Why is 9x + y − 6x + y equal to 3x + 2y and not 5xy?

Because x-terms and y-terms are unlike terms — they are different kinds of thing, like apples and bananas. You can only add terms that share exactly the same letters. The x-pile is 9x − 6x = 3x, and the y-pile is y + y = 2y, giving 3x + 2y. Mashing all four terms into 5xy glues different letters together, which changes the meaning of the expression.

How do you factorise 12a + 15b?

Find the highest common factor of 12 and 15, which is 3. Take 3 outside a bracket: 12a + 15b = 3(4a + 5b). Check by expanding: 3 × 4a = 12a and 3 × 5b = 15b. The common mistake is multiplying the terms together to get 27ab, but 27ab does not expand back to 12a + 15b.

Why is the simplified form of 3a + 5c − a + 6c equal to 2a + 11c and not 4a + 11c?

Because the term in front of the second a is negative: it is −a, which is the same as −1a. The a-pile is 3a − a = 3a − 1a = 2a, not 4a. Missing the minus sign is the sign-on-collected-term trap. The c-pile is 5c + 6c = 11c, so the answer is 2a + 11c.

GCSE Maths Foundation · AQA · Algebraic manipulation

Combining unlike terms: why 9x + y − 6x + y is 3x + 2y, not 5xy

Algebra simplification questions trip students who treat every number and letter as fair game to combine into a single term. Faced with $9x + y - 6x + y$ , a student mashes all four terms together and writes $5xy$ . The same root cause has a second face: asked to factorise $12a + 15b$ , a student multiplies the terms to give $27ab$ instead of pulling out the common factor.

The thirty-second fix: you can only add terms that share exactly the same letters. Sort into one pile per letter (and a pile for plain numbers), add within each pile, and keep different piles separate in the answer — never mash unlike terms into one.

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How to spot it in your own work

You wrote $5xy$ for $9x + y - 6x + y$ by combining all four terms into one.
You wrote $27ab$ when asked to factorise $12a + 15b$ , multiplying the terms together instead of pulling out the common factor.
You wrote $4a + 11c$ for $3a + 5c - a + 6c$ by reading the second $a$ as positive instead of negative.
You slid a plain number into a letter term, for example turning $4p + 5 + 2p + 3$ into $14p$ .

An exam question that triggers it

Here is the canonical AQA Foundation trigger, the shape of JUN22 Paper 3 Q12a:

Simplify $9x + y - 6x + y$ .

The misconception answer is $5xy$ , found by treating all four terms as one combined pile. But $x$ -terms and $y$ -terms are different kinds of thing and must stay in separate piles.

The correct answer is $3x + 2y$ . The $x$ -pile: $9x - 6x = 3x$ . The $y$ -pile: $y + y = 2y$ . Answer: $3x + 2y$ .

Why students fall for this

When a student sees several terms in a row they reach for the most heavily over-practised habit they own: add everything up into one answer. The expression $9x + y - 6x + y$ fires the same reflex as $9 + 1 - 6 + 1$ , so they squash the numbers ( $9 - 6 = 3$ , then add the stray $1$ s) and glue the letters on, producing $5xy$ . The arithmetic on the digits is right; the reading of the letters is wrong. Different letters represent different unknown quantities — $x$ and $y$ are no more combinable than apples and bananas.

The same flat reading wrecks factorising. Asked for the factorised form of $12a + 15b$ , a student sees two terms and "adds" them in the sense of merging — $12 + 15 = 27$ , glue the letters, $27ab$ . But factorising is the reverse of expanding: find the highest common factor and take it outside a bracket. The mash answer $27ab$ does not expand back to the original expression.

The fix: One pile per letter — collect within each pile, keep piles separate

Collect only like terms: sort into one pile per letter, plus a pile for plain numbers. Add or subtract within each pile. Keep the piles separate in the answer — never merge unlike piles into one term. For $9x + y - 6x + y$ : the $x$ -pile gives $9x - 6x = 3x$ , and the $y$ -pile gives $y + y = 2y$ , so the answer is $3x + 2y$ .

Factorise by pulling the common factor out — the reverse of expanding. For $12a + 15b$ : the highest common factor of $12$ and $15$ is $3$ , so $12a + 15b = 3(4a + 5b)$ . Check by expanding: $3 \times 4a = 12a$ and $3 \times 5b = 15b$ . Never multiply the terms together.

Worked example

Simplify $9x + y - 6x + y$ .

Sort the x-terms into one pile. $9x$ and $-6x$ are like terms (same letter). $9x - 6x = 3x$ .
Sort the y-terms into a second pile. $y$ and $y$ are like terms. $y + y = 2y$ .
Write the answer with both piles separate.
$9x + y - 6x + y = 3x + 2y$

The trap answer $5xy$ comes from treating all four terms as one pile. $x$ and $y$ are unlike terms — they represent different unknowns and stay in separate piles.

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Common questions

How do I simplify an expression with two different letters?: Sort into one pile per letter. For $8m + 3n - 2m + n$ : the $m$ -pile is $8m - 2m = 6m$ , and the $n$ -pile is $3n + n = 4n$ . Answer: $6m + 4n$ . Never merge the two piles into $9mn$ .
Why isn't $12a + 15b$ equal to $27ab$ when factorised?: Because $27ab$ is a multiplication of the two terms, not a factorisation. Factorising means pulling a common factor outside a bracket: $12a + 15b = 3(4a + 5b)$ . Check by expanding: $3 \times 4a = 12a$ and $3 \times 5b = 15b$ . The product $27ab$ does not expand back to $12a + 15b$ .
Why is the answer 2a + 11c, not 4a + 11c, for 3a + 5c − a + 6c?: The second $a$ -term is $-a$ , not $+a$ . A single $a$ on its own is $1a$ , so the $a$ -pile is $3a - 1a = 2a$ , not $3a + 1a = 4a$ . Always carry the sign in front of a term into its pile.
Can plain numbers combine with letter terms?: No — plain numbers and letter terms are unlike. In $4p + 5 + 2p + 3$ : the $p$ -pile is $4p + 2p = 6p$ , and the numbers-pile is $5 + 3 = 8$ , giving $6p + 8$ . The $5$ and $3$ never join the $p$ -pile to make $14p$ .

Related misconceptions

Expanding brackets: multiplying by the first term onlyThe outside factor must multiply every term inside — the same care with each separate term that collecting like terms demands.
Fraction of an amount: why 2/5 of 1020 is 408, not 40A related reading error: treating an instruction as a label to copy, instead of an operation to apply to each part.
Function machines: why 12 → [−4] → [×5] is 40, not −8Another spot where reading the structure of an expression — sequence of operations — matters as much as the arithmetic.

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