplained jumps are hard to audit, while unnecessary long prose can obscure the algebra. A three-pupil tutor should ask the child to explain one chosen multiplier rather than complete every line for them.

A useful demonstration: Keep each transformed equality on a new line and circle all terms affected by the multiplier.

Common error: A final correct number found by accidental cancellation does not demonstrate a reliable method.

20. Changing the problem tests transfer

A student may memorise that x=5 solves (x+1)/3=(x−1)/2. Change the right numerator or denominator and ask for a new balanced equation. If they can choose the correct least common multiple and distribute without prompts, they understand the method. If they can only repeat the same four lines, identify which underlying fraction or sign operation is still insecure.

A useful demonstration: Replace the right numerator x−1 with x−2 and solve from the beginning.

Common error: Copying the previous root is not a valid check on a newly worded equation.

Thirty worked examples with original-equation checks

Worked example 1: One fractional term and an integer

Question: Solve x/2+3=7.

Working: Multiply through by two: x+6=14, so x=8.

Original or structural check: Original: 8/2+3=4+3=7.

Avoid this mistake: The integer three must also be multiplied when denominators are cleared.

New variation: Change seven to nine.

Worked example 2: Fractions on both sides

Question: Solve (x+1)/3=(x−1)/2.

Working: Multiply through by six: 2(x+1)=3(x−1), giving 2x+2=3x−3 and x=5.

Original or structural check: Left6/3=2; right4/2=2.

Avoid this mistake: Brackets must include each numerator’s constant.

New variation: Change right numerator to x−2.

Worked example 3: An x term on every side

Question: Solve x/2+1/3=x/3+1/2.

Working: Multiply by six: 3x+2=2x+3, hence x=1.

Original or structural check: Left1/2+1/3=5/6; right1/3+1/2=5/6.

Avoid this mistake: Multiply all four terms by six.

New variation: Try replacing 1/2 on right with 2/3.

Worked example 4: A contradiction

Question: Solve (2x−3)/4=(x+1)/2.

Working: Multiply by four: 2x−3=2x+2, which implies −3=2, impossible. No solution.

Original or structural check: The left is always five-fourths below the right after common scaling.

Avoid this mistake: Do not invent an x when variable terms cancel to unequal constants.

New variation: Change −3 to +2 and classify the result.

Worked example 5: An identity

Question: Solve (x+4)/2=(3x+12)/6.

Working: Multiply by six: 3(x+4)=3x+12, identical for every x. All real x solve.

Original or structural check: At x=0 both sides equal two; at x=2 both equal three.

Avoid this mistake: Do not treat x=0 as the only solution.

New variation: Replace right constant12 with15.

Worked example 6: Brackets across two denominators

Question: Solve (x+2)/5+(x−1)/2=6/5.

Working: Multiply by ten: 2(x+2)+5(x−1)=12. Then 2x+4+5x−5=12, so 7x=13 and x=13/7.

Original or structural check: The original sides agree at 13/7 under exact fraction arithmetic.

Avoid this mistake: Do not round x to two and claim exact equality.

New variation: Change right side6/5 to 13/10.

Worked example 7: Difference of fractional coefficients

Question: Solve x/3−x/4=2.

Working: With LCD twelve, 4x−3x=24, so x=24.

Original or structural check: 24/3−24/4=8−6=2.

Avoid this mistake: Use LCD, not subtraction of denominators.

New variation: Try right side three.

Worked example 8: A whole-number term on each side

Question: Solve 3x/4−2=x/2+1.

Working: Multiply by four: 3x−8=2x+4, so x=12.

Original or structural check: 9−2=7 and 6+1=7.

Avoid this mistake: The constants −2 and +1 also scale by four.

New variation: Change −2 to −3.

Worked example 9: Three denominator terms

Question: Solve (x−1)/4+(x+3)/6=5/6.

Working: Multiply by twelve: 3(x−1)+2(x+3)=10. This yields 5x+3=10, x=7/5.

Original or structural check: At x=7/5, original sum equals5/6.

Avoid this mistake: Do not multiply a numerator only by the visible denominator.

New variation: Find a changed right side2/3.

Worked example 10: A minus before a whole fraction

Question: Solve 2−(x+1)/3=(x−2)/2.

Working: Multiply by six: 12−2(x+1)=3(x−2). Thus 10−2x=3x−6 and x=16/5.

Original or structural check: At x=16/5, left 2−(21/5)/3=3/5; right (6/5)/2=3/5.

