EDEXCEL GCSE MATHS · HIGHER
Exam Practice: Linear and Quadratic Simultaneous Equations
Solving linear and quadratic simultaneous equations · Equations and inequalities · Lesson 7 of 7 · 28 marks · 35 minutes
|
Name |
|
Date |
|
Instructions
• Answer all the questions.
• Write your answers in the spaces provided.
• The marks for each question are shown in brackets - use this as a guide to how much to write.
• The answers are on separate pages at the back. Attempt every question before you look at them.
• Answer all questions. Show your working.
Question 1 NON-CALCULATOR (5 marks)
Solve the simultaneous equations y = x² − 2x − 3 and y = x + 1.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 1 = 5 marks)
Question 2 NON-CALCULATOR (5 marks)
The circle x² + y² = 25 and the line y = x + 1 intersect at two points. Find the coordinates of the two points.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 2 = 5 marks)
Question 3 NON-CALCULATOR (4 marks)
Solve the simultaneous equations x² + y² = 20 and y = 2x.
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 3 = 4 marks)
Question 4 SHOW THAT (4 marks)
Show that the line y = 2x − 1 is a tangent to the curve y = x².
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 4 = 4 marks)
Question 5 NON-CALCULATOR (5 marks)
Solve the simultaneous equations x + y = 7 and x² + y² = 25.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 5 = 5 marks)
Question 6 NON-CALCULATOR (5 marks)
Solve the simultaneous equations y = x² + 1 and y = 3x − 1.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 6 = 5 marks)
TOTAL FOR PAPER = 28 MARKS
|
|
Answers and mark scheme
Check your answer only once you have written one.
Question 1 (5 marks)
x² − 2x − 3 = x + 1, so x² − 3x − 4 = 0 and (x − 4)(x + 1) = 0. x = 4 or x = −1. The solutions are (4, 5) and (−1, 0).
• Equating M1
• Rearranging to zero M1
• Factorising M1
• Both x values A1
• Both correct pairs A1
Question 2 (5 marks)
x² + (x + 1)² = 25, so 2x² + 2x − 24 = 0, x² + x − 12 = 0, (x + 4)(x − 3) = 0. The points are (3, 4) and (−4, −3).
• Substituting M1
• Simplifying to a quadratic M1
• Solving M1
• Both x values A1
• Both correct pairs A1
Question 3 (4 marks)
x² + 4x² = 20, so x² = 4, x = ± 2. The solutions are (2, 4) and (−2, −4).
• Substituting M1
• 5x² = 20 M1
• x = 2 and x = −2 A1
• Both pairs A1
Question 4 (4 marks)
x² = 2x − 1, so x² − 2x + 1 = 0 and (x − 1)² = 0. There is only one solution, x = 1, so the line touches the curve at one point and is a tangent.
• Equating M1
• Rearranging to x² − 2x + 1 = 0 M1
• (x − 1)² = 0 M1
• Conclusion C1
Question 5 (5 marks)
y = 7 − x, so x² + (7 − x)² = 25, 2x² − 14x + 24 = 0, x² − 7x + 12 = 0, (x − 3)(x − 4) = 0. The solutions are (3, 4) and (4, 3).
• Rearranging and substituting M1
• Expanding (7 − x)² M1
• Solving M1
• Both x values A1
• Both correct pairs A1
Question 6 (5 marks)
x² + 1 = 3x − 1, so x² − 3x + 2 = 0 and (x − 1)(x − 2) = 0. The solutions are (1, 2) and (2, 5).
• Equating M1
• Rearranging to zero M1
• Factorising M1
• Both x values A1
• Both correct pairs A1