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Simultaneous equations worksheet — GCSE Higher

Fifteen questions on "simultaneous equations" — DfE statement A19. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.

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Simultaneous equations worksheet — GCSE Higher

Maths · GCSE Higher · A19 · Simultaneous equations · Non-calculator · 15 questions

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1.Solve the simultaneous equations 2x + y = 7 and x + 2y = 8.
(a)x = 2, y = 5
(b)x = 3, y = 2
(c)x = 2, y = 3
(d)x = −2, y = 11
2.Solve the simultaneous equations 2x + y = 7 and x − y = 2.
(a)x = 1, y = 3
(b)x = 3, y = 2
(c)x = 3, y = 1
(d)x = 3, y = −1
3.A circular pond has equation x² + y² = 20, with lengths in metres from the centre of the garden. A straight path runs along the line y = 2x, entering the pond and leaving it again. Work out the coordinates of the two points where the path meets the edge of the pond.
y = 2x
(a)(2, 4)
(b)(2, 4) and (−2, −4)
(c)(2, 1) and (−2, −1)
(d)(2, −4) and (−2, 4)
4.The simultaneous equations kx + 2y = 4 and 3x + y = 5 have no solution. Work out the value of k.
(a)k = −6
(b)k = 6
(c)k = 2/3
(d)k = 3
5.A gym charges a joining fee plus a monthly fee. Anna paid £100 in total after 3 months of membership. Ben paid £160 in total after 6 months of membership (same joining fee and monthly fee as Anna). Work out the monthly fee.
(a)£26.67
(b)£40
(c)£6.67
(d)£20
6.Work out the values of x and y that satisfy both x + y = 10 and x − y = 4.
(a)x = 14, y = −4
(b)x = 7, y = 3
(c)x = 5, y = 5
(d)x = 7, y = 4
7.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
(a)Right — x² − 3x + 4 = 0 is correct.
(b)Right — subtracting is the same as adding.
(c)Wrong — the correct equation is x² − 5x + 4 = 0.
(d)Wrong — it should be x² − 5x − 4 = 0.
8.The curve y = x² − 1 and the line y = 3x − 3 meet at two points. Work out the x-coordinates of those two points.
y = 3x − 3y = x² − 1
(a)x = −1 and x = −2
(b)x = −1 and x = 4
(c)x = 1 and x = −1
(d)x = 1 and x = 2
9.The curve y = x² − 6 and the line y = 2x − 3 intersect at two points. Which pair of points is correct?
y = 2x − 3y = x² − 6
(a)(3, 9) and (−1, 1)
(b)(3, 3) and (−1, −5)
(c)(3, 3)
(d)(3, 3) and (1, −1)
10.There are 50 adults on a coach to the Lake District. Every adult is either a man or a woman. There are 10 more men than women. Work out the number of men.
(a)20
(b)40
(c)35
(d)30
11.Solve the simultaneous equations x + y = 1 and 2x + y = 5.
(a)x = 4, y = −3
(b)x = −2, y = 3
(c)x = 6, y = −5
(d)x = 4, y = 5
12.Solve the simultaneous equations y = 2x − 1 and y = x² − x − 1, giving both pairs of solutions.
y = 2x − 1y = x²
(a)x = 0, y = −1 and x = −3, y = −7
(b)x = 0, y = −1
(c)x = −2, y = −5 and x = 1, y = 1
(d)x = 0, y = −1 and x = 3, y = 5
13.Solve the simultaneous equations 4x + 3y = 25 and 4x − y = 1. Work out the value of y.
(a)6
(b)4
(c)12
(d)−6
14.Solve the simultaneous equations 5x − 2y = 16 and 3x + 2y = 16. Work out the value of x.
(a)8
(b)4
(c)2
(d)0
15.A student solves the simultaneous equations 2x + y = 11 and x − y = 1 by elimination, adding the two equations together. Which of these is the correct result of that step?
(a)3x = 12
(b)3x = 10
(c)3x + 2y = 12
(d)x = 12

