Why is c ÷ 4 = 8 solved by c = 32 and not c = 2?

Because solving means undoing each operation with its inverse. The c has been divided by 4, so undo it by multiplying by 4: c = 8 × 4 = 32. Dividing by 4 again gives 8 ÷ 4 = 2, but that repeats the operation instead of reversing it. Substitute back: 32 ÷ 4 = 8 ✓, while 2 ÷ 4 = 0.5, not 8.

What does it mean to move a term across the equals sign?

When a term crosses the equals sign it flips its sign: a +2 becomes −2, a −7 becomes +7. For −2 + y = 10, moving the −2 across gives y = 10 + 2 = 12. This is the same as adding 2 to both sides. Carrying the sign across unchanged is the classic error.

How do I solve an equation with the unknown on both sides?

Eliminate the smaller x-term by applying its inverse to both sides. For 7x − 22 = 4x + 29, subtract 4x from both sides to get 3x − 22 = 29, add 22 to get 3x = 51, then divide by 3 to get x = 17. Check: 7(17) − 22 = 97 = 4(17) + 29.

GCSE Maths Foundation · AQA · Solving equations

Solving equations: why c ÷ 4 = 8 means c = 32, not 2

On the non-calculator paper, "solve the equation" trips students who undo an operation by repeating it rather than using its inverse. Faced with $c \div 4 = 8$ , they divide by 4 again and write $c = 8 \div 4 = 2$ . The partner error is moving a term across the equals sign without flipping its sign, so $-2 + y = 10$ becomes $y = 8$ instead of $12$ . One root cause, two surfaces: the operation is repeated, not reversed.

The thirty-second fix: undo with the inverse, on both sides. $\div$ is reversed by $\times$ , $-$ by $+$ . A term that crosses the equals sign flips its sign. Then substitute your answer back to confirm the equation balances.

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How to spot it in your own work

You solved $c \div 4 = 8$ by dividing again, writing $c = 2$ instead of $32$ .
You undid a subtraction by subtracting again (or an addition by adding again).
You moved a term across the equals sign but kept its sign the same.
You did not substitute your answer back to check the equation balances.

An exam question that triggers it

Here is a canonical AQA Foundation trigger, identical in shape to JUN23 Paper 3 Q1c:

Solve $c \div 4 = 8$ .

The misconception answer is $2$ , found by dividing by 4 again ( $8 \div 4 = 2$ ). Substitute it back: $2 \div 4 = 0.5$ , which is not 8 — so it cannot be right.

The correct answer is $32$ . The $c$ has been divided by 4, so undo it with the inverse — multiply both sides by 4: $c = 8 \times 4 = 32$ . Check: $32 \div 4 = 8$ ✓.

Why students fall for this

The student has a wrong but internally consistent model of "undo": they assume the operation already printed in the equation is the move that isolates the unknown. They see $\div 4$ , so they reach for $\div 4$ . No arithmetic slip is made — the wrong operation is applied confidently, which is exactly why pointing at the working never fixes it.

The same model produces the sign error on two-sided equations. To move $4x$ from the right of $7x - 22 = 4x + 29$ , the student writes $7x + 4x$ , carrying the sign across unchanged, instead of subtracting $4x$ from both sides. A term that crosses the equals must flip; keeping its sign repeats the operation rather than reversing it.

The fix: Undo with the inverse, on both sides

To undo an operation, apply its inverse to both sides. The pairs are $+ \,/\, -$ and $\times \,/\, \div$ : add undoes subtract, multiply undoes divide, and vice versa. Equivalently, a term flips its sign when it crosses the equals sign.

For $c \div 4 = 8$ : the inverse of dividing by 4 is multiplying by 4, so $c = 8 \times 4 = 32$ . For $-2 + y = 10$ : add 2 to both sides (or move the $-2$ across, flipping it to $+2$ ), giving $y = 12$ . Always finish by substituting back to check the equation balances.

Worked example

Solve $7x - 22 = 4x + 29$ (AQA JUN24 Paper 1 Q24, adapted).

Eliminate the smaller x-term with its inverse. The right side has $+4x$ , so subtract $4x$ from both sides: $3x - 22 = 29$ .
Undo the −22. Add 22 to both sides: $3x = 51$ .
Undo the ×3. Divide both sides by 3:
$x = \dfrac{51}{3} = 17$
Check. $7(17) - 22 = 119 - 22 = 97$ and $4(17) + 29 = 68 + 29 = 97$ — both sides equal 97 ✓. The no-flip shortcut $7x + 4x = 51$ gives $11x = 51$ , which does not check.

Every step removed something by doing the opposite to both sides. The term that crossed the equals flipped its sign.

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Common questions

How do I undo dividing by a number when solving?: Multiply by the same number, on both sides — multiplying is the inverse of dividing. So $c \div 4 = 8$ gives $c = 8 \times 4 = 32$ . Never divide again; that repeats the operation instead of reversing it.
Why does a term change sign when it crosses the equals sign?: Because moving a term across is shorthand for doing the inverse to both sides. In $-2 + y = 10$ , removing the $-2$ means adding 2 to both sides, so it appears on the right as $+2$ : $y = 10 + 2 = 12$ . The sign flips from minus to plus.
How do I solve an equation with x on both sides?: Get all the x-terms on one side by subtracting the smaller one from both sides. For $7x - 22 = 4x + 29$ , subtract $4x$ : $3x - 22 = 29$ , then $3x = 51$ , so $x = 17$ . Subtract — do not add — because $+4x$ flips when it crosses.
How do I make a letter the subject of a formula?: Invert every operation attached to that letter, on both sides. To make $w$ the subject of $y = w - 1$ , add 1 to both sides: $w = y + 1$ . The $-1$ inverts to $+1$ ; leaving $w = y - 1$ never reversed it.

Related misconceptions

Fraction of an amount: why 2/5 of 1020 is 408, not 40Another 'read the instruction, not the surface' trap: 'of' means divide by the bottom, then multiply by the top.
Decimal place valuePlace-column reasoning underpins the arithmetic you lean on when checking a solved equation.
Error intervals and boundsRounding and bounds: another place where reversing the instruction correctly decides the answer.

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