Secondary 1 G3 Mathematics Tutor Bukit Timah | Solving Multi-Step Equations with Brackets

Did you know? A student can solve 3x = 21 immediately but get stuck on 3(x − 2) + 5 = 20. The unknown is still x, and the question is still a linear equation. The real difference is that several smaller decisions—brackets, signs, like terms and equality—must be handled in the correct order.

For Secondary 1 G3 Mathematics students in Bukit Timah, multi-step equations are an excellent measure of whether the algebra foundation is stable. Many apparent “careless mistakes” have an identifiable cause: a number was not distributed through a bracket, a negative sign was lost, or one side of an equation was changed without applying the same valid operation to the other.

This guide gives short worked examples and a clear repair routine for one-variable linear equations involving brackets and negative numbers. Students can solve them methodically without relying on an unexplained “move the term across” shortcut.

Quick answer: four moves to solve a multi-step equation

• Expand brackets correctly, including negative multipliers.
• Combine like terms on each side without dropping signs.
• Use equal operations on both sides to isolate the unknown.
• Substitute the solution into the original equation to check it.

For 3(x − 2) + 5 = 20, expansion gives 3x − 6 + 5 = 20. Combine: 3x − 1 = 20. Add 1 to both sides: 3x = 21. Divide both sides by 3: x = 7.

Check using the original: 3(7 − 2) + 5 = 3(5) + 5 = 20. The equality is satisfied.

Why an equals sign is not an instruction to calculate

An equation says that the expression on its left has the same value as the expression on its right. It is a balance statement.

For example, 2x + 4 = 16 means that two copies of an unknown amount plus four equal sixteen. To preserve equality, subtract four from both sides, obtaining 2x = 12. Then divide both sides by two, obtaining x = 6.

The reason is not that “the 4 moves to the other side and changes sign.” The short classroom phrase may sometimes match the result, but it hides the principle that we applied the same subtraction on both sides.

That principle remains dependable when the equation contains negative numbers, fractions or several brackets.

Worked example 1: one bracket and a constant

Solve 4(x + 2) − 3 = 21.

Expand the bracket: 4x + 8 − 3 = 21. Simplify: 4x + 5 = 21.

Subtract five from both sides: 4x = 16. Divide both sides by four: x = 4.

Check: left side = 4(4 + 2) − 3 = 4(6) − 3 = 21. The answer agrees with the right side.

If the learner mistakenly writes 4x + 2 − 3, identify the first wrong line: the coefficient four must multiply both x and 2.

Worked example 2: a negative multiplier

Solve −2(3x − 4) = 14.

Expanding the negative multiplier gives −6x + 8 = 14. Subtract eight from each side: −6x = 6.

Divide both sides by −6: x = −1.

Check carefully in the original: −2(3(−1) − 4) = −2(−3 − 4) = −2(−7) = 14.

The negative solution is not an error. It is the value that makes the original equation true.

A common wrong line is −6x − 8 = 14. It comes from multiplying −2 by −4 incorrectly. The sign correction belongs in the distributive step, not in an extra unrelated worksheet.

Worked example 3: a minus sign in front of an entire bracket

Solve 4 − (2x + 3) = −9.

The minus sign outside the bracket means subtract the entire expression 2x + 3. Expand as 4 − 2x − 3 = −9.

Combine constants: 1 − 2x = −9. Subtract one from both sides: −2x = −10. Divide by −2: x = 5.

Check: 4 − (2(5) + 3) = 4 − 13 = −9.

If a student writes 4 − 2x + 3, they have changed the subtraction of +3 into addition. Drawing a large bracket around the entire subtracted quantity can make the error easier to see.

Worked example 4: the unknown appears on both sides

Solve 2(x + 3) = 3x − 4.

Expand: 2x + 6 = 3x − 4.

Subtract 2x from both sides: 6 = x − 4. Add four to both sides: x = 10.

