Common Mistakes with Identities in GCSE Maths

A student looks at an identity question and thinks it seems familiar. There are brackets, letters and an equals sign. They expand, rearrange and solve for $x$. The algebra may even be correct, yet the marks disappear because an identity is not an ordinary equation.

The most common mistakes with identities in GCSE Maths usually come from that single misunderstanding. An equation is true for particular values of a variable. An identity is true for every permitted value. To answer identity questions reliably, expand both sides, collect like terms and compare matching coefficients.

In the Edexcel GCSE Mathematics specification, students are expected to know the difference between an equation and an identity and use algebra to show that expressions are equivalent. The same core skills are useful across Edexcel, AQA, OCR and Eduqas questions.

The identities checklist

Before following the student through the errors, keep this method beside you:

• Look for the identity symbol $\equiv$ or wording such as "for all values of $x$".
• Expand every bracket carefully.
• Collect like terms on each side.
• Write missing terms with coefficient $0$ if needed.
• Compare the coefficients of matching powers, such as $x^2$, $x$ and the constant term.
• Solve the resulting equations.
• Substitute your values back into the original identity to check them.

Secure expanding and collecting like terms first. The Expanding and Factorising practice booklet provides question-by-question practice with answers.

Mistake: treating an identity as an equation

The student sees

$$3(x+2)\equiv 3x+6$$

and tries to solve for $x$.

Why it happens

Most algebra questions at GCSE use an equals sign and ask for one or more solutions. That familiar routine can take over. But this statement is true for every value of $x$ because expanding the left-hand side produces the right-hand side exactly.

For example, when $x=1$, both sides equal $9$. When $x=10$, both sides equal $36$. These are not the only solutions. They are simply two checks from infinitely many possible values.

Correct method

Expand the left-hand side:

$$3(x+2)=3x+6$$

Both expressions are identical, so write:

$$3(x+2)\equiv 3x+6$$

The symbol $\equiv$ communicates that the expressions are equivalent for every value of $x$. By contrast,

$$3(x+2)=21$$

is an equation with the particular solution $x=5$.

This distinction matters because an identity question is usually asking you to compare expressions, prove equivalence or find unknown coefficients. It is not asking you to find a value of the variable.

Mistake: comparing terms before expanding

Now the student meets:

$$a(x+3)+b\equiv 4x+11$$

They compare what they can see and write $a=4$ and $b=11$.

Why it happens

The right-hand side is already simplified, while the left-hand side is not. The constant contribution from $a(x+3)$ is still hidden inside the bracket.

Correct method

Expand the left-hand side first:

$$a(x+3)+b\equiv ax+3a+b$$

Now compare it with:

$$4x+11$$

The coefficient of $x$ must match:

$$a=4$$

The constant terms must also match:

$$3a+b=11$$

Substitute $a=4$:

$$3(4)+b=11$$ $$12+b=11$$ $$b=-1$$

Check these values in the original expression:

$$4(x+3)-1=4x+12-1=4x+11$$

The expressions now match for every value of $x$.

You can use the Maths Genie Proof lesson and Question Bank to practise the same expand, compare and justify method without having to assemble the questions yourself.

A stick-figure algebra detective searches for a missing minus sign hiding behind a revision mug

Mistake: losing a negative sign

The student is asked to find $p$ and $q$ when

$$p(x-2)+q\equiv 5x-13$$

They expand the left-hand side as $px+2p+q$. That small sign error changes the entire problem.

Why it happens

Multiplication must apply to every term inside a bracket, including its sign. Here, multiplying $p$ by $-2$ gives $-2p$.

Correct method

Expand accurately:

$$p(x-2)+q\equiv px-2p+q$$

Compare the coefficients of $x$:

$$p=5$$

Compare the constant terms:

$$-2p+q=-13$$

Substitute $p=5$:

$$-2(5)+q=-13$$ $$-10+q=-13$$ $$q=-3$$

Finally, check:

$$5(x-2)-3=5x-10-3=5x-13$$

The student who checks by substitution catches many identities mistakes before moving on.

Mistake: matching the wrong coefficients

Consider:

$$(x+m)(x+n)\equiv x^2+7x+12$$

The student writes $m=7$ and $n=12$ because those numbers appear on the right.

Why it happens

The constants inside the brackets do not become the coefficients on the right individually. Expanding shows how they combine.

Correct method

Expand the left-hand side:

$$(x+m)(x+n)\equiv x^2+(m+n)x+mn$$

Now compare corresponding coefficients:

$$m+n=7$$

and

$$mn=12$$

Two numbers with sum $7$ and product $12$ are $3$ and $4$. Therefore:

$$(m,n)=(3,4)$$

or

$$(m,n)=(4,3)$$

Both orders work because the two brackets can be reversed. Expanding verifies the result:

$$(x+3)(x+4)=x^2+7x+12$$

For wider algebra revision, use the free GCSE Maths revision hub, where you can select your exam board and find relevant revision material. Edexcel students can also go directly to the Edexcel GCSE Maths revision page.

Mistake: ignoring a missing term

A question may state:

$$(x+r{)}^2\equiv x^2+6x+s$$

The student focuses only on the constant $s$ and forgets that the middle term determines $r$.

Expand the square:

$$(x+r{)}^2=x^2+2rx+r^2$$

Compare the coefficients of $x$:

$$2r=6$$ $$r=3$$

Then compare the constant terms:

$$s=r^2$$ $$s=9$$

A useful habit is to align expressions by descending powers:

$$x^2+2rx+r^2\equiv x^2+6x+9$$

If one side had no $x$ term, its coefficient would be $0$. Writing that zero explicitly can prevent you from comparing the wrong parts.

Exam-style practice

Try these without looking back at the examples. Show your expansion and coefficient comparisons because method marks may depend on visible working.

Question A

Find $a$ and $b$:

$$a(x+4)+b\equiv 6x+19$$

Question B

Find $p$ and $q$:

$$(x+p)(x+q)\equiv x^2+9x+20$$

Question C

Find $r$ and $s$:

$$(2x+r)(x+3)\equiv 2x^2+sx+12$$

Answers

For Question A, comparing coefficients gives $a=6$. Then $4a+b=19$, so $b=-5$.

For Question B, $p+q=9$ and $pq=20$. Therefore $(p,q)=(4,5)$ or $(5,4)$.

For Question C, the constant term gives $3r=12$, so $r=4$. Expanding gives

$$(2x+4)(x+3)=2x^2+10x+12$$

so $s=10$.

Turn careful working into dependable marks

The student’s progress does not come from memorising another isolated rule. It comes from slowing down at the right moments: identify the identity, expand fully, align like terms, compare coefficients and check the result. Knowing that routine is useful, but following it under exam pressure takes deliberate practice.

Start the Maths Genie Lessons and Question Bank and practise correcting the identity errors that could otherwise cost you marks.