Avoid this mistake: The outside minus affects both x and one.

New variation: Change initial two to three.

Worked example 11: Fifths, thirds and halves

Question: Solve x/5+2/3=x/2−1/6.

Working: Multiply by thirty: 6x+20=15x−5, hence25=9x and x=25/9.

Original or structural check: Substitution into the original confirms both sides.

Avoid this mistake: Least common multiple of 5,3,2,6 is30.

New variation: Try replace +2/3 with+1/3.

Worked example 12: A negative numerical solution

Question: Solve (3x+2)/6−(x−1)/4=1/3.

Working: Multiply by twelve: 2(3x+2)−3(x−1)=4, so 3x+7=4 and x=−1.

Original or structural check: At x=−1: −1/6−(−2/4)=1/3.

Avoid this mistake: Negative x is valid for an abstract algebra problem.

New variation: Change the right to2/3.

Worked example 13: Two similar numerator brackets

Question: Solve (x−4)/3+(x+4)/3=8.

Working: Combine the same denominators: (2x)/3=8, so 2x=24 and x=12.

Original or structural check: (8/3)+(16/3)=24/3=8.

Avoid this mistake: The constants cancel only because signs are opposite.

New variation: Change second numerator to x+2.

Worked example 14: Decimal coefficient and a quarter

Question: Solve 0.5x+1/4=3/4.

Working: Write 0.5=1/2; multiply by four:2x+1=3, hence x=1.

Original or structural check: 0.5+0.25=0.75.

Avoid this mistake: Decimal-to-fraction conversion can make exact reasoning clearer.

New variation: Change right side to5/4.

Worked example 15: A signed fraction plus an x term

Question: Solve 5/6−x/3=(x+1)/2.

Working: Multiply by six:5−2x=3x+3, so2=5x and x=2/5.

Original or structural check: Original left5/6−2/15=7/10; right(7/5)/2=7/10.

Avoid this mistake: Keep the minus on x throughout.

New variation: Change five-sixths to one.

Worked example 16: A bracket on both sides

Question: Solve (x+6)/4=(2x−1)/6.

Working: Multiply by twelve:3(x+6)=2(2x−1); 3x+18=4x−2; x=20.

Original or structural check: Left26/4=6.5; right39/6=6.5.

Avoid this mistake: Both numerator brackets need full multiplication.

New variation: Change the right numerator to2x+1.

Worked example 17: Another identity

Question: Solve (x−3)/5=(2x−6)/10.

Working: Multiply by ten:2(x−3)=2x−6. This is true for every real x.

Original or structural check: At x=3 both sides zero, and at x=8 both sides one.

Avoid this mistake: One matching x does not mean x=3 is the only root.

New variation: Change right constant−6 to−5.

Worked example 18: An extension with variable denominator

Question: Solve (x+1)/x=2, with x≠0.

Working: Multiply by nonzero x: x+1=2x, giving x=1.

Original or structural check: Original (1+1)/1=2.

Avoid this mistake: Record the original x≠0 restriction before multiplying.

New variation: Try (x+2)/x=3.

Worked example 19: Another variable-denominator extension

Question: Solve (x−3)/(x+1)=2, x≠−1.

Working: Multiply by x+1: x−3=2x+2, so x=−5.

Original or structural check: Original (−8)/(−4)=2; −5 is permitted.

Avoid this mistake: Do not accept x=−1, which would give a zero denominator.

New variation: Change target to three.

Worked example 20: Fraction of an unknown length

Question: A ribbon has one-half its length plus 3 cm equal to 11 cm. Find original length.

Working: Let x be length in cm. x/2+3=11 gives x=16 cm.

Original or structural check: Half of16 plus3 is11.

Avoid this mistake: A measured length must be positive.

New variation: Change the extra length to5 cm.

Worked example 21: Compare two hypothetical plans

Question: Plan A has cost (x+2)/3 dollars, Plan B (x−1)/2 dollars for a model variable x. Find equality x.

Working: Set (x+2)/3=(x−1)/2. Multiply by six:2x+4=3x−3, so x=7.

Original or structural check: At x=7 each equals3.

Avoid this mistake: These invented algebraic plans do not represent real prices or available services.

New variation: State what additional domain assumptions a practical plan needs.

Worked example 22: An inequality-looking but exact count question

Question: One-third of the full stock plus one-half of the stock equals 25 units. Find stock x.

Working: x/3+x/2=25. Multiply by six:2x+3x=150, so x=30.

