Did you know? An equation such as x/4 + x/6 = 5 looks difficult mainly because its denominators hide a very simple relationship. Multiply every term on both sides by 12 and it becomes 3x + 2x = 60. Now the equation is 5x = 60, so x = 12. The fractions have disappeared, but the equality has not changed.
For Secondary 1 G3 Mathematics students in Bukit Timah, equations with fractional coefficients are an important bridge between arithmetic and algebra. The successful method is not “shift numbers across and hope.” It is to preserve the balance of both sides while making the working easier to read.
This guide begins with short examples, then moves through brackets, negative signs, fraction bars and word problems. Every worked answer is checked by substitution so parents and students can see what a complete solution looks like.
Skip to a worked example, a quick practice set or a parent checklist.
- Quick answer: three steps to clearer equations
- Why balancing matters more than “moving across”
- Worked example 1: one fraction and one whole number
- Worked example 2: two fractions sharing a variable
- Worked example 3: multiplying a bracket by a fraction
- Worked example 4: the denominator covers a whole expression
- Worked example 5: a minus before a fraction
- Worked example 6: an equation with a negative answer
- Worked example 7: fractional coefficients on both sides
- Worked example 8: an equation from a simple story
- Five mistakes to diagnose separately
- Ten-minute self-test: eight equations
Quick answer: three steps to clearer equations
First identify denominators and choose a convenient common multiple. Next multiply *every term on both sides* by that number, maintaining equality. Then solve the simplified equation and substitute the answer into the original.
Consider x/3 + 4 = 9. Subtract 4 from both sides to obtain x/3 = 5. Multiply both sides by 3: x = 15.
Check: 15/3 + 4 = 5 + 4 = 9. Correct.
There was no need to use a complicated method. The operation on each side had a reason, and the final check demonstrated that the original equation is satisfied.
Why balancing matters more than “moving across”
An equation states that two expressions have equal values. Adding the same number to both sides preserves equality. Subtracting the same number also preserves it. Multiplying both sides by the same nonzero number preserves equality.
Students often learn the phrase “move the three across and multiply.” That may help them remember a step, but it can obscure why the new line remains valid.
Take x/3 = 5. If you multiply only the left side by 3, you get x = 5, which changes the equation. Multiplying both sides by 3 gives x = 15.
This way of thinking survives more complicated questions involving negative fractions and brackets. A valid algebraic transformation should not create or lose a solution.
Worked example 1: one fraction and one whole number
Solve (2x − 1)/3 = 5.
Multiply both sides by 3: 2x − 1 = 15. Add 1 to both sides: 2x = 16. Divide by 2: x = 8.
Check in the original expression: (2 × 8 − 1)/3 = (16 − 1)/3 = 15/3 = 5.
A common wrong line is 2x − 1 = 5/3. That divides when the equation requires multiplication by 3 to clear the denominator.
Notice how the fraction bar groups the entire numerator 2x − 1. The expression means both terms together are divided by three.
Worked example 2: two fractions sharing a variable
Solve x/4 + x/6 = 5.
The lowest common multiple of 4 and 6 is 12. Multiply every term by 12.
The first term becomes 12(x/4) = 3x. The second becomes 12(x/6) = 2x. The right side becomes 12 × 5 = 60.
Thus 3x + 2x = 60. So 5x = 60 and x = 12.
Check: 12/4 + 12/6 = 3 + 2 = 5.
If the child writes 3x + 2x = 5, they multiplied the left side but forgot the right. The answer may look algebraically tidy, but it no longer describes the original equality.
Worked example 3: multiplying a bracket by a fraction
Solve (2/3)(x + 3) = 10.
Multiply both sides by 3: 2(x + 3) = 30. Divide both sides by 2: x + 3 = 15. Subtract 3: x = 12.
Check: (2/3)(12 + 3) = (2/3) × 15 = 10.
Another valid route is to expand first: (2/3)x + 2 = 10. Then (2/3)x = 8, giving x = 12.
Both routes are correct. The first may be shorter when a whole bracket shares the same fractional multiplier. The second reveals how the distributive law works.
Ask the learner to compare the routes. The method is flexible because the mathematics is understood.
Worked example 4: the denominator covers a whole expression
Solve (x + 2)/4 = (x − 1)/3.
Multiply both sides by 12, the lowest common multiple of 4 and 3: 3(x + 2) = 4(x − 1).
Expand: 3x + 6 = 4x − 4. Subtract 3x from both sides: 6 = x − 4. Add 4: x = 10.
Check: left side = (10 + 2)/4 = 3; right side = (10 − 1)/3 = 3. They match.
This is a situation where cross multiplication works because the equation equates two fractions with nonzero constant denominators. But the deeper explanation is that both sides are multiplied by 12. A child should know why that operation is valid rather than apply the cross-product pattern indiscriminately to expressions involving sums.
Worked example 5: a minus before a fraction
Solve x/2 − (x − 1)/3 = 4.
Multiply all terms by 6: 3x − 2(x − 1) = 24.
Expand with care: 3x − 2x + 2 = 24. So x + 2 = 24 and x = 22.
Check: 22/2 − (22 − 1)/3 = 11 − 7 = 4.
A frequent mistake is to write 3x − 2x − 2 = 24. The negative multiplier was not distributed correctly across the −1 in the bracket.
This mistake belongs to signed bracket expansion, not to the fraction denominators. The fractional-coefficient expression guide revisits that prerequisite.
Worked example 6: an equation with a negative answer
Solve (−x + 5)/2 = 4.
