A6 - Equivalence, identities and algebraic arguments

A few matching answers can suggest a pattern, but algebra can explain whether it always works. Learn to distinguish equations from identities, show expressions are equivalent and assess a claim using a general argument or a counterexample.

Equations, identities and permissible values

Two calculators use $(x+4)^2-8x$ and $x^2+16$ . At $x=1$ , both return $17$ . Will they agree for every real input, or was that just a lucky choice?

One matching case cannot settle it. For instance, $x^2$ and $x$ agree at both $0$ and $1$ , but disagree at $2$ : $4\ne2$ .

You already know

Expansion and collecting like terms preserve an expression's value. Keep brackets around a complete expression when subtracting it, and keep original non-zero denominator restrictions.

Decide what kind of equality is being claimed before trying to solve it.

Equation and identity

An equation states that two expressions are equal. Solving it means finding the values making that equality true. An identity is an equality true for every permissible value of its letters; $\equiv$ can emphasise this stronger claim.

The equation $2x+3=11$ has the solution $x=4$ . An identity may use $=$ too: the difference is truth across the domain, not simply the printed symbol. The distributive and collection laws used to test these statements are not on the formula sheet: learn them.

Worked example 1: an equality true for every input

Question. Is $2(x+3)=2x+6$ an identity for real $x$ ? Justify your answer algebraically.

Plan. Expand the left side and compare both sides without selecting a value of $x$ .

\begin{aligned} 2(x+3)&=2x+6 &&\text{distribute the factor}\\ 2x+6&=2x+6 &&\text{both sides are identical}\\ 6&=6 &&\text{subtract }2x\text{ from both sides} \end{aligned}

Why this step? The final true statement imposes no restriction on $x$ , so every real input works.

Answer. Yes, it is an identity for all real $x$ .

Check. At $x=-1$ , both original sides are $4$ . This checks a case; the algebra establishes all cases.

Change only the right-hand constant from $6$ to $7$ . The letter terms still cancel, but the conclusion changes.

Worked example 2: an equality with no solutions

Question. Is $2(x+3)=2x+7$ an identity for real $x$ ? If not, can any $x$ satisfy it?

Plan. Expand, then subtract the matching letter terms.

\begin{aligned} 2x+6&=2x+7 &&\text{expand the left side}\\ 6&=7 &&\text{subtract }2x\text{ from both sides} \end{aligned}

Why this step? The false statement does not depend on $x$ , so changing $x$ cannot repair it.

Answer. It is not an identity and has no solutions.

Check. The left side is always exactly one less than the right side; at $x=-1$ , they are $4$ and $5$ .

“Permissible” matters. The identity $x/x=1$ holds for every $x\ne0$ , but the original fraction is undefined at zero. Cancellation never makes that input permissible. Expressions made only from sums, products and non-negative integer powers have no such denominator exclusion for real inputs.

Watch out

Wrong: “It works at $x=0$ , so it is an identity.” The equation $x^2=x$ works there, yet fails at $x=2$ . Establish every permitted case with algebra; a numerical check alone cannot do that.

Change only the right-hand constant once more.

Showing expressions are equivalent

Equivalent expressions represent the same value throughout the stated domain.

You can transform one expression into the other, or simplify both independently to the same form. Keep each equality honest: $2(x+3)=2x=2x+6$ is a false chain because $2x$ omits the constant. Write $2(x+3)=2x+6$ instead.

Equivalent expressions

Two expressions are equivalent on a stated domain if they have the same value for every input in that domain. A valid chain of expansions, collections or factorisations can establish equivalence.

For example, a linear chain can use all three operations:

\begin{aligned} 4(x+2)-2(x-1)&=4x+8-2x+2 &&\text{expand both signed brackets}\\ &=2x+10 &&\text{collect like terms}\\ &=2(x+5) &&\text{factorise} \end{aligned}

Starting with one side and reaching the other makes the argument clear. A show-that task needs these transformations, not an unsupported repetition of the given result.

Worked example 3: the calculators agree for all inputs

Question. Show algebraically that $(x+4)^2-8x\equiv x^2+16$ for every real $x$ .

Plan. Expand the squared bracket, then collect the linear terms.

\begin{aligned} (x+4)^2-8x&=(x+4)(x+4)-8x &&\text{write the square as a product}\\ &=x^2+4x+4x+16-8x &&\text{multiply every pair}\\ &=x^2+16 &&\text{the linear terms cancel} \end{aligned}

Why this step? Both cross products are needed to produce the $8x$ that cancels the outside subtraction.

Answer. The expressions are equivalent for every real $x$ , answering the opening calculator question.

Check. At $x=2$ , $6^2-16=20$ , and $2^2+16=20$ .

Change only the operation before the outside $8x$ from subtraction to addition. The cross products now add to it instead of cancelling.

Worked example 4: a changed sign changes the identity

Question. Show algebraically that $(x+4)^2+8x\equiv x^2+16x+16$ for every real $x$ .

Plan. Use the same bracket expansion and collect all linear terms.

\begin{aligned} (x+4)^2+8x&=x^2+4x+4x+16+8x &&\text{expand the square fully}\\ &=x^2+16x+16 &&\text{add the linear coefficients} \end{aligned}

Why this step? Changing the outside sign changes the coefficient from $4+4-8=0$ to $4+4+8=16$ .

Answer. The required identity holds for all real $x$ .

