Mathematics 2 ESO – Unit 8 SYSTEMS OF LINEAR EQUATIONS page 87 activity 45
a) Which of these pairs of values is a solution to the equation: 2x – 3y = –7
- I) x = 4 ; y = 1
- II) x = –2 ; y = 1
- III) x = –1 ; y = 0
- IV) x = 0 ; y = 0
b) Which of these pairs of values is a solution to equation –x + y = 1
- I) x = 4 ; y = 1
- II) x = –2 ; y = 1
- III) x = –1 ; y = 0
- IV) x = 0 ; y = 0
c) Give five possible solutions to the equation 2x + 3y = 5
d) Complete the table with solutions to equation: x – 3y = 2
| x | 10 | 9 | –4 | 0 | 1 | |||||
| y | 6 | 7 | –8 | 12 | 10 |
e) Complete the table with solutions to equation x – 3y = 2 and draw the graph.
| x | –4 | –3 | –2 | –1 | 0 | 1 | 2 | 3 | 4 |
| y |
f) Watch the video on “How to Graph a Linear Equation by Finding the Intercepts“. Then graph (draw) these equations:
- I) 4x – 4y = 6
- II) –x – 9y = 0
- III) –3x + 7y = 3
- IV) x + 15y = 8
g) Watch the video on “Solving Systems of Equations by Graphing”. Draw the graph and write the solution:
- I) 2x – y = 6 ; –x + y = 3
- II) –x – 4y = 5 ; 6x – 7y = 7
- III) –2x + 7y = 9 ; –2x + 3y = 0
- IV) x + 4y = 6 ; –5x + 5y = 15
h) Watch the video on “Simultaneous Equations – simple, linear, worked example“. Then solve by substitution and check the solution:
- I) x – 4y = 3 ; x + 8y = 3
- II) – x – 4y = -5 ; 2x – 4y = –2
- III) 2x + y = 4 ; – 5x + 3y = 1
- IV) 3x + y = 9 ; – 12x + y = –21
i) Solve by substitution:
- I) x – 2y = –1 ; x + 8y = 9
- II) – x – 4y = 6 ; x – 4y = 10
- III) – 2x + 7y = 0 ; – 5x + 3y = 0
- IV) x + 4y = –7 ; – 2x + 3y = –8
j) Watch the video on “Method of Elimination Steps to Solve Simultaneous Equations“. Then solve by elimination and check the solution:
- I) 2x – 6y = -4 ; –x + 2y = 1
- II) – 3x – 4y = –14 ; x – 7y = –12
- III) – 2x + 7y = 6 ; – 2x + 3y = 0
- IV) 2x + 4y = –8 ; – 2x + y = –2