Original or structural check: 10+15=25.

Avoid this mistake: This equation describes fractions of the same whole, not independent stocks.

New variation: What if the total allocated is20?

Worked example 23: A mixed-sign check

Question: Solve (x−2)/3−(x+1)/2=−5/6.

Working: Multiply by six:2(x−2)−3(x+1)=−5, giving2x−4−3x−3=−5, so −x−7=−5 and x=−2.

Original or structural check: Original (−4)/3−(−1)/2=−4/3+1/2=−5/6.

Avoid this mistake: The minus before the second bracket changes both terms.

New variation: Change right side to+1/6.

Worked example 24: One-sided simplification before solving

Question: Solve x/2+x/3=5.

Working: Combine coefficients: (3/6+2/6)x=5x/6=5, so x=6.

Original or structural check: 3+2=5.

Avoid this mistake: No need to introduce a second variable.

New variation: Solve when the right side is10.

Worked example 25: A deliberate original-check task

Question: A student claims x=2 solves (x+1)/3=(x−1)/2. Test it.

Working: Left3/3=1 and right1/2=0.5, so x=2 fails. The actual root is5.

Original or structural check: Checking the original disproves an incorrect guess immediately.

Avoid this mistake: A result can look neat while breaking the equality.

New variation: Try x=5 to show both sides equal two.

Worked example 26: Compare solutions after a constant change

Question: Solve (x+1)/3=(x−2)/2.

Working: Multiply by six:2x+2=3x−6, giving x=8.

Original or structural check: Left9/3=3 and right6/2=3.

Avoid this mistake: Changing one constant changes the root; do not reuse x=5.

New variation: Explain why cross-multiplication is valid here.

Worked example 27: Another contradiction

Question: Solve x/2+1=x/2+3.

Working: Subtract x/2 from both sides, leaving 1=3. No solution.

Original or structural check: The two expressions differ by constant two for all x.

Avoid this mistake: Do not divide by x or guess a root.

New variation: Make both constants one and classify it as identity.

Worked example 28: A variable coefficient comparison

Question: Solve 3x/4−x/6=7/3.

Working: Multiply by twelve:9x−2x=28, so 7x=28, x=4.

Original or structural check: 3−2/3=7/3.

Avoid this mistake: Combine exact fractions only after a valid common denominator.

New variation: Change right to7/6.

Worked example 29: Two different bracket fractions

Question: Solve (2x−1)/3=(x+5)/4.

Working: Multiply by twelve:4(2x−1)=3(x+5), giving8x−4=3x+15, so 5x=19 and x=19/5.

Original or structural check: Left(38/5−1)/3=11/5; right(19/5+5)/4=11/5.

Avoid this mistake: Do not round 19/5 until after the exact check.

New variation: Try x in equation after changing right constant to three.

Worked example 30: Classify rather than blindly solve

Question: Decide whether (x+2)/3=(2x+4)/6 has one, none or infinitely many solutions.

Working: Clearing by six gives2(x+2)=2x+4, an identity. Every real x is a solution.

Original or structural check: Both sides are the same expression for any real x.

Avoid this mistake: The goal is to classify an equality, not necessarily produce a number.

New variation: Change the right constant4 to5; then there is no solution.

Ten errors worth diagnosing rather than just marking

1. An integer term is not multiplied

How it happens: x/2+3=7 becomes x+3=14. The correct multiplication by two yields x+6=14. The integer has an implicit denominator one and must be included.

Precise repair: Write a bracket around each entire side and distribute the multiplier to every additive term.

Independent evidence: The learner solves a changed mixed-fraction equation with all constants scaled.

2. The numerator’s bracket disappears

How it happens: (x+1)/3 becomes2x+1 after multiplying by six. The whole numerator should become2(x+1)=2x+2.

Precise repair: Rewrite the fraction bar with parentheses, then expand only after clearing.

Independent evidence: A new two-term numerator is scaled correctly.

3. An outside minus changes only x

How it happens: −(x+1)/3 becomes−x−1/3 under a common multiplier without respecting the full quotient. Keep the sign attached to the entire bracket.

Precise repair: Write −2(x+1) before expanding to−2x−2.

Independent evidence: The pupil completes one negative-numerator example without a sign slip.

4. The wrong common multiplier is chosen

How it happens: Denominators four and six are handled with six, which is not divisible by four and therefore leaves fractional terms. Use twelve as a common multiple, preferably the least one.

