Printable · GCSE Foundation · ages 14-16
Expressions, equations, formulae, identities and inequalities worksheet — GCSE Foundation
Fifteen questions on "expressions, equations, formulae, identities and inequalities" — DfE statement A3. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Answer key: Expressions, equations, formulae, identities and inequalities worksheet — GCSE Foundation
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- (a) 5 — A factor of the whole expression must divide every term exactly. 15x + 20 = 5(3x + 4), so 5 is a factor. Distractor origins: 3 divides 15x exactly but does not divide 20 exactly; 4 divides 20 exactly but does not divide 15x exactly; 10 also divides 20 exactly but does not divide 15x exactly, so it is a factor of only one term, not of the whole expression.
- (d) 2n + 7 > 15 — An inequality compares two expressions using a symbol other than an equals sign, such as < or >. "2n + 7" is just an expression — it has no equals or inequality sign at all, so a candidate mistaking any bare expression for an inequality would choose this. "2n + 7 = 15" is an equation: it uses an equals sign and is only true for one particular value of n, so a candidate confusing an equation with an inequality would choose this. "P = 2n + 7" is a formula, since it relates two different letters, P and n, with an equals sign, so a candidate confusing a formula with an inequality would choose this. "2n + 7 > 15" is an inequality: it compares an expression with a number using a > sign.
- (c) 3x + 5 = 17 — An equation contains an equals sign and is true only for particular value(s) of the unknown — solving 3x + 5 = 17 gives the single value x = 4. 3x + 5 is an expression: it has no equals sign, so it cannot be solved, only simplified or evaluated. A = πr² is a formula: it shows the general relationship between different quantities (area and radius), rather than asking for one unknown value. 3(x + 5) ≡ 3x + 15 is an identity: the ≡ sign shows it is true for every value of x, not just one. The equation is 3x + 5 = 17.
- (b) £115, and C = 40 + 25h is a formula — Substitute h = 3 into the rule, multiplying before adding. The hours cost 25 × 3 = 75, and adding the call-out fee gives 40 + 75 = 115, so the charge is £115. The rule itself links two different quantities, C and h, and lets one be worked out from the other, so it is a formula; an expression would have no equals sign in it. Adding the fee before multiplying gives 65 × 3 = 195, which charges the call-out fee three times over, and stopping at 40 + 25 leaves £65, the charge for a single hour.
- (c) 5y − 2 — An expression is a collection of terms with no equals sign and no inequality sign in it: nothing is being claimed about two sides. Here 5y − 2 on its own fits that description. The line with an equals sign and one unknown letter to find is an equation; the line joined by a comparison sign is an inequality; the line that links two different letters so that one can be worked out from the other is a formula.
- (c) 5x² — A term is one part of an expression, separated from the others by a + or − sign. The expression 5x² − 3x + 7 has three terms: 5x², −3x and 7. Distractor origins: 5x² − 3x combines two terms together, not a single term; −3x + 7 also combines two terms together; 5x² − 3x + 7 gives the whole expression, not a single term.
- (a) 5x − 3 = 12 — 5x − 3 = 12 is an equation with exactly one solution: adding 3 and dividing by 5 gives x = 3, and no other value works. 5x − 3 = 5x − 3 is true for every value of x, since both sides are identical — it has infinitely many solutions, not one. 5x − 3 > 12 is an inequality: any value of x greater than 3 satisfies it, so it has a whole range of solutions, not a single one. 5x − 3 = 5x + 2 has no solution at all, since subtracting 5x from both sides leaves −3 = 2, which is never true. The equation with exactly one solution is 5x − 3 = 12.
- (d) −4 — The coefficient of a letter is the number multiplying it, taken with the sign written in front of that number. The term containing p is being subtracted, so the term is −4p and the number multiplying p is −4. Quoting 4 drops the sign; 7 is a constant term with no letter attached to it; 9 is the number multiplying q, which is a different letter.
- (d) 66 — 6 and (4x + 3) are correct factors of 24x + 18. — Substituting x=2 into 24x+18: 24×2+18=48+18=66. Checking the factorisation: 6×4x=24x and 6×3=18, so 6(4x+3)=24x+18, matching the original expression — 6 and (4x+3) are correctly identified as factors. A candidate who substitutes x=2 into (4x+3) but forgets to multiply by the factor 6 afterwards would compute just 4×2+3=11. A candidate who adds 24 and 18 together before multiplying by x, instead of multiplying 24 by x first, would compute (24+18)×2=84. A candidate who distributes the 6 over only the first term inside the brackets, forgetting the second, would check the factorisation as 6×4x=24x but leave the 3 unmultiplied, getting 24x+3 instead of 24x+18 — and would wrongly conclude that 6 and (4x+3) are not correct factors.
- (a) The equation has one solution; the inequality has many. — Method: solve each statement. From 2x + 3 = 11, 2x = 8, so x = 4 — a single value. From 2x + 3 > 11, 2x > 8, so x > 4 — every number greater than 4 makes the inequality true, so there are many solutions. So the equation has one solution and the inequality has many. Distractor origins: swapping the two round gives the range to the equation and the single value to the inequality; saying both have exactly one solution treats the > sign as if it were an = sign; saying both have many solutions treats the equation as if it were an inequality.
- (a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
- (b) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
- (c) 3 and 2n + 5 are both factors of 6n + 15 — Factors are the parts multiplied together to make an expression; terms are the parts added together. In 3(2n + 5) the 3 and the bracket 2n + 5 are multiplied, so both of them are factors of 6n + 15. The parts added together are 6n and 15, and those are its two terms, not its factors. A factor need not be a number: a bracket containing letters is a factor in exactly the same way.
- (b) 3w − 5 ≥ 16 — "Three times w, minus 5" translates to 3w − 5, and "is at least 16" means it must be 16 or more, giving 3w − 5 ≥ 16. A candidate who reads "at least" as a strict inequality writes 3w − 5 > 16. A candidate who misreads the wording and applies the subtraction before the multiplication writes 3(w − 5) ≥ 16. A candidate who reverses the direction of the inequality writes 3w − 5 ≤ 16.
- (d) V = lwh — V = lwh is a formula: it relates one quantity, the volume V, to others, the length, width and height, in a way that is true generally. A candidate who picks lwh has chosen the expression, not a full statement relating two quantities. A candidate who picks lwh = 60 has chosen an equation, since it is only true for particular values of l, w and h that multiply to 60. A candidate who picks 2(l + w) ≡ 2l + 2w has chosen an identity, mistaking the identity symbol ≡ for a sign that makes a statement a formula.
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