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.Which expression is equivalent to 4(x − 2) + 10?
- 2.The line y = 3x + 6 meets the x-axis at one point. Work out the coordinates of that point.y = 3x + 6
- 3.The simultaneous equations x = 2 and x + y = 9 are given. Work out the value of y.
- 4.Which of these statements is an identity?
- 5.The diagram shows a distance–time graph for a cyclist travelling at a constant speed, for the first 6 minutes of a journey. Distance is in kilometres and time is in minutes. Work out how far the cyclist would travel in 20 minutes at the same speed.
- 6.A number machine multiplies its input by 2 and then subtracts 5. Work out the output when the input is 6.
- 7.Which expression is equivalent to 2(x + 6)?
- 8.Solve the simultaneous equations x + y = 1 and 2x + y = 5.
- 9.A curve has equation y = x³ − 4x. Work out the coordinates of the point where the curve crosses the y-axis.y = x
- 10.Solve 3x + 14 = 8x − 6
- 11.The formula connecting speed, distance and time is v = s ÷ t. Work out v, in km/h, when s = 180 and t = 4.
- 12.A company's profit was £2000 in its first year. Each following year, the profit increases by £800. Work out the first year in which the profit is more than £7000.
- 13.A sequence has the position-to-term rule n² − 3, where n is the position number. Work out the difference between the 6th term and the 5th term.
- 14.Leo is three times as old as his brother. His brother is x years old. Write down an expression for Leo's age.
- 15.Which of these equations is equivalent to 5x − 2 = 3x + 8?
Answer key
- (a) 4x + 2 — Expand the bracket: 4(x − 2) = 4x − 8. Then add the 10 that follows the bracket: 4x − 8 + 10 = 4x + 2. The option 4x − 8 comes from expanding the bracket correctly but then forgetting to add the 10. The option 4x + 8 comes from forgetting to multiply the 2 inside the bracket by 4 (treating it as 4x − 2 instead of 4x − 8), then adding 10. The option 4x + 18 comes from a sign error when expanding — treating 4 × (−2) as +8 instead of −8, giving 4x + 8, then adding 10.
- (b) (−2, 0) — Method: a graph meets the x-axis where the y-value is 0, so setting y = 0 turns the equation into a linear equation in x. Working: 0 = 3x + 6 gives 3x = −6, so x = (−6) ÷ 3 = −2 and the meeting point is (−2, 0). Answer: (−2, 0). The distractors: (2, 0) comes from solving 3x = −6 and then dropping the minus sign from the result; (0, 6) is the y-axis crossing, found by substituting x = 0 instead of y = 0; (6, 0) comes from reading the constant 6 straight off as the x-coordinate, without dividing by 3 and without changing its sign.
- (b) y = 7 — Method: one of the equations already gives the value of x, so that value is substituted into the other equation. Working: replacing x with 2 in x + y = 9 gives 2 + y = 9, so y = 9 − 2 = 7. Answer: y = 7, and the pair x = 2 with y = 7 satisfies both equations. The distractors: y = 11 comes from rearranging x + y = 9 as y = 9 + x and adding instead of subtracting; y = 9 comes from ignoring the x term and reading the total as the value of y; y = 4.5 comes from halving 9, as though x and y had to be equal.
- (c) 2(x + 4) ≡ 2x + 8 — 2(x + 4) ≡ 2x + 8 is an identity because expanding the bracket on the left gives exactly the right-hand side for every value of x — try any number and both sides match. 2(x + 4) = 20 is an equation: expanding gives 2x + 8 = 20, which is only true for the single value x = 6. 2(x + 4) = 2x + 4 is not true for any value of x at all: expanding the left side gives 2x + 8, which can never equal 2x + 4 since 8 ≠ 4. x + 4 = 2x is also an equation, true only for the single value x = 4. The identity is 2(x + 4) ≡ 2x + 8.
- (b) 10 km — Method: find the constant speed from the graph (distance ÷ time for any point on the line), then multiply that speed by 20 minutes. Working: the line passes through (4 minutes, 2 km), so the speed is 2 ÷ 4 = 0.5 km per minute; in 20 minutes the cyclist travels 0.5 × 20 = 10 km. Answer: 10 km. Distractor refutation: 3 km comes from reading off the distance shown at the end of the plotted section (6 minutes) and stopping there, instead of extending the line to 20 minutes. 20 km comes from misreading the speed as 1 km per minute instead of 0.5 km per minute, doubling the true rate. 40 km comes from dividing 20 by the speed instead of multiplying by it, a reciprocal mix-up.
- (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
- (c) 2x + 12 — Multiply every term inside the bracket by 2: 2 × x = 2x and 2 × 6 = 12, so 2(x + 6) = 2x + 12. Writing 2x + 6 comes from multiplying only the x-term and leaving the 6 unchanged. Writing x + 12 comes from multiplying only the 6 and leaving the x unchanged. Writing 2x + 8 comes from adding 2 and 6 instead of multiplying them.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
- (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).
- (c) x = 4 — Method: with an unknown on both sides, subtract the smaller x term from both sides so that all the x is on one side, then collect the numbers on the other. Working: subtracting 3x from both sides gives 14 = 5x − 6; adding 6 to both sides gives 20 = 5x; dividing both sides by 5 gives x = 4. Checking: 3 × 4 + 14 = 26 and 8 × 4 − 6 = 26. Answer: x = 4. The distractors: x = 2.5 comes from taking 3x off the left-hand side only, leaving 14 = 8x − 6 and so 8x = 20; x = −1.6 comes from changing the sign of the 8x when it is moved across but leaving the sign of the 14 unchanged, giving 3x − 8x = −6 + 14 and so −5x = 8; x = 15 comes from reaching 5x = 20 correctly and then subtracting 5 instead of dividing by 5.
- (d) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (a) 11 — Method: work out each term separately using the rule n² − 3, then subtract. Working: 6th term = 6² − 3 = 36 − 3 = 33. 5th term = 5² − 3 = 25 − 3 = 22. Difference: 33 − 22 = 11. Answer: 11. 8 comes from subtracting the constant −3 once at the end instead of it already being included in both terms, (36 − 25) − 3. 1 comes from working out (6 − 5)² instead of finding 6² and 5² separately and then subtracting. −11 comes from subtracting in the wrong order, the 5th term minus the 6th term instead of the 6th minus the 5th.
- (a) 3x — Method: "three times as old" is a multiplication by 3, applied to the age that is known in terms of the letter. Working: the brother's age is x, and Leo's age is 3 × x, which is written as 3x. Answer: 3x. The distractors: x + 3 comes from reading "three times as old" as "three years older" and adding instead of multiplying; x − 3 comes from reading it as "three years younger" and subtracting; x/3 comes from dividing by 3 instead of multiplying, which describes the brother in terms of Leo rather than Leo in terms of the brother.
- (d) 2x = 10 — Method: collect the x-terms on one side and the constants on the other, keeping the equals sign balanced. Working: 5x − 3x = 8 + 2, giving 2x = 10. Answer: 2x = 10. 2x = 6 comes from a sign error moving the constant, subtracting instead of adding: 8 − 2 = 6. 8x = 6 comes from adding the x-terms instead of subtracting them (5x + 3x = 8x), alongside the constant sign error. 2x = −10 comes from a sign error moving the constant the other way: −2 − 8 = −10.
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