Precise repair: List each denominator and test divisibility before starting.

Independent evidence: A three-denominator exercise uses a valid common multiplier independently.

5. Variables are moved by intuition rather than equality

How it happens: A student changes3x−8=2x+4 into5x=−4 by changing terms inconsistently. Subtract2x from both sides, then add eight.

Precise repair: Explain each legal operation on the two complete sides.

Independent evidence: An unfamiliar rearrangement preserves equality at every line.

6. An exact fractional answer is rounded improperly

How it happens: x=13/7 is reported as two because the learner expects an integer. Exact algebra can have a non-integer solution; rounding changes the original equality.

Precise repair: Check x=13/7 by substitution and compare with x=2.

Independent evidence: The final answer preserves the requested exact form and context.

7. A contradiction is mistaken for x=0

How it happens: After variables cancel, −3=2 appears and the learner concludes x=0. An impossible statement means there is no solution, regardless of x.

Precise repair: Substitute x=0 and x=10 to demonstrate failure in the original.

Independent evidence: The learner distinguishes contradiction from identity.

8. An identity is mistaken for one root

How it happens: 3x+12=3x+12 is reduced to0=0 and reported as x=0. In fact every real input satisfies the equation.

Precise repair: Test two distinct values and explain why both work.

Independent evidence: The student correctly states infinitely many solutions in a new identity.

9. Variable denominator restriction is omitted

How it happens: When x appears below a fraction bar, multiplying through by x may introduce forbidden inputs if x=0. State the original domain before transformations.

Precise repair: Compare (x+1)/x=2 with an all-numeric-denominator equation.

Independent evidence: No excluded root is accepted in an extension example.

10. A transformed equation is checked instead of the original

How it happens: A student verifies a later line but not the source, so an early missed sign can escape notice. Substitute the final root into the exact original fractional equation.

Precise repair: Rewrite both original sides at the candidate x value and calculate them independently.

Independent evidence: The next solution includes a correct original-equation check.

Forty-eight short retrieval questions

Quick check 1: x/2+3=7

Answer: x=8. Reason: x+6=14.

Checking habit: preserve the original equation and ask whether the candidate makes both complete sides equal. When all x terms cancel, classify the remaining statement as an identity or contradiction instead of guessing a root.

Quick check 2: (x+1)/3=(x−1)/2

Answer: x=5. Reason: 2x+2=3x−3.

Quick check 3: x/2+1/3=x/3+1/2

Answer: x=1. Reason: 3x+2=2x+3.

Quick check 4: (2x−3)/4=(x+1)/2

Answer: No solution. Reason: −3≠2.

Quick check 5: (x+4)/2=(3x+12)/6

Answer: All real x. Reason: Identity.

Quick check 6: x/3−x/4=2

Answer: x=24. Reason: x/12=2.

Quick check 7: 3x/4−2=x/2+1

Answer: x=12. Reason: 3x−8=2x+4.

Quick check 8: (x−1)/4+(x+3)/6=5/6

Answer: x=7/5. Reason: 5x+3=10.

Quick check 9: 2−(x+1)/3=(x−2)/2

Answer: x=16/5. Reason: 10−2x=3x−6.

Quick check 10: x/5+2/3=x/2−1/6

Answer: x=25/9. Reason: 6x+20=15x−5.

Quick check 11: (3x+2)/6−(x−1)/4=1/3

Answer: x=−1. Reason: 3x+7=4.

Quick check 12: (x−4)/3+(x+4)/3=8

Answer: x=12. Reason: 2x/3=8.

Quick check 13: 0.5x+1/4=3/4

Answer: x=1. Reason: 2x+1=3.

Quick check 14: 5/6−x/3=(x+1)/2

Answer: x=2/5. Reason: 5−2x=3x+3.

Quick check 15: (x+6)/4=(2x−1)/6

Answer: x=20. Reason: 3x+18=4x−2.

Quick check 16: (x−3)/5=(2x−6)/10

Answer: All real x. Reason: Identity.

Quick check 17: (x+1)/x=2, x≠0

Answer: x=1. Reason: x+1=2x.

Quick check 18: (x−3)/(x+1)=2

Answer: x=−5. Reason: x−3=2x+2.

Quick check 19: x/2+3=11

Answer: x=16. Reason: Half x=8.

Quick check 20: (x+2)/3=(x−1)/2

Answer: x=7. Reason: 2x+4=3x−3.

Quick check 21: x/3+x/2=25

Answer: x=30. Reason: 5x=150.

