Algebra

Solving linear equations without losing marks

By Mr Gan · O-Level E Maths · Updated August 2026 · 7 min read

To solve a linear equation, clear any brackets and fractions first, then gather all the unknown terms on one side and all the number terms on the other, and finally divide by the coefficient of the unknown. The marks are lost not from wrong method but from a single sign slip when moving a term across the equals sign, so writing every step out, rather than jumping ahead, is what actually protects your score.

Why students lose marks on questions they "know how to do"

Most students can solve 2x + 3 = 11 in their head. The trouble starts when the unknown appears on both sides, when there are brackets to expand, or when a fraction sits in the equation. Under exam pressure, students skip steps to save time, and that is exactly where a sign gets dropped or a term gets moved without flipping its sign.

O-Level markers award method marks for each correct step shown, even if the final answer is wrong. Rushing to a one-line answer throws away marks you could have kept even after a slip.

The method

Step 1

Clear brackets and fractions

Step 2

Gather x terms on one side

Step 3

Gather number terms on the other side

Step 4

Divide by the coefficient of x

Watch the whole method in about a minute.

Worked example 1: unknown on both sides

Solve 5x − 4 = 2x + 11.

Solution

No brackets or fractions here, so move straight to gathering terms. Subtract 2x from both sides to bring the x terms together:
5x − 2x − 4 = 11

Simplify the left side:
3x − 4 = 11

Add 4 to both sides to move the number term across:
3x = 15

Divide both sides by 3, the coefficient of x:
x = 5

Quick check: substitute x = 5 back into the original equation. Left side: 5(5) − 4 = 21. Right side: 2(5) + 11 = 21. Both sides match, so the answer is correct. This 10-second check catches sign errors before they cost marks.

Worked example 2: brackets on both sides

Solve 3(2x − 1) = 5(x + 4). Expanding accurately is the whole game here, so if that step still feels slow, work through our guide on simplifying algebraic expressions first.

Solution

Expand both brackets first:
6x − 3 = 5x + 20

Subtract 5x from both sides to gather the x terms on the left:
6x − 5x − 3 = 20

Simplify:
x − 3 = 20

Add 3 to both sides:
x = 23

The harder version of the same question puts a minus sign in front of a bracket. Solve 4(x + 3) − 2(x − 5) = 30.

Solution

Expand the first bracket:
4(x + 3) = 4x + 12

Expand the second bracket, multiplying both terms inside by −2:
−2(x − 5) = −2x + 10

Put the two expansions together and simplify:
4x + 12 − 2x + 10 = 30, so 2x + 22 = 30

Subtract 22 from both sides, then divide by 2:
2x = 8, so x = 4

Check by substituting x = 4: 4(7) − 2(−1) = 28 + 2 = 30 ✓

The step students get wrong

When a bracket has a negative number in front, such as −(2x − 5), both terms inside the bracket must have their sign flipped on expansion: −2x + 5, not −2x − 5. Losing this one sign changes the entire answer. Expand slowly, one term at a time, and never skip writing the expanded line.

Worked example 3: a fraction in the equation

Solve (x + 1) / 3 = (x − 2) / 4. The same clearing move is the first step whenever you meet algebraic fractions.

Solution

Multiply both sides by 12, the lowest common multiple of 3 and 4, to clear both fractions in one step:
4(x + 1) = 3(x − 2)

Expand both brackets:
4x + 4 = 3x − 6

Subtract 3x from both sides:
x + 4 = −6

Subtract 4 from both sides:
x = −10

Fraction shortcut: always multiply by the lowest common multiple of every denominator in one move, rather than clearing one fraction at a time. It is faster and leaves fewer opportunities for an arithmetic slip.

When there is no solution or infinite solutions

Occasionally the x terms cancel out completely during simplification. If the numbers left behind are unequal, such as 0 = 7, the equation has no solution. If the numbers left behind are equal, such as 0 = 0, every value of x satisfies the equation. Neither outcome means you have made a mistake, so check your working once and then state the conclusion plainly rather than hunting for a number that is not there.

Frequently asked questions

Why do I need to show every step instead of just the answer?

O-Level E-Maths awards method marks for each correct step of working, separate from the mark for the final answer. A student who shows correct working but makes one final arithmetic slip can still collect most of the marks. A student who writes only the final answer, and gets it wrong, collects nothing.

What is the difference between solving a linear equation and a linear inequality?

The steps are almost identical, with one important exception: when you multiply or divide both sides of an inequality by a negative number, the inequality sign must flip direction. This rule does not apply to equations, since there is no direction to flip. See our guide on solving linear inequalities for the full method.

How is this different from changing the subject of a formula?

Solving a linear equation finds a single numerical value of x. Changing the subject of a formula rearranges an equation with several letters so that one letter is expressed in terms of the others, and no numerical answer is produced. The rearranging steps, moving terms and dividing by a coefficient, are the same skill applied to a different goal. See our guide on changing the subject of a formula.

Do I need to clear fractions before or after expanding brackets?

Clear the fractions first by multiplying every term on both sides by the lowest common multiple of the denominators. This turns the equation into one with only whole-number coefficients, which is far easier to expand and simplify correctly.

— Mr Gan Math Tuition

Still losing marks on equations you know how to solve?

Mr. Gan works with students who want a method that holds up under exam pressure, not just when there is time to think.

Chat with Mr. Gan