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GCSE · Algebra

5.11D Simultaneous Equations: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Solve the simultaneous equations:
x + y = 15
x - y = 3

(Total for Question 1 is 3 marks)

Solve the simultaneous equations:
2x + y = 11
2x - y = 5

(Total for Question 2 is 3 marks)

Solve the simultaneous equations:
y = 3x - 2
2x + y = 13

(Total for Question 3 is 4 marks)

Show that x = 5, y = 4 is not a solution of the simultaneous equations:
3x + 2y = 17
x - y = 1

(Total for Question 4 is 3 marks)

Solve the simultaneous equations:
2x + 3y = 16
x + y = 6

(Total for Question 5 is 4 marks)

Solve the simultaneous equations:
3x + 2y = 16
2x + 3y = 19

(Total for Question 6 is 5 marks)

Solve the simultaneous equations:
5x - 2y = 11
3x + 2y = 13

(Total for Question 7 is 5 marks)

Solve the simultaneous equations:
2x + y = -1
x - y = 4

(Total for Question 8 is 5 marks)

Solve the simultaneous equations:
4x + 3y = 25
3x - 2y = 6

(Total for Question 9 is 5 marks)

Two pens and three pencils cost 2.20 pounds in total. One pen and one pencil cost 0.90 pounds in total.

(a)Using p for the cost of a pen in pounds and c for the cost of a pencil in pounds, write down a pair of simultaneous equations for this information.(2)

(b)Solve your equations to find the cost of a pen and the cost of a pencil.(3)

(Total for Question 10 is 5 marks)

Priya is x years old and her brother Kwame is y years old. The sum of their ages is 31. Kwame is 5 years older than Priya.
x + y = 31
y = x + 5
Find the ages of Priya and Kwame.

(Total for Question 11 is 5 marks)

At a summer fair, adult tickets cost a pounds each and child tickets cost c pounds each. 3 adult tickets and 2 child tickets cost 28 pounds in total. 5 adult tickets and 1 child ticket cost 42 pounds in total.

(a)Write down a pair of simultaneous equations for this information.(2)

(b)Solve your equations to find the price of an adult ticket and the price of a child ticket.(4)

(Total for Question 12 is 6 marks)

The sum of two numbers is 24. Three times the smaller number is 4 more than the larger number. Let x be the larger number and y be the smaller number. Form and solve a pair of simultaneous equations to find the two numbers.

(Total for Question 13 is 5 marks)

Line L1 has equation y = 3x - 4. Line L2 has equation y = -2x + 11. Find the coordinates of the point where L1 and L2 intersect.

(Total for Question 14 is 4 marks)

A rectangle has length x cm and width y cm. Its perimeter is 54 cm. The length is 3 cm more than twice the width.

(a)Write down a pair of simultaneous equations for this information.(2)

(b)Solve your equations to find the length and width of the rectangle.(3)

(c)Hence work out the area of the rectangle.(1)

(Total for Question 15 is 6 marks)

Mobile phone Tariff A costs 15 pounds plus 2p per minute of calls. Tariff B costs 9 pounds plus 5p per minute of calls. Let m be the number of minutes of calls in a month.

(a)Write an expression, in pence, for the total monthly cost of Tariff A and for the total monthly cost of Tariff B.(2)

(b)Find the number of minutes of calls for which the total cost of Tariff A equals the total cost of Tariff B.(3)

(c)State the total cost, in pounds, at this number of minutes.(1)

(Total for Question 16 is 6 marks)

Solve the simultaneous equations:
x/2 + y/3 = 7
x/3 + y/2 = 8

(Total for Question 17 is 6 marks)

Solve the simultaneous equations:
y = x + 2
x² + y² = 20
Give your answers as coordinate pairs.

(Total for Question 18 is 6 marks)

Solve the simultaneous equations:
y = x² - 2x + 1
y = 3x - 3

(Total for Question 19 is 6 marks)

Show that the line y = x + 6 does not intersect the curve x² + y² = 10. You must show your working, including the value of the discriminant.

(Total for Question 20 is 5 marks)

Mark scheme · 5.11D Simultaneous Equations: Fluency and Exam Drill

Question 1

M1 adds the two equations to eliminate y (or equivalent valid elimination method)
A1 x = 9
A1 y = 6 cao
Answer: x = 9, y = 6

Question 2

M1 adds the two equations to eliminate y
A1 x = 4
A1 y = 3 cao
Answer: x = 4, y = 3

Question 3

M1 substitutes y = 3x - 2 into the second equation
M1 forms and simplifies a single linear equation in x, e.g. 5x - 2 = 13
A1 x = 3
A1 y = 7 cao
Answer: x = 3, y = 7

Question 4

B1 shows x - y = 1 is satisfied (5 - 4 = 1)
B1 substitutes into 3x + 2y to obtain 23
B1 correct conclusion with reason, e.g. 'since 23 is not equal to 17, x = 5, y = 4 does not satisfy both equations, so it is not a solution'
Answer: x - y = 1 is satisfied, but 3x + 2y = 23, not 17, so x = 5, y = 4 is not a solution.

