Spaceship Battle Solution Question 1 Here we have two linear equations y = ‐3x + 14 and y = 2x – 1. We need to find the point at which the two lines intersect. If the two lines intersect, they will have the same y value at the intersection point. So we can equate the two linear equations to calculate what the x value is: 3
3
14
2
1 2
14
5
15 1 3 (1mark) This gives you the x coordinate where the beams will intersect. Now you can substitute the x value back into either of the two equations to find out the y value: 2
2
1 3
1 5 (1 mark) Now we have the coordinates of the intersection point: (3,5). Students can check the solution by graphing the two lines and seeing where they intersect on a graph: (1 mark) From the graph we see that the two lines intersect at (3, 5). All material copyright Math Thrills Pty Ltd www.MathThrills.com Question 2 For this question we have the coordinates of the intersection point (2, y). We also know that the linear equations are y = ‐3x ‐ 9 and y = 2x + n. As in the first task, at the intersection point we have the same y value so students should equalize the right sides of equalities: 3
9
3
2
2
9
5
9
(1 mark) Now in this equation we can substitute the x coordinate for the intersection point provided in the question x=2: 5∗2
9
10
9
19 (1 mark) Question 3 For this question the linear equations are y = kx + 2 and y = 4x + 1, and the intersection point has the coordinates (x, 5). Students can put these coordinates into the second equation to find the x coordinate of the intersection point: 5
4
4
1 4
1 5
4
1 4 1 (1 mark) Student can substitute the coordinates of the intersection point into the first equation to find the value k. 5
1∗
5
2 2 3 (2 marks) All material copyright Math Thrills Pty Ltd www.MathThrills.com