Check: left side = 2(10 + 3) = 26. Right side = 3(10) − 4 = 26.

The solution is confirmed by two independent calculations. This is a useful example for showing why the unknown does not need to begin on only one side of an equation.

A tutor should encourage a student to choose one consistent way to collect variable terms, rather than switching direction unpredictably mid-solution.

Worked example 5: two brackets with different signs

Solve 5(x − 2) − 2(x + 3) = 8.

Expand the first bracket: 5x − 10. The negative multiplier on the second bracket gives −2x − 6.

Combine: 5x − 10 − 2x − 6 = 8, so 3x − 16 = 8.

Add sixteen to both sides: 3x = 24. Divide by three: x = 8.

Check: 5(8 − 2) − 2(8 + 3) = 5(6) − 2(11) = 30 − 22 = 8.

The second bracket is a frequent trouble spot. Ask the learner to point to both terms it multiplies, then say the resulting signs aloud.

Worked example 6: a bracket with x on the right

Solve 3 − 2(x − 4) = x + 2.

Expand the left: 3 − 2x + 8 = x + 2. Combine constants: 11 − 2x = x + 2.

Add 2x to both sides: 11 = 3x + 2. Subtract two: 9 = 3x. Therefore x = 3.

Check in the original: left = 3 − 2(3 − 4) = 3 − 2(−1) = 5. Right = 3 + 2 = 5.

Even when the final answer is a small positive integer, the working passes through a negative number. Signed-number control is therefore part of algebraic reliability.

Worked example 7: a fractional multiplier

Solve (1/2)(4x − 6) = 5.

Distribute one half across the bracket: 2x − 3 = 5. Add three to both sides: 2x = 8. Divide by two: x = 4.

Check: (1/2)(4(4) − 6) = (1/2)(16 − 6) = (1/2)(10) = 5.

Another correct route is to multiply both sides of the original equation by two, giving 4x − 6 = 10, followed by 4x = 16. Both routes lead to x = 4.

These are equivalent methods, not rival tricks. The child should choose a route that preserves the equation’s meaning and gives easily checked steps.

Why the original equation is the best final check

Suppose a child wrongly expands −3(x − 2) as −3x − 6, solves the altered equation perfectly and obtains a numerical value. Substituting into the altered equation will not reveal the original expansion error.

Checking against the original question does. That is why a reliable solution ends by returning to the expression as printed.

For every linear equation, write the candidate value, evaluate both original sides separately and confirm equality. If the two sides differ, retrace the working to the first invalid line.

This is a stronger habit than simply writing a tick beside the answer.

The four common error families

Bracket error: 3(x + 2) is rewritten as 3x + 2 instead of 3x + 6.

Sign error: −2(x − 4) is rewritten as −2x − 8 instead of −2x + 8.

Equality error: a student subtracts five from one side but not the other.

Collection error: 4x − x is rewritten as 4 rather than 3x.

These mistakes call for different teaching. If the child repeatedly misexpands negative brackets, return to signed multiplication. If terms disappear while rearranging, revisit equality. If letters are combined incorrectly, practise like terms separately.

The aim is to correct the first wrong mathematical decision, not merely count how many equations were completed.

What if both sides simplify to the same expression?

Not every apparent linear equation has a single numerical solution.

Consider 2(x + 3) = 2x + 6. Expanding gives 2x + 6 = 2x + 6, which is true for every value of x. There are infinitely many solutions in the permitted number domain.

Now consider 2(x + 3) = 2x + 7. Expanding gives 2x + 6 = 2x + 7, which would imply 6 = 7. That is impossible, so there is no solution.

These are useful extension examples because they discourage the belief that every equation must end with one value for x. The equality itself decides the outcome.

From a sentence to a multi-step equation

A fictional puzzle says: “Three times the result of decreasing a number by four, then adding five, equals twenty.” Find the number.

Let x be the unknown number. “Decrease by four” gives x − 4. Triple that result and add five: 3(x − 4) + 5 = 20.