Answer key

(c) x = 2, y = 3 — Method: no letter cancels straight away, so make one letter the subject of one equation and substitute that expression into the other. Working: 2x + y = 7 gives y = 7 − 2x, so x + 2y = 8 becomes x + 2(7 − 2x) = 8, that is x + 14 − 4x = 8, so −3x = −6 and x = 2; substituting back gives y = 7 − 4 = 3. Answer: x = 2, y = 3, which checks in 2 + 6 = 8. The distractors: x = 3, y = 2 comes from adding the equations to x + y = 5 and then subtracting them the wrong way round, writing x − y = 1 instead of x − y = −1; x = 2, y = 5 comes from substituting x = 2 into 2x + y = 7 as 2 + y = 7, dropping the coefficient; x = −2, y = 11 comes from dividing −3x = −6 as though two minus signs gave a negative answer.
(c) x = 3, y = 1 — Method: the y terms are +y and −y, so adding the two equations removes y and leaves an equation in x alone. Working: adding 2x + y = 7 and x − y = 2 gives 3x = 9, so x = 3; substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. Answer: x = 3, y = 1, which also satisfies 2 × 3 + 1 = 7. The distractors: x = 1, y = 3 comes from finding the two values correctly and then writing them against the wrong letters; x = 3, y = 2 comes from substituting x = 3 into 2x + y = 7 as 2 + 3 + y = 7, adding the coefficient instead of multiplying by it; x = 3, y = −1 comes from substituting into x − y = 2 as though it read x + y = 2.
(b) (2, 4) and (−2, −4) — Substitute y = 2x into x² + y² = 20: x² + (2x)² = 20, which gives x² + 4x² = 20, so 5x² = 20, x² = 4, and x = 2 or x = −2. Using y = 2x for each x-value: x = 2 gives y = 4; x = −2 gives y = −4. The path meets the pond's edge at (2, 4) and (−2, −4). Distractor routes: (2, −4) and (−2, 4) swaps the sign pairing, matching each x-value with the wrong sign of y instead of keeping each x with its own correctly-signed y. (2, 1) and (−2, −1) comes from using y = x/2 instead of y = 2x when finding the y-coordinates. (2, 4) alone stops after the positive square root of x² = 4 and never finds the second point from x = −2.
(b) k = 6 — Method: two simultaneous linear equations have no solution when the lines they describe are parallel, so write each equation in the form y = mx + c and make the gradients equal. Working: kx + 2y = 4 rearranges to y = −(k/2)x + 2, so its gradient is −k/2, and 3x + y = 5 rearranges to y = −3x + 5, so its gradient is −3; setting −k/2 = −3 gives k = 6, and the first equation is then 6x + 2y = 4, which simplifies to 3x + y = 2 and can never agree with 3x + y = 5. Answer: k = 6. The distractors: k = −6 comes from reading the gradient of kx + 2y = 4 as +k/2 and solving k/2 = −3; k = 3 comes from making the x terms identical instead of making the gradients equal; k = 2/3 comes from writing the gradient of 3x + y = 5 upside down as −1/3 and solving −k/2 = −1/3.
(d) £20 — Let f be the joining fee and m the monthly fee: f + 3m = 100 and f + 6m = 160. Subtracting the first equation from the second eliminates f: 3m = 60, so m = 20. A candidate who finds the joining fee instead of the monthly fee would get f = 100 − 3(20) = £40. A candidate who divides Ben's total by his number of months, ignoring that part of the cost is a fixed joining fee, would get 160 ÷ 6 ≈ £26.67. A candidate who divides the difference in cost by the total number of months instead of the difference in months would get (160 − 100) ÷ 9 ≈ £6.67.
(b) x = 7, y = 3 — Method: one equation contains +y and the other −y, so adding them removes y; the value found is then substituted back to get the other letter. Working: adding x + y = 10 and x − y = 4 gives 2x = 14, so x = 7; substituting into x + y = 10 gives 7 + y = 10, so y = 3. Answer: x = 7, y = 3, and 7 − 3 = 4 as required. The distractors: x = 7, y = 4 comes from finding x correctly and then taking the 4 in x − y = 4 to be the value of y; x = 5, y = 5 comes from splitting the total of 10 equally and never using the difference; x = 14, y = −4 comes from adding the equations to 2x = 14 and forgetting to halve, so that x is taken as 14 and y as 10 − 14.
