Printable · GCSE Higher · ages 14-16
Identities, equivalence and algebraic proof worksheet — GCSE Higher
Fifteen questions on "identities, equivalence and algebraic proof" — DfE statement A6. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Identities, equivalence and algebraic proof worksheet — GCSE Higher
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- 1.A market stall's cost of hiring n tables is modelled by two formulas: Formula A: C = 3(2n + 5); Formula B: C = 6n + 15, where C is in pounds. A stallholder says the two formulas always give the same cost. Work out the cost given by each formula when n = 4, and use your results to decide whether the stallholder is correct.
- 2.Which expression is equivalent to 7x − 3(2x − 6)?
- 3.(2x + 3)(x + a) ≡ 2x² + 11x + 12 is an identity. Work out the value of a.
- 4.A student attempts to prove that the product of two consecutive integers is always even: (i) Let the two consecutive integers be n and n + 1. (ii) Since n(n + 1) is even, one of n and n + 1 must be an even number. (iii) Therefore, n(n + 1) is even. At which statement does the proof first assume the very fact it is trying to prove?
- 5.A student claims: 'For every positive integer n, n² + n + 1 is a prime number.' Which value of n shows that this claim is false?
- 6.Which expression is equivalent to 6x − (2x − 5)?
- 7.A student says that (x + 4)² is equivalent to x² + 16. For which value of x do the two expressions give the SAME result, making it look (misleadingly) like the student could be right?
- 8.A proof that (n + 3)² − (n − 3)² is always a multiple of a certain number begins: Line 1: (n + 3)² − (n − 3)² = (n² + 6n + 9) − (n² − 6n + 9). Which expression correctly completes Line 2?
- 9.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
- 10.Two expressions are 4(x + 3) and 4x + 3. A student checks whether they are equivalent by substituting x = 2. Which statement correctly interprets the result?
- 11.Which expression is equivalent to 0.5(4x + 6) − x?
- 12.A rectangular garden has width w metres and length (w + 3) metres. A gardener writes its perimeter as 2w + 3. Which statement corrects the gardener's mistake?
- 13.Which expression is equivalent to 3(2x − 5) + 4x?
- 14.n is an integer. Which of these expressions is always an even number?
- 15.Which expression is equivalent to 3(x + 4) − 2(x − 1)?
Answer key
- (d) Formula A gives £39 and Formula B gives £39, and since 3(2n + 5) expands to 6n + 15 for every value of n, the stallholder is correct. — Formula A: 3(2 × 4 + 5) = 3 × 13 = £39. Formula B: 6 × 4 + 15 = 24 + 15 = £39. Expanding Formula A algebraically gives 3(2n + 5) = 6n + 15, which is identical to Formula B for every value of n, not just n = 4, so the stallholder is correct — this is an identity, not a coincidence. The option giving £29 for Formula A comes from multiplying only the 2n by 3 and forgetting to also multiply the 5, then adding the unmultiplied 5: 3 × 2 × 4 = 24, + 5 = 29. The two options that reach the correct numbers but reject the stallholder's claim both use faulty reasoning — matching values at one value of n, or counting terms, does not decide whether two expressions are identical for every n; expanding the bracket does.
- (c) x + 18 — Expand −3(2x − 6) by multiplying both terms by −3: −3 × 2x = −6x and −3 × (−6) = 18, giving 7x − 6x + 18 = x + 18. Writing x − 18 comes from not flipping the sign of the −6 inside the bracket, so −3 × (−6) is treated as −18 instead of +18. Writing x + 6 comes from forgetting to multiply the −6 by 3, only carrying its sign. Writing 13x − 18 comes from treating the whole bracket as being added rather than subtracted, so 3(2x − 6) = 6x − 18 is added to 7x.
- (d) a = 4 — Expand the left-hand side: (2x + 3)(x + a) = 2x² + 2ax + 3x + 3a = 2x² + (2a + 3)x + 3a. For this to match 2x² + 11x + 12 for every value of x, the x-coefficients must be equal and the constants must be equal: 2a + 3 = 11 and 3a = 12. Both give a = 4, so a = 4. Writing a = 12 comes from the constant-term equation 3a = 12: reading it as saying a itself is 12, rather than dividing both sides by 3. Writing a = 8 comes from the x-coefficient equation 2a + 3 = 11: working out 11 − 3 = 8 correctly but then stopping, without dividing by the 2 in front of a. Writing a = −4 comes from rearranging 2a + 3 = 11 the wrong way round, as 2a = 3 − 11 = −8, which gives a = −4 instead of a = 4.
- (c) Statement (ii) — Statement (ii) opens with 'Since n(n + 1) is even', treating the very fact the proof is meant to establish as if it were already known — that is circular reasoning, assuming the conclusion to help derive itself. Statement (i) only names the two consecutive integers as n and n + 1; it makes no claim about whether their product is even, so it introduces nothing circular. Statement (iii) states the conclusion, and would be a valid final step if statement (ii) had reached 'one of n and n + 1 is even' by a genuine argument, such as considering the cases where n is even or odd separately. Saying the proof assumes nothing circular is wrong, because statement (ii)'s opening clause is exactly that assumption.
