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.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?
- 2.Which line of algebra shows that the sum of two consecutive odd numbers is always a multiple of 4?
- 3.Which expression is equivalent to 3(x + 4) − 2(x − 1)?
- 4.A student says 4(2x − 3) is equivalent to 8x − 3. Which statement gives the correct verdict and reason?
- 5.(2x + 3)(x + a) ≡ 2x² + 11x + 12 is an identity. Work out the value of a.
- 6.Which of these is an identity?
- 7.An equation has exactly one value of x that makes it true, but an identity is true for every value of x. Which of these best explains why 3x + 5 = 20 is an equation rather than an identity?
- 8.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.
- 9.Which expression is equivalent to 0.5(4x + 6) − x?
- 10.n is an integer. Which of these expressions is always an even number?
- 11.A photo printing service has two adverts for its price. Advert A: cost in pounds = 3(2n + 4) for n photos. Advert B: cost in pounds = 6n + 12. A customer says the two adverts always charge the same amount. Is the customer correct?
- 12.A student is proving that (n + 1)² − n² is always an odd number. Which of these correctly completes the first line of algebra?
- 13.Which expression is equivalent to 7x − 3(2x − 6)?
- 14.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
- 15.A student attempts to prove that the sum of any three consecutive integers is a multiple of 3. Line 1: Let the three consecutive integers be n, n + 1 and n + 2. Line 2: Their sum is n + (n + 1) + (n + 2) = 3n + 2. Line 3: 3n + 2 leaves a remainder of 2 when divided by 3, so it is not a multiple of 3. Line 4: So the sum of three consecutive integers is not always a multiple of 3. Which line contains the FIRST error?
Answer key
- (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.
- (a) (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1) — Two consecutive odd numbers can be written as 2n + 1 and 2n + 3, for a whole number n. Adding them: (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), which is a multiple of 4 for every whole number n, proving the general result. Using 2n + 1 twice does not represent two different numbers, so it proves nothing about a sum of two numbers; writing n + (n + 2) drops the +1 that makes the numbers odd in the first place, and only shows a multiple of 2; and check every constant term is added correctly — 1 + 3 is 4, not 3.
- (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.
- (a) False — 4(2x − 3) = 8x − 12, not 8x − 3. — Expand the bracket by multiplying both terms by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12, which is not 8x − 3 — the student is wrong. Saying 4(2x − 3) = 8x − 3 comes from multiplying only the 2x by 4 and copying the −3 across unchanged. Saying 4(2x − 3) = 2x − 12 comes from multiplying only the −3 by 4 and leaving 2x unmultiplied. Claiming it is true because both expressions are linear ignores that equivalence depends on the actual coefficients, not the type of expression.
- (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.
- (a) 2(3x + 1) = 6x + 2 — Expanding 2(3x + 1) = 6x + 2 gives an expression that matches the right-hand side exactly for every value of x — it is an identity. 4x − 3 = 3x + 5 is an ordinary equation with one solution, x = 8. 7 − x = x − 7 is also an ordinary equation with one solution, x = 7. 5x + 1 = 5(x + 1) never holds for any value of x at all, since expanding the right-hand side gives 5x + 5, and 5x + 1 = 5x + 5 would require 1 = 5, which is impossible.
- (c) Only x = 5 satisfies 3x + 5 = 20, not every value of x. — 3x + 5 = 20 is only true when x = 5, since 3 × 5 + 5 = 20; for any other value of x the two sides are not equal, so it is an equation, not an identity. Saying it cannot be simplified confuses simplifying with the equation/identity distinction, which is about how many values of x make it true. Saying it has an = sign is not a valid test, since identities are also written with an = or ≡ sign. A number on the right-hand side does not decide it either — what matters is whether both sides match for every value of x, not the form of the right-hand side.
- (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 + 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) 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) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
- (b) (n + 1)² − n² = 2n + 1 — (n + 1)² = n² + 2n + 1, so (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1, which is odd because it is one more than the even number 2n. Expanding (n + 1)² as n² + 1 uses the false rule (a + b)² = a² + b², and subtracting n² from that leaves just 1 — always expand (a + b)² as a² + 2ab + b². Writing n² + 2n + 1 expands correctly but never carries out the subtraction of n². Writing 2n forgets the constant term left after subtracting.
- (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.
- (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.
- (d) line 2 — Line 1 correctly represents three consecutive integers using n. Line 2 adds them: n + (n + 1) + (n + 2). Collecting terms: the n-terms give 3n, and the constants give 1 + 2 = 3, so the correct sum is 3n + 3, not 3n + 2 as Line 2 states — this is the first error, an arithmetic slip in collecting the constant terms. Lines 3 and 4 both follow correctly from Line 2's incorrect result, but that result itself is wrong: the true sum, 3n + 3 = 3(n + 1), is a multiple of 3 for every whole number n. Check the working of each line against what came before it, in order, rather than judging whether the final conclusion feels right — an error that flips the conclusion can sit several lines before the line that states it.
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