Quick check 22: (x−2)/3−(x+1)/2=−5/6

Answer: x=−2. Reason: −x−7=−5.

Quick check 23: x/2+x/3=5

Answer: x=6. Reason: 5x/6=5.

Quick check 24: (x+1)/3=(x−2)/2

Answer: x=8. Reason: 2x+2=3x−6.

Quick check 25: x/2+1=x/2+3

Answer: No solution. Reason: 1≠3.

Quick check 26: 3x/4−x/6=7/3

Answer: x=4. Reason: 7x=28.

Quick check 27: (2x−1)/3=(x+5)/4

Answer: x=19/5. Reason: 5x=19.

Quick check 28: (x+2)/3=(2x+4)/6

Answer: All real x. Reason: Equivalent expressions.

Quick check 29: LCD of2 and3

Answer: 6. Reason: Smallest common multiple.

Quick check 30: LCD of4,6,12

Answer: 12. Reason: All divide twelve.

Quick check 31: LCD of5,3,2,6

Answer: 30. Reason: All divide thirty.

Quick check 32: What does multiplication by2 do to x/2+3?

Answer: x+6. Reason: Every term scaled.

Quick check 33: Expand2(x+1)

Answer: 2x+2. Reason: Distribution.

Quick check 34: Expand−2(x+1)

Answer: −2x−2. Reason: Negative distribution.

Quick check 35: Expand3(x−1)

Answer: 3x−3. Reason: Distribution.

Quick check 36: Expand(x+2)²

Answer: x²+4x+4. Reason: Square bracket.

Quick check 37: If x=5, (x+1)/3

Answer: 2. Reason: 6/3.

Quick check 38: If x=5, (x−1)/2

Answer: 2. Reason: 4/2.

Quick check 39: Is x=2 a root of above?

Answer: No. Reason: 1 versus0.5.

Quick check 40: Can a fraction result be exact?

Answer: Yes. Reason: Do not round without need.

Quick check 41: What means 0=0 after valid simplification?

Answer: Identity. Reason: Every permitted x.

Quick check 42: What means 1=3 after valid simplification?

Answer: Contradiction. Reason: No solution.

Quick check 43: In (x+1)/x, excluded x

Answer: 0. Reason: Zero denominator.

Quick check 44: In (x−3)/(x+1), excluded x

Answer: −1. Reason: Denominator zero.

Quick check 45: What validates proposed root?

Answer: Original substitution. Reason: Check both sides.

Quick check 46: Is cross-multiplication safe for sums of fractions directly?

Answer: Not as a shortcut. Reason: All additive terms must be included.

Quick check 47: What is 0.5 as a fraction?

Answer: 1/2. Reason: Exact terminating decimal.

Quick check 48: If x/2+x/3 simplified

Answer: 5x/6. Reason: Combine coefficients.

Quick check 49: Does simplifying alone ask for x?

Answer: No. Reason: Equivalent expression.

Four weeks of deliberate practice

Week 1: meanings of fractions and the balancing rule

Use number examples to show why multiplying both sides by the same positive integer preserves equality. Practise x/2+3=7 and one fraction on each side, keeping whole numerator brackets visible.

Independent progress check: The child can name the least common multiple and scale every term on a changed problem without prompting.

Week 2: signed brackets and different numeric denominators

Practise denominator sets 2,3,4,6 and 12. Introduce a minus in front of an entire fraction and ask pupils to expand negative brackets separately before collecting x terms.

Independent progress check: An unfamiliar two-sided equation has correct distribution and the original check.

Week 3: compare one, no and infinitely many solutions

Use two equations that become contradictions, two identities and several unique roots. Ask the student to classify the remaining equality when x terms cancel rather than expect a numerical answer each time.

Independent progress check: The learner distinguishes 0=0 from 1=3 and explains the outcome in words.

Week 4: mix contextual and extension questions

Solve fractional equations from lengths or abstract contexts, preserving exact fractional roots. Introduce variable-denominator exclusions only when assigned by the school. Finish with a short mixed set whose method is not announced by its heading.

Independent progress check: Independent new work includes full original-equation checks and sensible interpretation.

A ninety-minute lesson with three pupils

The original eduKateSG Clementi tutorial reference describes three-pupil small-group Mathematics lessons lasting 90 minutes. A practical illustration uses ten minutes of independent diagnosis, fifteen minutes reconstructing equation balance, twenty minutes of guided denominator clearing, twenty-five minutes of changed equations with both-side fractions, ten minutes comparing contradictions and identities, and ten minutes reviewing the earliest wrong step. This does not claim any particular current class availability.

