Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Algebra worksheet — GCSE Higher
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- 1.Line C passes through (0, 0) and (4, 2). Line D passes through (0, 0) and (2, 2). Write down which of the two lines is the steeper.
- 2.A curve has equation y = x³ − 4x. Work out the coordinates of the point where the curve crosses the y-axis.y = x
- 3.Solve the simultaneous equations 5x − 2y = 16 and 3x + 2y = 16. Work out the value of x.
- 4.A candidate solves the simultaneous equations y = x − 2 and y = x² − 4x + 2 by substitution. They write: "x − 2 = x² − 4x + 2, so x² − 3x + 4 = 0." Which of these is a correct comment on the candidate's working?y = x − 2y = x² − 4x + 2
- 5.Which expression is equivalent to 6x − (2x − 5)?
- 6.The graph of y = f(x) has a root (an x-intercept) at x = 5. Work out the x-coordinate of the corresponding root on the graph of y = f(x + 2).
- 7.Which of these quadratic graphs does NOT cross the x-axis at all?
- 8.Which expression means 'add 3 to n, then multiply the result by 4'?
- 9.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.
- 10.Solve 7x − 5 = 3x + 11
- 11.Solve 2x² − 32 = 0.
- 12.Solve 5x² − 15x = 0.
- 13.A plumber charges a call-out fee of £30 plus £25 per hour worked. Work out the total charge for a job that takes 3 hours.
- 14.A hall is set out in rows of chairs for an assembly. Row 1 has 50 chairs, row 2 has 45 chairs, row 3 has 40 chairs, and row 4 has 35 chairs, with each row after the first having 5 fewer chairs than the one before. Work out an expression, in terms of n, for the number of chairs in row n.
- 15.Solve the inequality 4x + 1 > 2x + 9.
Answer key
- (c) Line D — Method: the steeper of two lines is the one whose gradient is greater, and the gradient of a line through two points is the change in y divided by the change in x. Working: line C rises 2 for a run of 4, so its gradient is 2 ÷ 4 = 1/2; the other line rises 2 for a run of 2, so its gradient is 2 ÷ 2 = 1; since 1 is greater than 1/2, it is line D that is the steeper. Answer: Line D. The distractors: Line C is chosen by comparing the x-coordinates and calling the line that reaches further along the x-axis the steeper one; They are equally steep comes from comparing only the y-coordinates, which are both 2, without dividing by the two different runs; There is not enough information comes from believing that a gradient can only be measured off a drawn graph, when two points on a line are enough to work it out.
- (c) (0, 0) — A curve crosses the y-axis where x = 0. Substituting x = 0 into y = x³ − 4x gives y = 0³ − 4(0) = 0 − 0 = 0, so the curve crosses the y-axis at (0, 0). A candidate who reads off the coefficient of x as the y-intercept, instead of working out the constant term, might write (0, −4). A candidate who swaps the coordinates might write (4, 0). A candidate who takes the coefficient of x but drops its sign might write (0, 4).
- (b) 4 — Adding the two equations: the y-terms, −2y and +2y, cancel, and the x-terms combine to 5x + 3x = 8x; the right-hand sides add to 16 + 16 = 32. This gives 8x = 32, so x = 4. A candidate who adds only one of the right-hand sides, instead of both, would get 8x = 16, so x = 2. A candidate who divides 32 by 4 instead of 8 would get x = 8. A candidate who subtracts the equations instead of adding them, getting 2x − 4y = 0, and then wrongly assumes y = 0, would get x = 0.
- (c) Wrong — the correct equation is x² − 5x + 4 = 0. — Rearranging x − 2 = x² − 4x + 2 by moving every term across: 0 = x² − 4x + 2 − x + 2 = x² − 5x + 4. The candidate's x-term is wrong: combining −4x with the moved −x gives −4x − x = −5x, not −3x. So the correct equation is x² − 5x + 4 = 0. Distractor routes: "x² − 3x + 4 = 0 is correct" simply agrees with the candidate's sign error without checking it. "It should be x² − 5x − 4 = 0" correctly spots that something is wrong but blames the wrong term — the constant, 2 + 2 = 4, is correct; the error is in the x-term. "Subtracting is the same as adding" states something false: subtracting a term and adding it give different signs.
- (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.
- (a) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
- (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
- (c) 4(n + 3) — 'Add 3 to n' must happen before 'multiply the result by 4', so the addition needs brackets to show it happens first: 4(n + 3). Writing 4n + 3 multiplies n by 4 immediately and only adds the 3 afterwards, which reverses the order the words describe. Writing n + 3 × 4 only multiplies the 3 by 4, and adds n as a separate, unmultiplied term — it treats 'the result' as just the 3, not the whole of n + 3. Writing 3(n + 4) keeps the correct structure but swaps which number is added and which is multiplied. The expression for 'add 3 to n, then multiply the result by 4' is 4(n + 3).
- (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) 4 — Method: collect the x terms on one side and the number terms on the other. Working: subtract 3x from both sides: 4x − 5 = 11. Add 5 to both sides: 4x = 16. Divide by 4: x = 4. Answer: 4. 0.6 comes from adding the x terms instead of subtracting when collecting them, 7x + 3x = 10x, and also subtracting the constants the wrong way round, 11 − 5 = 6, giving 10x = 6. 1.5 comes from correctly collecting the x terms as 4x but subtracting the constants the wrong way round, 11 − 5 instead of 11 + 5. −4 comes from moving the x terms to the wrong side, giving 3x − 7x instead of 7x − 3x, along with a matching sign error on the constants.
- (d) x = 4 or x = −4 — Method: divide both sides by 2 to get x² = 16, then take the square root of both sides: x = 4 or x = −4. Distractor origins: x = 4 forgets the negative root; x = 8 comes from halving 16 instead of taking its square root; x = 16 stops at x² = 16 without ever taking the square root.
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (a) 55 − 5n — The number of chairs decreases by 5 in each row after the first, so the common difference is d=−5, and the first term is a=50. The nth term is a+(n−1)d = 50+(n−1)(−5) = 50−5n+5 = 55−5n. A candidate who uses the common difference as the constant term instead of correctly finding 55, giving the constant as −5 instead, would write −5n−5. A candidate who uses the first term, 50, as the coefficient of n instead of the common difference, would write 50n−5. A candidate who does not multiply the common difference by n at all, treating the nth term as n+d instead of dn+c, would write n−5.
- (d) x > 4 — Method: collect the x terms on one side and the numbers on the other, then divide by the coefficient of x; dividing by a positive number leaves the sign as it is. Working: subtracting 2x from both sides of 4x + 1 > 2x + 9 gives 2x + 1 > 9; subtracting 1 from both sides gives 2x > 8; dividing both sides by 2 gives x > 4. Answer: x > 4. The distractors: x < 4 comes from turning the sign round while dividing by 2; x > 5 comes from adding the 1 to the 9 instead of subtracting it, giving 2x > 10; x < 5 comes from making both of those mistakes together.
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