Expand: 3x − 12 + 5 = 20, so 3x − 7 = 20. Add seven: 3x = 27. Thus x = 9.

Check the story: nine decreased by four is five; triple five is fifteen; add five to obtain twenty.

This example links equation solving to reading comprehension. If the equation is formed incorrectly, accurate arithmetic will not rescue the interpretation.

A ten-question self-test

• A. Solve 3(x − 2) + 5 = 20.
• B. Solve −2(3x − 4) = 14.
• C. Solve 4 − (2x + 3) = −9.
• D. Solve 2(x + 3) = 3x − 4.
• E. Solve 5(x − 2) − 2(x + 3) = 8.
• F. Solve 3 − 2(x − 4) = x + 2.
• G. Solve −(x − 6) = 2x.
• H. Solve 6 − (x + 2) = −4.
• I. Solve 2x − (3 − x) = 12.
• J. Solve (1/2)(4x − 6) = 5.

Answer key with the essential decision

A. x = 7; expand to 3x − 1 = 20.

B. x = −1; multiply −2 by both bracket terms.

C. x = 5; the full bracket is subtracted.

D. x = 10; collect x terms while preserving equality.

E. x = 8; expansion gives 3x − 16 = 8.

F. x = 3; the simplified equation is 11 − 2x = x + 2.

G. x = 2; −x + 6 = 2x gives 6 = 3x.

H. x = 8; 6 − x − 2 = −4 gives 4 − x = −4.

I. x = 5; 2x − 3 + x = 12 gives 3x = 15.

J. x = 4; distributing one half gives 2x − 3 = 5.

If a child answers incorrectly, ask for the line where the written equation first stopped being equivalent to the one before it.

A fifteen-minute practice sequence

Spend three minutes expanding four brackets, including two negative multipliers. Use the algebraic expressions and brackets guide if this step is unreliable.

Use the next five minutes to solve two equations, one with x on both sides and one with a bracket preceded by a minus sign. Require a check using the original equation.

Then show a wrong worked example and ask the child to explain the exact line where equality or a sign was lost. Finish with one fresh word problem that the learner must translate before solving.

Revisit similar questions later with different coefficients. Independence on a changed example is stronger evidence of understanding than copying a familiar page.

What parents can ask a Bukit Timah Mathematics tutor

A useful Secondary 1 G3 tutor diagnoses *which* algebra step is causing the problem. The correct support for weak brackets differs from the correct support for fraction arithmetic or equation balance.

The eduKateSG Secondary 1 Mathematics small-group tutorial reference describes close feedback in three-student classes near Sixth Avenue MRT. The Bukit Timah G3 Mathematics tuition overview provides local programme context; current places, timing and fit should be checked directly.

The MOE G2 and G3 Mathematics syllabus includes algebraic expressions and solving simple linear equations within the lower-secondary G3 foundation. Individual schools may arrange the topic order differently.

Frequently asked questions

### Why does a minus sign outside a bracket affect every term?

Because subtracting the entire bracket is equivalent to multiplying it by −1. Every term must be multiplied.

### Is “move it across and change the sign” always safe?

It can be a shorthand for valid operations, but it is safer to state what was added, subtracted, multiplied or divided on both sides.

### What should I do if the solution is negative?

Check it in the original equation. Negative values can be legitimate solutions unless the word-problem context excludes them.

### Why can some equations have no solution?

If simplification leads to an impossible statement such as 6 = 7, no value of the unknown can make the original equation true.

### Is solving an equation different from factorising?

Yes. Factorising rewrites an expression in an equivalent multiplicative form. Solving an equation finds values that make two sides equal.

Read the connected guides

• Linear equations containing fractions
• Algebraic expressions and brackets
• Extracting common factors
• Translating word phrases into algebra

A student improves when every new line means the same thing as the previous line. That is the real skill behind a multi-step equation.