Exemplar for Internal Achievement Standard Mathematics and Statistics Level 3 This exemplar supports assessment against: Achievement Standard Apply systems of simultaneous equations in solving problems An annotated exemplar is an extract of student evidence, with a commentary, to explain key aspects of the standard. It assists teachers to make assessment judgements at the grade boundaries. New Zealand Qualifications Authority To support internal assessment
Grade Boundary: Low Excellence 1. For Excellence, the student is required to apply systems of simultaneous equations, using extended abstract thinking, in solving problems. This involves one or more of devising a strategy to investigate or solve a problem, identifying relevant concepts in context, developing a chain of logical reasoning, or proof, or forming a generalisation, and also using correct mathematical statements, or communicating mathematical insight. The student has correctly solved the system of simultaneous equations for all three methods which create the third equation. The student has also interpreted these solutions geometrically (1) (2) & (3). The student has identified relevant concepts in context by linking the way the third equation is created to parallel and identical planes and explaining the general case for the third situation (4). The student has formed a general solution for the second situation (5). Correct mathematical statements are used throughout the response. For a more secure Excellence, the student could find specific points that satisfy the second case.
2 x + 1 so = 1 y 4 z = 8 Method 1 The three equations are = 1 y 4 z = 8 3 x + 2y + 8z = 7 These equations represent planes in 3D. Using my calculator the answers are x=5, y=2 and z =-1.5.This means that there is a unique solution and the three planes that these equations represent intersect in a unique point (5,2,-1.5). Method 2 The three equations are = 1 y 4 z = 8 6 x 9y Because I get the third equation by multiplying the first one by three they are really the same equation. This means that, in 3D we are looking for points that lie on planes one and two only. Because these two are not parallel they will intersect in a line and any points on this line will satisfy all three equations. The equations are consistent and there are multiple solutions. If I take z = t, then the second equation becomes y 4t = 8. y = 4t + 8. Substitute these into the first equation 2x 3(4t + 8) +2t = 1 2x 12t -24 +2t =1 2x = 25 + 10t x = ½(25 + 10t) So any point with coordinates (½(25 + 10t), 4t + 8, t) will provide a solution that satisfies this system of equations. Method 3 The three equations are = 1 y 4 z = 8 = 6 In this case the first and third equations represent parallel planes because the x,y and z numbers are all equal but the constants (1 and 6) are not. In 3D we are looking for the points that lie on two parallel planes and a third one that is not parallel. They will look like this if you look at them sideways on.
There are obviously no points that lie on all three planes. The planes are inconsistent and there are no solutions. This is a specific example of the general case. If two of the equations have the x,y and z numbers in proportion but the constants are not, then the two planes will be parallel, so the equations are inconsistent and there are no solutions
Grade Boundary: High Merit 2. For Merit, the student is required to apply systems of simultaneous equations, using relational thinking, in solving problems. This involves one or more of selecting and carrying out a logical sequence of steps, connecting different concepts or representations, demonstrating understanding of concepts, or forming and using a model, and also relating findings to a context, or communicating thinking using appropriate mathematical statements. The student has correctly solved the system of simultaneous equations for all three methods which create the third equation. The student has also interpreted these solutions geometrically (1) (2) & (3). The student has identified relevant concepts in context by linking the way the third equation is created to parallel and identical planes and explaining the general case for the third situation (4). The student has used appropriate mathematical statements in the response. To reach Excellence, the student needs to find a general solution to the second case.
2 x + 1 so = 1 y 4 z = 8 Method 1 The three equations are = 1 y 4 z = 8 3 x + 2y + 8z = 7 These equations represent planes in 3D. Using my calculator the answers are x=5, y=2 and z =-1.5.This means that there is a unique solution and the three planes that these equations represent intersect in a unique point (5,2,-1.5). Method 2 The three equations are = 1 y 4 z = 8 6 x 9y Because I get the third equation by multiplying the first one by three they are really the same equation. This means that, in 3D we are looking for points that lie on planes one and two only. Because these two are not parallel they will intersect in a line and any points on this line will satisfy all three equations. The equations are consistent and there are multiple solutions. Method 3 The three equations are = 1 y 4 z = 8 = 6 In this case the first and third equations represent parallel planes because the x,y and z numbers are all equal but the constants (1 and 6) are not. In 3D we are looking for the points that lie on two parallel planes and a third one that is not parallel. They will look like this if you look at them sideways on. There are obviously no points that lie on all three planes. The planes are inconsistent and there are no solutions. This is a specific example of the general case. If two of the equations have the x,y and z numbers in proportion but the constants are not, then the two planes will be parallel, so the equations are inconsistent and there are no solutions.
