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.
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Algebra worksheet — GCSE Foundation
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- 1.The equation 3(2x − 1) = 4x + 9 is rearranged by expanding the brackets. Which of these is the correctly expanded equation?
- 2.A table of values is being drawn for the graph of y = x³. Work out the value of y when x = −2.y = x
- 3.Write down the expression that means the same as w ÷ 6.
- 4.Write down the expression that means the same as m² × m × 3.
- 5.Which of these equations gives a straight-line graph when plotted?
- 6.Solve (2x + 1)/3 = 5
- 7.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
- 8.Solve the simultaneous equations x + y = 15 and x − y = 3. Work out the value of x.
- 9.Which of these rules generates the sequence 6, 11, 16, 21, …?
- 10.A gym charges a reduced price for the first month of membership and a standard price for every month after that. The total cost is £5 for 1 month, £12 for 2 months, £19 for 3 months and £26 for 4 months. Work out an expression, in terms of n, for the total cost, in pounds, of n months of membership.
- 11.Amelia's mother is three times as old as Amelia. Four years ago her mother was five times as old as Amelia was then. Work out Amelia's age now.
- 12.A ribbon of length L metres is cut into 5 equal pieces, and then 3 metres is removed from one piece. Write down an expression, in metres, for the length of that piece after 3 metres is removed.
- 13.A square has sides of length y cm. Write down an expression for the perimeter of the square.
- 14.The line y = 3x + 6 meets the x-axis at one point. Work out the coordinates of that point.y = 3x + 6
- 15.Isla says that the lines y = 2x + 1 and y = −2x + 3 are parallel. Write down the statement that gives the correct answer for the correct reason.y = 2x + 1y = -2x + 3
Answer key
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (c) −8 — (−2)³ = (−2) × (−2) × (−2) = −8, since multiplying three negative numbers gives a negative result. A candidate who forgets the sign of a negative number when cubing it might treat (−2)³ as if it were 2³ = 8. A candidate who multiplies −2 by 3 instead of cubing it might get −2 × 3 = −6. A candidate who combines both mistakes — multiplying by 3 and dropping the sign — might get 2 × 3 = 6.
- (c) w/6 — Method: a ÷ b is written as a fraction a/b, with the number being divided (w) on top. Working: w ÷ 6 = w/6. Answer: w/6. 6/w comes from writing the numbers the wrong way round, putting the 6 on top instead of w. 6w comes from reading the ÷ sign as ×, multiplying instead of dividing. w − 6 comes from reading ÷ as −, subtracting instead of dividing.
- (d) 3m³ — Method: multiply the powers of m by adding their indices, then bring the number coefficient to the front. Working: m² × m has indices 2 and 1; add them to get 3, giving m³, then × 3 gives 3m³. Answer: 3m³. 3m² comes from multiplying the indices instead of adding them: 2 × 1 = 2, giving m², then × 3 = 3m². m³ comes from correctly combining the m's but dropping the coefficient 3. m⁶ comes from multiplying the index by the coefficient instead of writing the coefficient in front: taking the 2 in m² and the 3 to give m raised to the power 2 × 3, which is m⁶, with the lone m left out.
- (b) y = 2x − 3 — A straight-line graph has an equation of the form y = mx + c, where x appears only to the power 1. Of the four equations, y = 2x − 3 fits this form, so it gives a straight line. y = x² − 3 has x squared, so it gives a curved (quadratic) graph, not a straight line. y = x³ − 3 has x cubed, so it gives a cubic curve. y = 3/x has x on the bottom of a fraction, so it gives a reciprocal curve with two branches.
- (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
- (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.
- (b) 9 — Add the two equations to eliminate y: (x + y) + (x − y) = 15 + 3, so 2x = 18, and x = 9 (then y = 15 − 9 = 6). A candidate who reports the value of y instead of x would give 6. A candidate who forgets to divide by 2 after adding would give 18. A candidate who simply subtracts the two totals (15 − 3) instead of adding the equations would give 12.
- (c) 5n + 1 — Method: find the common difference, then use it as the coefficient of n in the position-to-term rule, and find the constant by checking against the first term. Working: the common difference is 5, so the rule has the form 5n + c. Using the 1st term: 5(1) + c = 6, so c = 1. The rule is 5n + 1. Answer: 5n + 1. 5n − 1 uses the correct coefficient but the wrong sign for the constant. 6n comes from using the first term as the coefficient of n instead of the common difference — it matches the 1st term by coincidence but fails from the 2nd term onward. n + 5 swaps the coefficient and the constant around, using the common difference as the constant instead of the coefficient of n.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (d) 8 — Method: write both present ages in terms of one letter, take four years off each of them, and form an equation from the comparison four years ago. Working: let Amelia be y, so her mother is 3y; four years ago they were y − 4 and 3y − 4, giving 3y − 4 = 5(y − 4); expanding gives 3y − 4 = 5y − 20; adding 20 and subtracting 3y from both sides gives 16 = 2y, and dividing by 2 gives y = 8. Checking: Amelia is 8 and her mother 24; four years ago they were 4 and 20, and 20 = 5 × 4. Answer: 8. The distractors: 4 comes from solving correctly but giving Amelia's age four years ago instead of her age now; 16 comes from stopping at 2y = 16 and giving 16 as her age; 2 comes from collecting the y terms by adding the 5y instead of subtracting it, giving 3y + 5y = 20 − 4 and so 8y = 16.
- (c) L/5 − 3 — Each of the 5 equal pieces is L/5 metres long, and removing 3 metres from one piece gives L/5 − 3. Subtracting the 3 metres before dividing by 5, (L − 3)/5, divides the removed length between all 5 pieces instead of taking it from just one. Dividing only the 3 by 5 instead of dividing L by 5, L − 3/5, divides the wrong number. Writing 5/L − 3 inverts the fraction, swapping which number is the numerator.
- (d) 4y — Method: the perimeter is the total distance round the outside of the shape, and a square has four sides all of the same length. Working: the four sides are y, y, y and y, so the perimeter is y + y + y + y, which is written as 4y. Answer: 4y. The distractors: 2y comes from adding only two of the four sides, as if the square were being treated like a rectangle with one length and one width; y² comes from working out the area of the square, y × y, instead of its perimeter; y + 4 comes from adding 4 to the side length instead of multiplying the side length by 4.
- (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) No, because their gradients are 2 and −2 — Method: two lines are parallel exactly when their gradients are equal as signed numbers, so m is read from each equation written in the form y = mx + c and the two are compared. Working: y = 2x + 1 has gradient 2 and y = −2x + 3 has gradient −2; those are not equal, so the lines are not parallel, and indeed one slopes upwards while the other slopes downwards. Answer: No, because their gradients are 2 and −2. The distractors: saying yes because both gradients have size 2 comes from comparing the sizes of the gradients and ignoring their signs; saying yes because the gradients add to 0 comes from using a sum of zero as the test for parallel lines instead of equality of gradients; saying no because the y-intercepts are 1 and 3 reaches the right verdict by the wrong route, since the intercepts decide where the lines sit rather than whether they are parallel.
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