Lesson A19
A19 solve two simultaneous equations in two variables (linear/linear or linear/quadratic)... Quiz: Pearson Edexcel Maths, Unit 2
20 questions
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Lesson A19, solve two simultaneous equations in two variables (linear/linear or linear/quadratic)...: 20 multiple choice questions for the Pearson Edexcel GCSE Maths (1MA1), Unit 2: Algebra, written with Revision Ninja.
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The 20 questions
-
Simultaneous equations can be solved algebraically by
- differentiation
- drawing only
- elimination or substitution
- guessing and checking
-
Two simultaneous linear equations in two unknowns usually have
- infinitely many solutions
- one solution pair
- two solution pairs
- no solutions
-
The solution of x + y = 5 and x - y = 1 has x =
- 6
- 2
- 4
- 3
-
What does the graphical solution of two simultaneous equations represent?
- Intersection point
- Gradient point
- Y-intercept
- X-intercept
-
A linear and a quadratic equation together can have up to how many solutions?
- none ever
- three
- two
- one
-
What is the key action when solving simultaneous equations by substitution?
- Multiply by zero
- Replace a variable
- Graph the lines
- Add both equations
-
The solution of 2x + y = 7 and y = 3 has x =
- 5
- 2
- 3
- 4
-
The points where y = x and y = x^2 meet are
- (1, 1) and (-1, -1)
- (0, 0) only
- (1, 0) and (0, 1)
- (0, 0) and (1, 1)
-
The solution of x + 2y = 8 and x - y = -1 is
- x = 4, y = 2
- x = 2, y = 2
- x = 3, y = 2
- x = 2, y = 3
-
The solution of 3x + 2y = 17 and y = x + 1 is
- x = 4, y = 3
- x = 3, y = 5
- x = 3, y = 4
- x = 5, y = 2
-
The points where y = x + 2 and y = x^2 meet are
- (4, 2) and (-1, 1)
- (2, 4) and (1, -1)
- (2, 4) and (-1, 1)
- (-2, 0) and (1, 3)
-
The points where y = x + 1 and y = x^2 - 1 meet are
- (2, 3) and (-1, 0)
- (-2, -1) and (1, 2)
- (2, 1) and (-1, -1)
- (2, 3) and (1, 0)
-
Two numbers have sum 20 and difference 6. The larger number is
- 7
- 13
- 26
- 14
-
Two pens and three pencils cost £8.50. One pen and one pencil cost £3.50. A pen costs
- £1.50
- £3
- £2
- £2.50
-
The points where y = 2x and y = x^2 + x meet are
- (0, 0) and (2, 1)
- (1, 2) only
- (0, 2) and (1, 0)
- (0, 0) and (1, 2)
-
Using y = 2x + 3 and y = x^2, the positive x-value where they meet is
- x = 3
- x = -1
- x = 1
- x = 2
-
The solution of x/2 + y/3 = 4 and x + y = 10 is
- x = 2, y = 8
- x = 4, y = 4
- x = 6, y = 4
- x = 4, y = 6
-
The solution of 3x - 2y = 4 and 5x + 2y = 20 is
- x = 3, y = 2.5
- x = 2.5, y = 3
- x = 3, y = 5
- x = 4, y = 4
-
Substituting y = 2x into x^2 + y = 6 gives
- 3x^2 - 6 = 0
- x^2 + 2x - 6 = 0
- x^2 - 2x - 6 = 0
- x^2 + 2x + 6 = 0
-
The solution of 2x + 3y = 12 and x + y = 5 is
- x = 3, y = 2
- x = 3, y = 3
- x = 4, y = 1
- x = 2, y = 3
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