Grade Boundary: Low Merit 3. For Merit, the student is required to apply systems of simultaneous equations, using relational thinking, in solving problems. This involves one or more of selecting and carrying out a logical sequence of steps, connecting different concepts or representations, demonstrating understanding of concepts, or forming and using a model, and also relating findings to a context, or communicating thinking using appropriate mathematical statements. The student has connected different concepts and representations by solving and geometrically interpreting the solutions for the methods 1 (1) and 3 (2). The student has correctly recognised the dependent equations in method 2, but incorrectly interpreted these geometrically as a triangular shape (3). The student has used appropriate mathematical statements in the response. For a more secure Merit, the student needs to interpret the solutions for method 2 correctly and make a general statement about the solution set of each system of equations.
1. 2 x + 1 original y = 4 z + 8 2 x + 1 x3 y 4 z = 8 x13 3 x + 2y + 8z = 7 x2 6 x 9y 6 x + 4y + 16z = 14 (-) 13y 10z = 11 2. 2 x + 1 x3 y 4 z = 8 6 x 9y 6 x 9y 6 x 9y 0 = 0 3 lines of solutions dependent 3. = 1 y 4 z = 8 = 6 = 1 = 6 (-) 0 = 5 Inconsistent 13y 52z = 104 13y 10z = 11 62 z = 93 z = 1.5 y 4( 1.5) = 8 y + 6 = 8 y = 2 2x 3(2) + 2( 1.5) = 1 2x 6 3 = 1 2x = 10 x = 5 ( 5,2, 1.5) Unique solution consistent For the first set of equations, it is possible to solve resulting in a unique solution that is consistent. The planes will intersect at the point (5,2,-1.5) For the second set of equations, attempting to solve will result in two 6 x 9y equations (0=0). This will resemble planes always being cut by two planes making a triangular shape. This is dependent. For the third set of equations, attempting to solve will result in two equations with the same coefficients but different constants (0=-5). This is inconsistent and will resemble two parallel planes both being cut by a third plane.
Grade Boundary: High Achieved 4. For Achieved, the student is required to apply systems of simultaneous equations in solving problems. This involves selecting and using methods, demonstrating knowledge of concepts and terms and communicating using appropriate representations. This student has selected and solved the equations for methods 1 and 2 (1) and interpreted the nature of the solutions for method 1 geometrically (2). The student has communicated using appropriate representations. To reach Merit, the student could solve and provide the geometric nature of the solutions for both methods 2 and 3.
Method 1 Equation 1 2 x + 1 2x 3y = 1 Equation 2 y = 4 z + 8 y 4x = 8 Equation 3 3 x + 2y + 8z = 7 I used my graphical calculator to get x = 5, y = 2, and z= - 1.5 In method 1 as there is one set answer for x, y, z graphically the three planes have one point where they all cross. point of intersection Method 2 Equation 1 2 x + 1 = 2x 3y = 1 Equation 2 y = 4 z + 8 Equation 3 6 x = 9y + 3 = 6x 9y Graphical calculator gives MA error 2 x + 1 x3 6 x = 9y + 3 6 x 9y (-) 6 x 9y 0 = 0 Infinite solutions
Grade Boundary: Low Achieved 5. For Achieved, the student is required to apply systems of simultaneous equations in solving problems. This involves selecting and using methods, demonstrating knowledge of concepts and terms and communicating using appropriate representations. The student has selected and solved the equations for method 1 (1) and interpreted the nature of the solution to method 1 geometrically (2). The student has communicated using appropriate representations. For a more secure Achieved, the student would need solve or provide a geometric interpretation for another system of equations.
Original equation 1) 2 x + 1 2) y = 4 z + 8 Rearranged equations 1) = 1 2) y 4 z = 8 Method 1: = 1 y 4 z = 8 3 x + 2y + 8z = 7 x = 5 y = 2 z = -5 graphical calculator Method 1: One unique solution is given at the point where all three equations meet.
Grade Boundary: High Not Achieved 6. For Achieved, the student is required to apply systems of simultaneous equation in solving problems. This involves selecting and using methods, demonstrating knowledge of concepts and terms and communicating using appropriate representations. The student has solved the system of equations for method 1 (1). To reach Achieved, the student would need to provide a geometrical interpretation of this solution.
2 x + 1 y = 4 z + 8 3 x + 2y + 8z = 7 1 = 0 (1) y 4 z 8 = 0 (2) 3 x + 2y + 8z 7 = 0 (3) (1)x6 12 x 18y + 12z 6 = 0 (4) (3)x4 12 x 18y + 32z 28 = 0 (5) (4) (5) 26 y 20z + 22 = 0 (6) (2)x5 5 y 20z 40 = 0 (7) (6) (7) 31 y + 62 = 0 y = 2 Sub y = 2 in (2) (2) 4z 8 = 0 (2) 4z = 8-4z = 6 z = -1.5 sub y = 2, z = -1.5 to (5) 12x + 8(-2) + 32(-1.5) 28 = 0 12x + 16 48 28 = 0 12x 60 = 0 x = 5 therefore y = 2 z = - 1.5 x = 5