- (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
- (d) 4x + 5 — A minus sign directly before a bracket changes the sign of both terms inside it: 6x − (2x − 5) = 6x − 2x + 5 = 4x + 5. The option 4x − 5 comes from only changing the sign of the 2x term and not the −5, giving 6x − 2x − 5. The option 8x − 5 comes from adding 2x instead of subtracting it, as if the minus sign did not apply to the bracket, giving 6x + 2x − 5. The option 8x + 5 repeats that same addition mistake and also changes the sign of the −5 term.
- (c) x = 0 — Expand (x + 4)² correctly: (x + 4)² = x² + 8x + 16. This equals x² + 16 only when 8x is zero, i.e. when x = 0 — at every other value of x the two expressions differ by 8x. Choosing x = 4 confuses the constant inside the bracket with the value of x that makes the expressions match. Choosing x = −4 makes the same confusion but with the sign flipped. Choosing x = 8 mistakes the coefficient of the middle term, 8x, for the value of x itself.
- (b) 12n — Distributing the minus sign across the second bracket gives n² + 6n + 9 − n² + 6n − 9, and the n² terms and the +9/−9 cancel, leaving 6n + 6n = 12n. Writing 18 comes from only negating the first term of the second bracket, n², and treating the −6n and +9 as unchanged, which gives n² + 6n + 9 − n² − 6n + 9 = 18. Writing 2n² + 18 comes from adding the two brackets instead of subtracting them, (n² + 6n + 9) + (n² − 6n + 9) = 2n² + 18. Writing 6n comes from correctly negating the bracket but then only counting one of the two 6n terms, missing that they add rather than cancel.
- (b) It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true. — Expanding the left-hand side, 3(x − 4) = 3x − 12. Setting this equal to the right-hand side, 3x − 12 = 3x − 4, gives −12 = −4 once the 3x terms are removed from both sides — a statement that is never true, so no value of x satisfies the equation at all, and it is certainly not an identity. The option about substituting a specific value misunderstands algebraic expansion, which holds for every x, not one chosen value. The option matching the first term wrongly assumes that is enough to prove equivalence. The option about multiplying the 4 by 3 on both sides is nonsensical, since there is only one bracket to expand, on the left-hand side.
- (a) 4(x + 3) = 20 and 4x + 3 = 11 when x = 2, so the two expressions are not equivalent, because the bracket means the 3 must be added before multiplying by 4. — Substituting x = 2: 4(x + 3) = 4 × 5 = 20, and 4x + 3 = 8 + 3 = 11. The two values are different, and expanding 4(x + 3) algebraically gives 4x + 12, which can never equal 4x + 3 (that would require 12 = 3) — so the two expressions are never equivalent, for any value of x. The option claiming they become equal for a larger x is wrong: 4x + 12 = 4x + 3 has no solution at all. The option claiming they are equivalent because they share the terms 4x and 3 ignores that the bracket changes the constant term. The option that calculates 4(x + 3) as 11 ignores the bracket completely, applying the 4 only to the x term.
- (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
- (a) It is 2(w + (w + 3)) = 4w + 6, not 2w + 3. — The perimeter of a rectangle is twice the width plus twice the length: 2 × w + 2 × (w + 3) = 2w + 2w + 6 = 4w + 6, so the gardener's 2w + 3 is wrong. Writing w + (w + 3) = 2w + 3 forgets to double the sides at all, only adding one width and one length once. Writing 4(w + 3) = 4w + 12 wrongly treats all four sides as equal to the length, as if the garden were a square. Writing 3w + 6 comes from doubling the length correctly but adding the width only once instead of doubling it too.
- (c) 10x − 15 — Expand the bracket first: 3(2x − 5) = 6x − 15. Then add the 4x: 6x − 15 + 4x = 10x − 15. The option 10x − 5 comes from forgetting to multiply the 5 inside the bracket by 3 (treating it as 6x − 5), then adding 4x. The option 10x + 15 comes from a sign error when expanding, treating 3 × (−5) as +15 instead of −15, then adding 4x. The option 6x − 15 comes from expanding the bracket correctly but forgetting to add the 4x term at all.
- (a) n² + n — n² + n = n(n + 1), the product of two consecutive integers. One of any two consecutive integers is always even, so their product is always even, whatever whole number n is. n² − n + 1 = n(n − 1) + 1 is always ODD, since n(n − 1) is even and adding 1 makes it odd — the opposite of what's asked. 2n + 1 is always odd by definition, not even. n² + 1 is not always even at all: it depends on whether n is odd or even, and testing n = 2 gives 5, which is odd.
- (a) x + 14 — Expand each bracket separately: 3(x + 4) = 3x + 12, and −2(x − 1) = −2x + 2 (multiply −2 by both x and −1). Combine: 3x + 12 − 2x + 2 = x + 14. Writing x + 10 comes from taking −2(x − 1) as −2x − 2, not flipping the sign of the −1 inside the bracket. Writing 5x + 10 comes from treating the second bracket as +2(x − 1) instead of subtracting it, so the x-terms are added rather than subtracted. Writing x + 13 comes from only multiplying the 2 by the x, and carrying the −1 across unmultiplied.
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