Question 5

M1 multiplies x + y = 6 by 2 or 3 to match a coefficient, e.g. 3x + 3y = 18
M1 subtracts the equations correctly to eliminate a variable
A1 x = 2
A1 y = 4 cao
Answer: x = 2, y = 4

Question 6

M1 multiplies one equation to match a coefficient, e.g. first equation x3
M1 multiplies the other equation to match the same coefficient, e.g. second equation x2
M1 subtracts or adds correctly to eliminate one variable
A1 x = 2
A1 y = 5 cao
Answer: x = 2, y = 5

Question 7

M1 identifies that adding the equations eliminates y
M1 adds the equations correctly to obtain 8x = 24
A1 x = 3
A1 y = 2 cao
B1 correct check shown in both original equations (ft their x and y)
Answer: x = 3, y = 2

Question 8

M1 adds the two equations to eliminate y
M1 obtains 3x = 3
A1 x = 1
A1 y = -3 cao
B1 correct check shown, e.g. 2(1) + (-3) = -1
Answer: x = 1, y = -3

Question 9

M1 multiplies each equation to match the coefficient of one variable, e.g. eq1 x2, eq2 x3
M1 adds the resulting equations correctly to eliminate y
A1 x = 4
A1 y = 3 cao
B1 correct check shown in the unused equation
Answer: x = 4, y = 3

Question 10

(a) B1 2p + 3c = 2.20 oe
(a) B1 p + c = 0.90 oe
(a) Answer: 2p + 3c = 2.20 and p + c = 0.90
(b) M1 rearranges p + c = 0.90 to p = 0.90 - c (or equivalent elimination step)
(b) M1 substitutes to form a single equation in one variable, e.g. 2(0.90 - c) + 3c = 2.20
(b) A1 pen = 0.50 pounds and pencil = 0.40 pounds cao
(b) Answer: Pen = 0.50 pounds, pencil = 0.40 pounds

Question 11

M1 substitutes y = x + 5 into x + y = 31
M1 forms and simplifies 2x + 5 = 31
A1 x = 13 (Priya's age)
A1 y = 18 (Kwame's age) cao
B1 correct check shown, e.g. 13 + 18 = 31 and 18 - 13 = 5
Answer: Priya is 13, Kwame is 18

Question 12

(a) B1 3a + 2c = 28 oe
(a) B1 5a + c = 42 oe
(a) Answer: 3a + 2c = 28 and 5a + c = 42
(b) M1 rearranges 5a + c = 42 to c = 42 - 5a (or equivalent elimination approach)
(b) M1 substitutes to form a single equation in a, e.g. 3a + 2(42 - 5a) = 28
(b) M1 simplifies correctly to -7a = -56 (or equivalent) and solves for a
(b) A1 adult ticket = 8 pounds and child ticket = 2 pounds cao
(b) Answer: Adult ticket = 8 pounds, child ticket = 2 pounds

Question 13

B1 x + y = 24
B1 3y = x + 4 oe
M1 combines the equations to form a single equation in one variable, e.g. 3y = (24 - y) + 4
A1 y = 7
A1 x = 17 cao
Answer: Larger number = 17, smaller number = 7

Question 14

M1 sets the two expressions for y equal, 3x - 4 = -2x + 11
M1 solves to find x = 3
A1 y = 5 (ft their x substituted into either equation)
A1 states the coordinates (3, 5) cao
Answer: (3, 5)

Question 15

(a) B1 x + y = 27 oe (accept 2x + 2y = 54)
(a) B1 x = 2y + 3 oe
(a) Answer: x + y = 27 and x = 2y + 3
(b) M1 substitutes x = 2y + 3 into x + y = 27
(b) M1 simplifies to 3y + 3 = 27 and solves for y
(b) A1 width = 8 cm and length = 19 cm cao
(b) Answer: Length = 19 cm, width = 8 cm
(c) B1 152 cm² (ft their length x width)
(c) Answer: 152 cm²

Question 16

(a) B1 Tariff A: 1500 + 2m
(a) B1 Tariff B: 900 + 5m
(a) Answer: Tariff A = 1500 + 2m pence, Tariff B = 900 + 5m pence
(b) M1 sets 1500 + 2m = 900 + 5m
(b) M1 rearranges to 600 = 3m (or equivalent)
(b) A1 m = 200 cao
(b) Answer: 200 minutes
(c) B1 19.00 pounds (ft their m)
(c) Answer: 19.00 pounds

Question 17

M1 multiplies the first equation by 6 to clear fractions, obtaining 3x + 2y = 42
M1 multiplies the second equation by 6 to clear fractions, obtaining 2x + 3y = 48
M1 multiplies to match a coefficient, e.g. first eq x3, second eq x2
M1 eliminates correctly to obtain 5x = 30 (or equivalent)
A1 x = 6
A1 y = 12 cao
Answer: x = 6, y = 12

Question 18

M1 substitutes y = x + 2 into x² + y² = 20
M1 expands (x+2)² correctly to x² + 4x + 4
M1 forms the 3-term quadratic 2x² + 4x - 16 = 0 (or equivalent, e.g. x² + 2x - 8 = 0)
M1 factorises or uses the quadratic formula correctly, e.g. (x+4)(x-2) = 0
A1 x = 2 and x = -4
A1 y = 4 and y = -2 respectively, cao
Answer: (x, y) = (2, 4) or (x, y) = (-4, -2)

Question 19

M1 sets x² - 2x + 1 = 3x - 3
M1 rearranges to form x² - 5x + 4 = 0
M1 factorises correctly, (x-1)(x-4) = 0
A1 x = 1 and x = 4
A1 y = 0 (when x = 1)
A1 y = 9 (when x = 4) cao
Answer: (x, y) = (1, 0) or (x, y) = (4, 9)

Question 20

M1 substitutes y = x + 6 into x² + y² = 10
M1 expands (x+6)² correctly to x² + 12x + 36
M1 rearranges to the 3-term quadratic 2x² + 12x + 26 = 0 (or equivalent, e.g. x² + 6x + 13 = 0)
M1 correctly calculates the discriminant, b² - 4ac = 36 - 52 = -16
A1 correct conclusion with reason: since the discriminant is negative, there are no real solutions, so the line does not intersect the curve
Answer: Discriminant = -16, which is less than 0, so there are no real solutions and the line does not intersect the curve.

Simultaneous Equations: Fluency and Exam Drill - Worksheets, Questions and Revision