30S Applied Math Mr. Knight – Killarney School Slide 1 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Solving Systems Using Graphmatica Learning Outcome B-1 LP-L3 Objectives: To solve systems of linear inequalities using Graphmatica

30S Applied Math Mr. Knight – Killarney School Slide 2 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Graphmatica is a graphing utility that will allow you to graph equations and inequalities. Graphing lines and Entering Functions Open Graphmatica. The top line, the Menu Bar, contains various menus, e.g., File, Edit, Redraw, View, Labels, and so on. Theory – Graphing lines and Entering Functions

30S Applied Math Mr. Knight – Killarney School Slide 3 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica The middle line, the Button Bar, contains various icons. For example, the second icon is "Open" a file, the third icon is "Save" a file, and the sixth icon (a yellow pencil) is "Draw a graph." The third line is an open space where equations can be entered. This is called the Equation Editor. Below the graph area is the Status Bar that displays messages from the Help menu. Theory – Graphing lines and Entering Functions

30S Applied Math Mr. Knight – Killarney School Slide 4 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Change the background colour of the graph area: Click on View from the Menu Bar. Click on the submenu Colors… to select an appropriate colour scheme. Click on White, below the third grid to change the background colour to white. Click on OK to close the dialog box Theory – Change the background colour of the graph area

30S Applied Math Mr. Knight – Killarney School Slide 5 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Draw a graph: Enter y = 3x + 4 in the Equation Editor. Hit Enter. The graph of y = 3x + 4 shows in the graphing area. Notice: the Status Bar below the graph shows the equation that has been graphed. Theory – Draw a graph

30S Applied Math Mr. Knight – Killarney School Slide 6 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica You may want to change the part of the graph that you see. Click on View on the Menu Bar. Click on the submenu Grid Range… Adjust the view by entering new values for the horizontal Left: and Right: and the vertical Bottom: and Top:. For example, use -7 and 7 as the horizontal left and right, and use -4 and 8 as the vertical bottom and top. Click on OK to see the result of your choices. Theory – Change Viewing Area

30S Applied Math Mr. Knight – Killarney School Slide 7 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica To graph a new equation If you want to graph a new equation, simply edit the old one or delete it and enter a new one. Key in y = -2x + 1 into the Equation Editor. Click Enter to see the graph. The second graph is shown along with the first. Theory – To graph a new equation

30S Applied Math Mr. Knight – Killarney School Slide 8 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica An equation can be entered in any form. It does not have to have to be written in the form y = mx + b. For example, enter 3x - 2y = 4 directly into the Equation Editor. Press Enter to see all three graphs. You can delete the last equation. Click on Redraw from the Menu Bar. Click on Delete Equation from the sub-menu. The last equation, which is still highlighted, will be deleted. You can delete all equations and graphs. Click on Redraw from the Menu Bar. Click on Clear All from the sub-menu. All equations and graphs will be deleted. Theory – To graph a new equation

30S Applied Math Mr. Knight – Killarney School Slide 9 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Closing Graphmatica You can close Graphmatica by clicking on File from the Menu Bar. Then click on Exit from the submenu. Select No when the Save Untitled dialog box opens. You can also close Graphmatica by clicking on the x in the upper right corner of the main Graphmatica screen. Theory – Closing Graphmatica

30S Applied Math Mr. Knight – Killarney School Slide 10 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Try This Exercise Yourself: Graph the following equations on the same grid. y = 4x + 5 x + 2y = 6 y = 5 x = -2 x = 3y - 3 After graphing all the equations, change the Grid Range to Left (-10), Right (15), Bottom (-5), and Top (10) as shown. Practice

30S Applied Math Mr. Knight – Killarney School Slide 11 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Solution (The graph below is shown with a different grid range than specified above in order to have it fit on the screen.) Solution

30S Applied Math Mr. Knight – Killarney School Slide 12 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Graphing Inequalities Graphmatica can also graph inequalities. Example 1 Open Graphmatica. Input y > 2x + 3 into the Equation Editor. Use the ">" symbol above the "." symbol on your keyboard. Press Enter to see the graph. This graph has a dotted line because the expression does not have an equality (=). Delete this graph by clicking on Redraw > Delete Equation Theory – Graphing Inequalities

30S Applied Math Mr. Knight – Killarney School Slide 13 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Input y  3x - 2 as y <= 3x - 2 into the Equation Editor. Press Enter to see the graph. The graph has a solid line boundary because the equality (=) is included in the set. Verify that the region is correct by using (0, 0) as the test point in the inequality: y < 3x < 3(0) < -2. This is false and means that (0, 0) is not part of the region. The graph reflects this decision by shading the side of the line that does NOT include (0, 0). Theory – Graphing Inequalities Example 2

30S Applied Math Mr. Knight – Killarney School Slide 14 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Deselecting Equations and Inequalities Enter y = -2x + 1 into the Equation Editor and show the graph. Enter y = 0.5x into the Equation Editor and show the graph. Theory – Deselecting Equations and Inequalities

30S Applied Math Mr. Knight – Killarney School Slide 15 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica You can "turn off" one of the graphs in the following way. Click on the down arrow located at the end of the Equation Editor. This will show a "history" of the equations that you have entered. Theory – Deselecting Equations and Inequalities

30S Applied Math Mr. Knight – Killarney School Slide 16 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Click on the inequality y <= 3x - 2. This action highlights it and puts into the Equation Editor line. Click on Redraw from the Menu Bar. Then, click on Hide Graph from the sub-menu. The graph of this inequality will be deleted from the coordinate grid. You can show a graph that is hidden by selecting the equation from the "history" list in the Equation Editor. Then click on the Draw icon or press the Enter key. Theory – Deselecting Equations and Inequalities

30S Applied Math Mr. Knight – Killarney School Slide 17 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Try This Exercise Yourself Draw the following graphs in Graphmatica 1.-4  x x - 7y > x - 2y  y + 2x > 15 Practice Graphing Inequalities

30S Applied Math Mr. Knight – Killarney School Slide 18 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica -4  x - 6 2x - 7y > 21 -5x - 2y  14 3y + 2x > 15 Practice Graphing Inequalities

30S Applied Math Mr. Knight – Killarney School Slide 19 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Express each of the following regions as an inequality Practice Graphing Inequalities y  3x - 4 y  -1.5x + 3

30S Applied Math Mr. Knight – Killarney School Slide 20 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica Solving Systems of Linear Inequalities The solution of a system of inequalities is the common area or intersection where two or more half-planes overlap. Example 1 Graph the solution defined by the following inequalities: x + y > -1 and y > x - 2. The intersection of the two half-planes is the solution. The area where there is double shading is the solution. In this common region, the conditions of both inequalities are satisfied simultaneously. Theory – Solving Systems of Linear Inequalities

30S Applied Math Mr. Knight – Killarney School Slide 21 Unit: Linear Programming Lesson 3: Solving Systems Using Graphmatica A system can consist of more than two inequalities. Example 2 Graph the solution defined by the following inequalities: y  2, y  x - 1 and y  -x - 5. Solution Each inequality represents a half-plane, so there are three half-planes. The solution is the area that contains shading from all three inequalities and is a triangular region. Theory – Solving Systems of Linear Inequalities