Multiply by 2: −x + 5 = 8. Subtract 5: −x = 3. Hence x = −3.
Check: (−(−3) + 5)/2 = (3 + 5)/2 = 4.
The answer is negative even though the equation contains positive numbers. A student should not change −3 to +3 because the question “looks like it should be positive.”
If a word problem imposes a nonnegative real-world condition, the interpretation would need further attention. A purely algebraic equation can have a negative solution.
Worked example 7: fractional coefficients on both sides
Solve (3/4)x + 1 = (1/2)x + 5.
Multiply each term by 4: 3x + 4 = 2x + 20. Subtract 2x: x + 4 = 20. Subtract 4: x = 16.
Check the original: left side is (3/4)(16) + 1 = 12 + 1 = 13. Right side is (1/2)(16) + 5 = 8 + 5 = 13.
When the variable occurs on both sides, bring variable terms together with equivalent operations. Do not change signs through memory alone without ensuring equality remains true.
Worked example 8: an equation from a simple story
One third of a number, plus five, equals twelve. Find the number.
Let n represent the number. The phrase “one third of a number” means n/3. Adding five gives n/3 + 5, so the equation is n/3 + 5 = 12.
Subtract 5: n/3 = 7. Multiply by 3: n = 21.
Check: one third of 21 is 7 and 7 + 5 = 12.
There is a difference between “one third of a number plus five” and “one third of the sum of the number and five.” The second might be written (n + 5)/3. Those equations are not interchangeable. Read the grammatical grouping carefully.
Five mistakes to diagnose separately
- Forgetting to multiply the whole right side when clearing denominators.
- Multiplying only the first term on the left rather than every term.
- Dropping brackets around a numerator or a fraction-scaled group.
- Mishandling a negative sign during expansion.
- Failing to substitute the final value back into the original equation.
A tutor should identify the first invalid line. Giving twenty more questions without understanding the error may simply rehearse the same mistake.
Ten-minute self-test: eight equations
Try without reading the answer key. Write one reason for each major operation.
- A. x/3 + 4 = 9.
- B. (2x − 1)/3 = 5.
- C. x/4 + x/6 = 5.
- D. (2/3)(x + 3) = 10.
- E. (x + 2)/4 = (x − 1)/3.
- F. x/2 − (x − 1)/3 = 4.
- G. (−x + 5)/2 = 4.
- H. (3/4)x + 1 = (1/2)x + 5.
Answers with substitution checks
A. x = 15; 15/3 + 4 = 9.
B. x = 8; (16 − 1)/3 = 5.
C. x = 12; 12/4 + 12/6 = 5.
D. x = 12; (2/3)(12 + 3) = 10.
E. x = 10; both sides equal 3.
F. x = 22; 22/2 − 21/3 = 4.
G. x = −3; (3 + 5)/2 = 4.
H. x = 16; both sides equal 13.
The student should be able to explain why the denominator was cleared and why the answer satisfies the *original* equation. Merely reproducing x = 12 from a memory of the worked model is not enough.
A practical fifteen-minute lesson
Begin with two minutes on denominator multiples: find the lowest common multiple of 3 and 4, or 4 and 6. Then practise one fraction equation with the variable on one side and check it.
Spend the next five minutes on a grouped numerator such as (2x − 1)/3. Make the pupil identify what the fraction bar covers. Then introduce an equation with two fractional terms and ask for a common multiplier.
Finish with an intentionally wrong solution. For instance, show x/4 + x/6 = 5 becoming 3x + 2x = 5, and ask which operation was applied to only one side. The child should repair that line and explain the balance principle.
A later retest with changed denominators matters more than a promise of rapid improvement. The useful outcome is independent, correct decision-making.
What a Bukit Timah Mathematics tutor should look for
A tutor ought to distinguish a weakness in fractions, negative signs, algebraic grouping, equation balance or word-problem translation. These can produce similar-looking final errors but need different instruction.
The original eduKateSG Secondary 1 Mathematics small-group tutorial reference describes close feedback in three-student lessons near Sixth Avenue MRT. The Bukit Timah Secondary 1 G3 Mathematics tuition overview gives location-specific programme context. Parents should confirm the student’s current curriculum and actual class arrangements directly.
The MOE G2 and G3 Mathematics syllabus includes linear equations, algebraic expressions and operations in the Secondary 1 G3 learning programme. The school may change the order or grouping of topics in its assessments.
Frequently asked questions
Why is the lowest common multiple useful?
It supplies a multiplier that clears every constant fraction denominator without introducing unnecessarily large numbers. Any nonzero common multiple would work, but the lowest is usually simpler.
Is cross multiplication always allowed?
It is appropriate for an equality between two fractions with valid nonzero denominators. It should not be applied by pattern alone to equations involving sums such as x/4 + x/6 = 5.
What if the answer is a fraction or a negative number?
That may be perfectly valid. Substitute into the original equation to verify, and consider any real-world constraints if the problem describes physical quantities.
Can a calculator replace the algebra?
A calculator can check numerical arithmetic, but the learning goal includes understanding and showing valid transformations that preserve equality.
Where do I revise if brackets are still confusing?
Start with fractional coefficients and expansion and the broader algebraic expressions guide, then return to full equations.
Related Secondary 1 G3 Mathematics guides
- Rational and irrational numbers, fractions and number lines
- Algebraic expressions with fractional coefficients
- Linear equations from real-world word problems
- Weighted assessment revision and error-log practice
An equation is a balance, not a collection of instructions to move symbols around. Keep both sides equal, and the fractions gradually stop looking intimidating.