Check. At $x=2$ , $6^2+16=52$ , and $4+32+16=52$ .

A constant can also emerge from two products: $(x+2)(x+3)-x(x+5)=x^2+5x+6-(x^2+5x)=6$ for every real $x$ . Brackets around the subtracted expression make both cancellations visible.

Watch out

Wrong: $(x+4)^2=x^2+16$ . Squaring a sum includes the two cross products, so the correct expansion is $x^2+8x+16$ . It is the later $-8x$ that makes Worked example 3 simplify to $x^2+16$ .

Change only the outside coefficient in Worked example 4.

Supporting claims and finding counterexamples

A general algebraic argument explains a claim using quantities that can take any permitted values.

A rectangle makes the assumptions clear. Let its length $l$ and width $w$ both be positive. Increasing its length adds a strip of positive area; algebra measures that strip for any permitted dimensions.

Worked example 5: supporting a general area claim

Question. Explain algebraically why increasing a rectangle's length by $k>0$ , while keeping width $w>0$ fixed, increases its area by $kw$ .

Plan. Subtract the original area from the new area and simplify.

\begin{aligned} \text{increase}&=(l+k)w-lw &&\text{new area minus original area}\\ &=lw+kw-lw &&\text{distribute the width}\\ &=kw>0 &&k,w\text{ are positive} \end{aligned}

Why this step? The unchanged $lw$ cancels, leaving precisely the added strip's area.

Answer. The area increases by $kw$ for every positive $l,w,k$ .

Check. With $l=5$ , $w=3$ , $k=2$ , the area rises from $15$ to $21$ , an increase of $6=2\times3$ .

Change only the length increase from $k$ to $2k$ . The width remains fixed.

Worked example 6: doubling the change

Question. Explain algebraically why increasing the length by $2k$ , with $k,w>0$ , increases the area by $2kw$ .

Plan. Compare the new and original areas in the same order.

\begin{aligned} \text{increase}&=(l+2k)w-lw &&\text{subtract original from new}\\ &=lw+2kw-lw &&\text{expand}\\ &=2kw>0 &&\text{collect and use positivity} \end{aligned}

Why this step? Only the strip's length has doubled, so the area difference also doubles.

Answer. The increase is $2kw$ throughout the stated positive domain.

Check. For $l=5$ , $w=3$ , $k=2$ , the new area is $9\times3=27$ , so the increase is $12$ , twice the previous $6$ .

Testing examples is useful for finding a false claim, but successful examples leave untested inputs. One permitted failure is enough to disprove “always”.

Counterexample

A counterexample is a permitted case that makes a universal claim false. It must satisfy the claim's assumptions. A negative input cannot disprove a claim that explicitly allows only positive inputs.

For example, “ $(a+b)^2=a^2+b^2$ for all real $a,b$ ” fails at $a=b=1$ , where $4\ne2$ . The expansion $a^2+2ab+b^2$ explains the missing term. Agreement when $a=0$ or $b=0$ cannot establish the claim generally.

Worked example 7: a permitted counterexample

Question. A student says that adding two to any real number always makes its square larger. Use $x=-3$ to assess the claim.

Plan. Compare the original square with the square after adding two.

\begin{aligned} x^2&=(-3)^2=9 &&\text{find the original square}\\ (x+2)^2&=(-1)^2=1 &&\text{add before squaring}\\ 1&<9 &&\text{compare the results} \end{aligned}

Why this step? The claim permits all real inputs, so $-3$ is a valid case and its failure disproves “always”.

Answer. The claim is false: the square decreases from $9$ to $1$ .

Check. The algebraic difference is $(x+2)^2-x^2=4x+4$ , giving $-8$ at $x=-3$ , as expected.

Change only the input from $-3$ to $-2$ . It is still within the same stated domain.

Worked example 8: a second failing input

Question. Use $x=-2$ to assess the same claim.

Plan. Calculate both squares and compare them directly.

\begin{aligned} x^2&=(-2)^2=4 &&\text{original square}\\ (x+2)^2&=0^2=0 &&\text{square after adding two}\\ 0&<4 &&\text{compare} \end{aligned}

Why this step? A counterexample needs the comparison showing failure, not merely a number with no explanation.

Answer. This input also disproves the claim: the square decreases from $4$ to $0$ .

Check. $4x+4=-4$ , which matches the change $0-4$ .

The general difference $4x+4=4(x+1)$ explains the full pattern: it is positive when $x>-1$ , zero when $x=-1$ , and negative when $x<-1$ . Adding a positive number can make a square smaller if it moves a negative input nearer zero. The algebra explains exactly when.

Watch out

Wrong: “Five successful positive inputs establish a claim about every real input.” Negative inputs and infinitely many other cases remain untested. Algebra can establish a general relationship; one permitted counterexample can disprove it.

Change only the input from Worked example 8.

Quick Recap

An identity holds for every permitted input; an equation may hold for some inputs or none.
Show equivalence with a valid chain of expansions, collections or factorisations.
Algebra supports a general claim only under its stated assumptions.
One permitted counterexample disproves “always”; many successful numerical checks do not establish it.

The opening calculators agree for every real input because $(x+4)^2-8x=x^2+8x+16-8x=x^2+16$ .

Last Accessed

Papers 1, 2 and 3

A6 - Equivalence, identities and algebraic arguments

Equations, identities and permissible values

Showing expressions are equivalent

Supporting claims and finding counterexamples

Quick Recap

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