Printable · GCSE Higher · ages 14-16

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 4x + 3y = 25 and 4x − y = 1. Work out the value of y.
(a)6
(b)4
(c)12
(d)−6
3.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)
4.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
5.The simultaneous equations 2x + 3y = 12 and x − y = 1 are given. Work out the value of y.
(a)y = 2
(b)y = 2.2
(c)y = 3
(d)y = 10
6.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
7.A cafe sells coffees at £c each and pastries at £p each. Three coffees and two pastries cost £9.60. Two coffees and two pastries cost £7.60. Work out the price of one coffee.
(a)£0.20
(b)£2.00
(c)£3.20
(d)£1.80
8.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)
9.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
10.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
11.Solve the simultaneous equations y = 3 − x and x² + y² = 9, giving both pairs of solutions.
(a)x = 0, y = 3 and x = 6, y = −3
(b)x = 0, y = 3 and x = −3, y = 6
(c)x = 0, y = 3 and x = 3, y = 0
(d)x = 0, y = 3
12.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
13.The numbers x and y satisfy x + y = 10 and xy = 21. Work out the pair of values.
(a)3 and 7
(b)−3 and −7
(c)5 and 5
(d)1 and 21
14.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
(a)4.67
(b)−4
(c)2
(d)4
15.Solve the simultaneous equations 3x + 2y = 16 and x + y = 7. Work out the value of y.
(a)2
(b)5
(c)9
(d)−5

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.
(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) (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.
(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.
(a) y = 2 — Method: make x the subject of the simpler equation, substitute it into the other equation and then read off the letter the question asks for. Working: x − y = 1 gives x = y + 1, so 2x + 3y = 12 becomes 2(y + 1) + 3y = 12, that is 2y + 2 + 3y = 12, so 5y = 10 and y = 2. Answer: y = 2, and the matching value x = 3 checks in 2 × 3 + 3 × 2 = 12. The distractors: y = 3 comes from solving the pair correctly and then writing down the value of x; y = 2.2 comes from expanding 2(y + 1) as 2y + 1, which leaves 5y = 11; y = 10 comes from rearranging x − y = 1 as x = 1 − y, which turns the first equation into 2 + y = 12.
(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) £2.00 — Subtracting the second equation from the first eliminates the pastries: (3c + 2p) − (2c + 2p) = 9.60 − 7.60, so c = 2.00. A candidate who divides the first total by the number of coffees alone, ignoring the pastries, would get 9.60 ÷ 3 = £3.20. A candidate who finds the price of a pastry instead of a coffee — using c = 2.00 in 2c + 2p = 7.60 to get p = 1.80 — would answer £1.80. A candidate who reaches the correct difference of £2.00 but then mistakenly divides again or misplaces the decimal point would get £0.20.
(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.
(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.
(c) x = 0, y = 3 and x = 3, y = 0 — Substitute y = 3 − x into x² + y² = 9: x² + (3 − x)² = 9. Expanding (3 − x)² = 9 − 6x + x² gives x² + 9 − 6x + x² = 9, which simplifies to 2x² − 6x = 0, or 2x(x − 3) = 0, so x = 0 or x = 3. Using y = 3 − x: x = 0 gives y = 3; x = 3 gives y = 0. Distractor routes: x = 0, y = 3 alone stops after the factor 2x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = 3 and x = −3, y = 6 comes from factorising 2x² − 6x as 2x(x + 3), a sign error that gives a second root of −3 instead of 3. x = 0, y = 3 and x = 6, y = −3 comes from expanding (3 − x)² as 9 − 6x, dropping the x² term, which changes the quadratic to x² − 6x = 0 and its second root to 6.
(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.
(a) 3 and 7 — Method: one equation is linear and the other is not, so rearrange the linear equation and substitute it into the other to leave a single quadratic in one letter. Working: x + y = 10 gives y = 10 − x, so xy = 21 becomes x(10 − x) = 21, which rearranges to x² − 10x + 21 = 0; factorising gives (x − 3)(x − 7) = 0, so the two values are 3 and 7, and 3 + 7 = 10 with 3 × 7 = 21. Answer: 3 and 7. The distractors: 1 and 21 comes from using only the product and taking the first factor pair of 21; −3 and −7 comes from factorising as (x + 3)(x + 7), which reverses the sign of both values; 5 and 5 comes from using only the sum and splitting 10 into two equal parts.
(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.
(b) 5 — From x + y = 7, x = 7 − y. Substituting into 3x + 2y = 16: 3(7 − y) + 2y = 16, so 21 − 3y + 2y = 16, giving 21 − y = 16, so y = 5 (then x = 2). A candidate who forgets to multiply the y-term inside the bracket by 3 would write 21 − y + 2y = 16, giving 21 + y = 16, so y = −5. A candidate who subtracts the two equations directly, (3x + 2y) − (x + y) = 16 − 7, gets 2x + y = 9, and if they wrongly treat this as giving y alone, ignoring the x term, they would answer 9. A candidate who reports the value of x instead of y would answer 2.

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