Printable · GCSE Foundation · ages 14-16
Algebra worksheet — GCSE Foundation
Fifteen questions across the algebra statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Algebra worksheet — GCSE Foundation
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- 1.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
- 2.Solve x² = 81.
- 3.Write down the y-intercept of the line with equation y = 3x + 8.y = 3x + 8
- 4.Grace has t books. Noah has 1 book fewer than Grace. Write down an expression for the number of books Noah has.
- 5.Explain why the equation 4x + 3 = 4x + 10 has no solution for x.
- 6.In a shop a shirt costs £x and a pair of trousers costs £(2x + 30). Together the two items cost at least £150. Work out the lowest possible price of the shirt.
- 7.Work out the value of 5a + 2 when a = 3
- 8.Solve 6x − 5 = 3x + 13.
- 9.Work out the value of m² + 3m when m = −2
- 10.To rearrange the formula y = 3x − 8 to make x the subject, Priya writes: 'Add 8 to both sides to get y + 8 = 3x, then divide both sides by 3 to get x = y + 8 ÷ 3.' Work out the correct expression for x.y = 3x − 8
- 11.Make x the subject of the formula y = 4x − 3.y = 4x − 3
- 12.A sequence has the position-to-term rule n² + 1, where n is the position number. Work out how much bigger the 5th term is than the 4th term.
- 13.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.
- 14.Work out the equation of the straight line that passes through (−2, 3) and (4, −9).
- 15.A taxi firm charges a fixed fee of £3.50 plus £2.20 per mile. Work out the total cost of a journey of 6 miles.
Answer key
- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (d) x = 9 or x = −9 — Taking the square root of both sides of x² = 81 gives x = ±9, i.e. x = 9 or x = −9, since both 9² and (−9)² equal 81. A candidate who forgets the negative square root gives only x = 9. A candidate who halves 81 instead of taking its square root gets x = 40.5. A candidate who squares 81 instead of taking its square root gets x = 6561.
- (d) 8 — Method: compare the equation with y = mx + c, where c is the y-intercept. Working: in y = 3x + 8, the constant term is 8. Answer: the y-intercept is 8. 3 comes from confusing the y-intercept with the gradient. −8 comes from a sign error, treating the constant term as negative. 11 comes from wrongly adding the gradient and the constant term together.
- (b) t − 1 — Method: "fewer than" means a subtraction, and the amount taken away is taken away from the quantity being compared with. Working: Grace has t books and Noah has one book fewer, so Noah has t with 1 taken off, written t − 1. Answer: t − 1. The distractors: t + 1 comes from reading "fewer" as "more" and adding instead of subtracting; 1 − t comes from writing the subtraction the wrong way round, taking Grace's books from the 1 rather than the 1 from Grace's books; t comes from giving the number of books Grace has, which is not the person the question asks about.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (d) £40 — Method: add the two prices to make one expression in x, turn 'at least' into ≥, solve the inequality, then read the lowest possible price off the boundary of the solution set. Working: the two items cost x + (2x + 30) = 3x + 30 pounds, so 3x + 30 ≥ 150; subtracting 30 from both sides gives 3x ≥ 120; dividing both sides by 3 gives x ≥ 40, and the smallest value the shirt price is allowed to take is the boundary, £40. Answer: £40. The distractors: £50 comes from leaving the £30 out of the total and solving 3x ≥ 150; £60 comes from using the trousers expression on its own, 2x + 30 ≥ 150; £120 comes from stopping at 3x ≥ 120 and reading the 120 as the price of the shirt without dividing by 3.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
- (d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
- (d) −2 — Method: substitute the value into both terms, working out the index and the multiplication before the addition. Working: m² = (−2) × (−2) = 4 and 3m = 3 × (−2) = −6, so the expression becomes 4 + (−6), which is −2. Answer: −2. The distractors: −10 comes from squaring −2 as −4, giving −4 + (−6); 10 comes from working out 3m as +6 and losing the minus sign, giving 4 + 6; −14 comes from working from left to right instead of multiplying first, giving (4 + 3) × (−2).
- (d) x = (y + 8) / 3 — Method: add 8 to both sides, then divide the whole side by 3. Working: y = 3x − 8, so adding 8 to both sides gives y + 8 = 3x, then dividing both sides by 3 gives x = (y + 8) / 3. The value x = y + 8 / 3 is Priya's version, which comes from dividing only the 8 by 3 instead of dividing the whole expression y + 8 by 3. The value x = (y − 8) / 3 comes from a sign error, keeping −8 instead of moving it to +8. The value x = 3(y + 8) comes from multiplying by 3 instead of dividing.
- (d) x = (y + 3)/4 — Two operations have been applied to x: it has been multiplied by 4 and then 3 has been subtracted. Undo them in the opposite order. Adding 3 to both sides gives y + 3 = 4x. Dividing both sides by 4 then gives (y + 3)/4 = x, so x = (y + 3)/4. The bracket matters: writing y/4 + 3 divides only the y by 4 and leaves the 3 untouched. Writing (y − 3)/4 subtracts the 3 instead of adding it, and writing 4(y + 3) multiplies by 4 rather than dividing.
- (a) 9 — Method: the gaps in this sequence are not constant, so work out each of the two terms named from the rule and then subtract the earlier from the later. Working: the 5th term is 5² + 1 = 25 + 1 = 26 and the 4th term is 4² + 1 = 16 + 1 = 17, so the difference is 26 − 17. Answer: 9. The distractors: 7 comes from using the 3rd and 4th terms, one position too early, 17 − 10; 11 comes from using the 5th and 6th terms, one position too late, 37 − 26; 1 comes from subtracting the position numbers, 5 − 4, instead of the terms themselves.
- (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.
- (a) y = −2x − 1 — Method: the gradient is the change in y divided by the change in x with both differences taken in the same order, and the constant then comes from substituting either point into y = mx + c. Working: m = (−9 − 3) ÷ (4 − (−2)) = (−12) ÷ 6 = −2, so the line is y = −2x + c; substituting (−2, 3) gives 3 = −2 × (−2) + c = 4 + c, so c = 3 − 4 = −1. Answer: y = −2x − 1. The distractors: y = −2x + 1 comes from rearranging 3 = 4 + c the wrong way round and taking the constant as 4 − 3; y = 2x + 7 comes from losing the minus sign when −12 is divided by 6 and then substituting correctly, 3 = 2 × (−2) + c; y = −(1/2)x + 2 comes from writing the gradient upside down as the change in x over the change in y, 6 ÷ (−12).
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
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