(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) x = 1 and x = 2 — Method: where a line meets a curve the two expressions for y are equal, so set them equal and solve the quadratic that results. Working: x² − 1 = 3x − 3 collects to x² − 3x + 2 = 0; factorising gives (x − 1)(x − 2) = 0, so x = 1 or x = 2, and each value gives the same y on both graphs. Answer: x = 1 and x = 2. The distractors: x = −1 and x = −2 come from factorising as (x + 1)(x + 2) and so reversing the sign of both roots; x = −1 and x = 4 come from moving the −3 across the equals sign without changing its sign, which gives x² − 3x − 4 = 0; x = 1 and x = −1 come from setting each expression equal to zero separately instead of equal to each other.
(b) (3, 3) and (−1, −5) — Set the two expressions for y equal: x² − 6 = 2x − 3, which rearranges to x² − 2x − 3 = 0. Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1. Substitute into the linear equation, y = 2x − 3: x = 3 gives y = 3; x = −1 gives y = −5. The points are (3, 3) and (−1, −5). Distractor routes: (3, 3) and (1, −1) comes from mis-factorising x² − 2x − 3 as (x − 3)(x − 1), giving a second root of 1 instead of −1. (3, 9) and (−1, 1) comes from finding the y-coordinate from y = x² instead of substituting back into the given line equation y = 2x − 3. (3, 3) alone stops after finding only the first root of the quadratic.
(d) 30 — Method: give each group a letter, turn each sentence of the stem into an equation in those letters, then eliminate one letter by adding the two equations. Working: let m be the number of men and w the number of women. The 50 adults give m + w = 50, and 10 more men than women gives m − w = 10. Adding the two equations term by term cancels w: (m + w) + (m − w) = 50 + 10, so 2m = 60 and m = 30. Substituting back into m + w = 50 gives w = 20, and 30 − 20 = 10 as required. Answer: 30. The distractors: 20 comes from subtracting the equations instead of adding them, which cancels m and gives 2w = 40, so w = 20 — the other unknown, not the one asked for; 40 comes from forming the second equation as w = 10, reading '10 more men than women' as '10 women', and then getting m = 50 − 10 from the first equation; 35 comes from using m + w = 50 alone with m = w, giving 25 each, and then adding the whole difference of 10 to the men instead of splitting it between the two groups, which leaves a difference of 20 rather than 10.
(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.
(d) x = 0, y = −1 and x = 3, y = 5 — Set the two expressions for y equal: 2x − 1 = x² − x − 1. Rearranging, subtracting 2x and adding 1 to both sides: 0 = x² − x − 1 − 2x + 1 = x² − 3x, so x² − 3x = 0. Factorise: x(x − 3) = 0, giving x = 0 or x = 3. Using y = 2x − 1: x = 0 gives y = −1; x = 3 gives y = 5. Distractor routes: x = 0, y = −1 alone stops after the factor x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = −1 and x = −3, y = −7 comes from mis-factorising x² − 3x as x(x + 3), a sign error that gives a second root of −3 instead of 3. x = −2, y = −5 and x = 1, y = 1 comes from adding 2x to both sides instead of subtracting it when rearranging, giving x² + x − 2 = 0 instead of x² − 3x = 0.
(a) 6 — Subtracting the second equation from the first: the x-terms, 4x and 4x, cancel; the y-terms combine as 3y − (−y) = 4y; and the right-hand sides give 25 − 1 = 24. This gives 4y = 24, so y = 6. A candidate who subtracts in the wrong order would get 4y = 1 − 25 = −24, so y = −6. A candidate who forgets the sign on the −y term, treating 3y − y as 2y, would get 2y = 24, so y = 12. A candidate who divides 24 by 6 instead of 4 would get y = 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.
(a) 3x = 12 — Adding the two equations: the y-terms, +y and −y, have opposite signs, so they cancel; the x-terms combine to 2x + x = 3x; and the right-hand sides add to 11 + 1 = 12. This gives 3x = 12. A candidate who forgets that the y-terms cancel, and instead adds them as if they had the same sign, would write 3x + 2y = 12. A candidate who subtracts the right-hand sides instead of adding them would get 3x = 10. A candidate who correctly reaches 3x = 12 but then treats 12 itself as the value of x, skipping the final division, would write x = 12.

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