A6 - Algebraic arguments and proof | AQA GCSE Mathematics

Algebra lets you explain why a pattern works, rather than just check a few examples. You will distinguish equations from identities, show that expressions are equivalent, and use algebra to test claims. The final Higher part builds complete proofs about whole numbers.

Equations and identities

Think of a number, multiply it by $5$ , add $15$ , divide by $5$ , then subtract your starting number. Will everyone get $3$ ? Trying a few numbers gives a clue. Algebra will settle the question for every starting number at once.

You already know

An expression, such as $3x+6$ , has no equality sign. Expanding a bracket multiplies every term inside: $3(x+2)=3x+6$ .

Compare $x+3=7$ with $3(x+2)=3x+6$ . The first is true at $x=4$ , but false at $x=5$ . The second is true for every value of $x$ : both sides perform the same calculation in different forms.

Equation and identity

An equation states that two expressions are equal. It may be true for some values, all values, or no values of its variables. An identity is an equation true for every allowed value of its variables.

For example, $3(x+2)\equiv3x+6$ is an identity. The symbol $\equiv$ emphasises that it is always true. The equation $x+3=7$ is not an identity.

“Allowed” matters when a denominator is present. An identity involving $1/x$ cannot include $x=0$ . For ordinary polynomial expressions, every real value of $x$ is allowed. These definitions and the identity symbol are not on the formulae sheet: learn them.

Key idea

An identity describes the same value in two forms. Solving an equation asks which values make it true; showing an identity asks why it is true for every allowed value.

Even a long run of examples can fool you. The expression $n^2+n+41$ gives a prime for every whole number $n$ from $0$ to $39$ . Yet at $n=40$ , it gives $1681=41^2$ , which is not prime. Forty successes do not prove the next case.

Showing expressions are equivalent

To show that two expressions are equivalent, transform one into the other using valid algebra. You can also simplify them separately until they have exactly the same form.

Equivalent means equal for every allowed value. Substituting one value gives a check, but it does not show that the expressions agree for all the others.

Equivalent expressions

Expressions with the same value for every allowed value of their variables. For example, $4x+8$ and $4(x+2)$ are equivalent. The expressions $x+2$ and $2x$ agree at $x=2$ , but are not equivalent.

The expansion and collection rules used here are not on the formulae sheet: learn them. In a “Show that” question, write the algebraic steps even though the answer is supplied. Both examples are non-calculator.

Worked example 1: addition outside a bracket

Question. Show that $3(x+4)+2x$ is equivalent to $5x+12$ .

Plan. Expand the bracket, then collect the terms containing $x$ .

\begin{aligned} 3(x+4)+2x&=3x+12+2x&&\text{multiply both bracket terms}\\ &=5x+12&&\text{collect like terms} \end{aligned}

Why this step? Expansion exposes the two $x$ -terms, which can then be combined. No particular value of $x$ has been chosen.

Answer. Both expressions are equal for every $x$ , so they are equivalent.

Check. At $x=2$ , the original gives $18+4=22$ , and $5x+12$ also gives $22$ .

Change only the operation before $2x$ to subtraction. Collect with its sign attached.

Worked example 2: subtraction outside a bracket

Question. Show that $3(x+4)-2x$ is equivalent to $x+12$ .

Plan. Expand, then subtract the like terms.

\begin{aligned} 3(x+4)-2x&=3x+12-2x&&\text{expand the bracket}\\ &=x+12&&\text{combine }3x-2x \end{aligned}

Why this step? The minus belongs to the $2x$ -term. It remains a subtraction after the bracket is expanded.

Answer. The expressions are equivalent for every $x$ .

Check. At $x=2$ , $18-4=14$ , agreeing with $2+12=14$ .

Equivalence is about the complete expressions. Expanding only part of a bracket can create a plausible-looking but different expression.

Watch out

Wrong: $4(x+3)=4x+3$ . The factor $4$ multiplies both terms. Right: $4(x+3)=4x+12$ . At $x=0$ , the wrong expression gives $3$ instead of $12$ .

Change the multiplier and show every step of the same argument.

Using algebra to test a claim

Choose a letter for the starting number and follow the instructions exactly. A simplified result can explain whether the starting number still matters.

Let the starting number be $n$ . Multiplying it by $2$ gives $2n$ ; adding $6$ gives $2n+6$ . If the next step divides the result by $2$ , it divides that whole expression.

Dividing a sum

\frac{an+b}{a}=n+\frac ba\qquad(a\ne0)

The denominator divides both terms in the numerator.

Not on the formulae sheet: learn it.

These non-calculator examples turn number instructions into an algebraic argument.

Worked example 3: a fixed result

Question. A number is multiplied by $2$ , then $6$ is added. The result is divided by $2$ , then the starting number is subtracted. Sam says the answer is always $3$ . Use algebra to show that Sam is correct.

Plan. Call the starting number $n$ and simplify the complete calculation.

\begin{aligned} \frac{2n+6}{2}-n&=n+3-n&&\text{divide both numerator terms}\\ &=3&&\text{the }n\text{-terms cancel} \end{aligned}

Why this step? The letter represents any starting number. Once it cancels, the result cannot depend on which number was chosen.

Answer. Sam is correct: the result is always $3$ .

Check. Starting with $4$ gives $(8+6)/2-4=7-4=3$ .

Now change the multiplier and matching divisor from $2$ to $3$ . Keep the added number $6$ , and see which fixed result survives.

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A6 - Algebraic arguments and proof

Equations and identities

Showing expressions are equivalent

Using algebra to test a claim

Continue this lesson.

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