One pupil may need help finding the least common multiple, another may drop a negative sign in a numerator bracket, and a third may incorrectly round a fractional root. A good tutor checks each student’s independent first line, makes the relevant rule visible, and tests that specific correction on new values. Three learners in one room do not automatically share the same misunderstanding.

Questions parents can ask without giving away the answer

Parent question	What it reveals
What does the fraction bar divide?	Correct whole-numerator interpretation.
Why did you choose twelve?	The number divides every numeric denominator.
Which terms receive the multiplier?	Every term on both sides.
What happens to a minus before a bracket?	It is distributed to every bracket term.
Can a root be a fraction?	Exact real-number answers may be non-integers.
What does 0=0 mean?	An identity, not necessarily x=0.
What does 2=5 mean after cancellation?	A contradiction with no solution.
Did you check the original statement?	The candidate really satisfies the source equation.

Frequently asked questions

1. How do I solve equations with fractions on both sides?

Identify all numeric denominators and multiply every term on both sides by their least common multiple. Keep brackets around numerators, simplify the resulting linear equation and substitute the candidate into the original. This is the same balance principle used for whole-number equations.

2. Must I use the least common denominator?

Any nonzero common multiple can clear numeric denominators if applied correctly to every term. The least common multiple is often most efficient because it keeps intermediate integers smaller.

3. Why do brackets matter when clearing fractions?

A fraction like (x+1)/3 contains two numerator terms divided as one expression. Multiplying by six produces2(x+1), not2x+1. Correct distribution preserves the original quantity.

4. Can the final answer be an improper fraction?

Yes. Exact roots such as13/7 are valid when the equation represents an abstract real-number quantity. Do not round unless instructed or a physical context imposes a separate whole-number restriction.

5. What if all x terms cancel?

Inspect the remaining statement. If it is a contradiction such as−3=2, there is no solution. If it is an identity such as12=12, every permitted x solves the equation.

6. What is the difference between simplifying and solving?

Simplifying rewrites an expression equivalently without finding a particular x. Solving an equation asks for the values of x that make an equality hold. The command word changes the task.

7. Are variable denominators handled the same way?

They require additional attention. Values making original denominators zero are excluded; multiplying by a variable expression must respect these restrictions. Such rational-equation questions can be treated as an extension based on school coverage.

8. What is cross-multiplication really doing?

For an equality of two fractions with nonzero denominators, it is multiplication of both sides by the product of denominators. For sums of several fractions, multiply every term by a common denominator instead of applying a shortcut to one pair.

9. Why should the original equation be checked?

A numerical candidate can satisfy an incorrectly rearranged equation even when the earlier fraction or sign handling was wrong. Substitution into the original is the strongest quick check of a proposed solution.

10. How can a parent identify the child’s specific error?

Look for the first line where a denominator, bracket or sign was handled incorrectly. Ask what common multiplier was used and whether every term was affected. Repair that decision, then use one changed example without the solution visible.

11. Are these exact equations required in every Secondary 2 G2 paper?

Schools sequence algebra and fraction work differently. Numeric-denominator linear equations are curriculum-connected, while variable-denominator extensions or more complex classifications should follow the learner’s current assigned scope.

12. Does completing this topic guarantee a grade?

No. It develops algebraic reasoning, but school results depend on broader learning and assessment. Independent correct work on new examples is a useful local measure of progress.

A final habit worth retaining

Did you know? Most mistakes in equations with fractions happen during an unjustified shortcut, not during the final division. Keep the balance visible, scale the complete numerators, preserve signs and return to the original equality after solving. A student who does these independently on unfamiliar fractions is developing a skill that carries into later algebra.

The original Clementi three-pupil Mathematics tutorial reference remains unchanged. For class suitability and current arrangements, contact eduKate Singapore directly. Grades, vacancies and subject-level progression cannot be guaranteed.

Explore the other three Secondary 2 G2 Mathematics guides in this batch

This batch connects exact fractional equations, irregular polygon angle reasoning, reliable statistical samples and percentage changes. These are different skills that share one central habit: define the mathematical quantity, use the correct relation and check the answer in its original context. Choose the guide that matches a current school problem and attempt a changed example after reading it.

Secondary 2 G2 Mathematics Tutor Bukit Timah | Algebraic Equations with Fractions on Both Sides