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Day 77 – Graphing system of inequalities

Day 77 – Graphing system of inequalities. Problem.

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Day 77 – Graphing system of inequalities

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  1. Day 77 – Graphing system of inequalities

  2. Problem Libby is making a window frame for etched glass. The frame will be for a window that is square on the bottom with an isosceles triangle on top. The perimeter of the window must be no more than 15 feet. What are some possible dimensions of the window?

  3. Let x represent the length of each side of the square, and let y represent the length of one of the two congruent sides of the isosceles triangle. • The perimeter of the window must be no more than 15 feet. Write an inequality for the perimeter using the variables x and y.

  4. Answer #1 and #2 1. 2.

  5. The sum of two sides of a triangle is always greater than the third side. Use this fact to write the second inequality. • Write the system of inequalities. Solve each inequality for y.

  6. Answer #3 and #4 3. 4.

  7. Recall that you can use a graph to find the solutions. All of the points that lie in the solution region of both inequalities are in the solution of the system. Remember that the dimensions of the windows must be positive, so the reasonable domain and range are contained in the first quadrant.

  8. Graph the boundary lines. Is one or both of the boundary lines solid or dashed? Why?

  9. Notice that the boundary lines divide the first quadrant into four regions. Points A, B, C and D are each placed in one of these four regions. In the following table the coordinates of each point in both inequalities, and complete the table.

  10. On your graph, shade the region containing the point that makes both inequalities true. Do either of the boundary lines contain points that are solutions to this system of linear inequalities? Explain 

  11. Answer to #7 The line 3x + 2y ≤ 15 contains points that are solutions to the system because of the (≤ ) equal.

  12. Solve by Graphing 1

  13. Solve by Graphing 1 Graph x + y > 10. Any point in half-plane E is a solution of x + y >10. Note that the graph of x + y = 10 is dashed to show that it is not included in the half-plane. E x + y > 10

  14. Solve by Graphing 1 Graph x - y > -4. Any point in half-plane F is a solution of x - y > -4. Note that the points on the graph of X – y = -4 are not included in the half-plane F x – y > - 4

  15. Solve by Graphing 1 Any point in the intersection of the half-planes E and F (double shading) is a solution of x + y > 10 and x – y > - 4 Thus, all points in the darkest region (but no points on the lines) are solutions of the system. E F

  16. Solve by Graphing 2

  17. Solve by Graphing 2 Graph y ≥ 2x - 4. Any point in half-plane E or on the line y = 2x – 4 (edge of half-plane E) is a solution of y ≥ 2x – 4. E y = 2x - 4

  18. Solve by Graphing 2 Graph x + y ≤ 5. Any point in half-plane F or on the line x + y = 5 (edge of half-plane F) is a solution of x + y ≤ 5. x + y = 5 F

  19. Solve by Graphing 2 Any point in the intersection of half-planes E and F (double-shading) and any point on the lines that border the intersection is a solution of y ≥ 2x – 4 and x + y ≤ 5 E F

  20. Consumerism Mrs. Fuentes wants to buy at least 10 books. Each paperback costs an average of $10, and each hardcover book costs an average of $20. Mrs. Fuentes is planning to spend less than $250 on books. • Write a system of linear inequalities that represents this situation. Then graph the solution of the system. • Based on you graph, name three possible combinations of paperback and hardcover books Mrs. Fuentes could buy.

  21. Consumerism a. Let x represent the number of paperback books and y represent the number of hardcover books. The number of books Mrs. Fuentes wants to buy is x + y ≥ 10. The amount she plans to spend is 10x + 20y < 250. So, the system of inequalities that represents the situation is

  22. Graph x + y = 10 with a solid line. Test points on either side of the line, such as (0,0) and (10,10), to see which half-plane to shade. Shade above the line, because the point (10,10) is above the line and makes x + y ≥ 10 true. Then graph 10x + 20y = 250 with a dashed line. Testing points above and below the line shows that the half-plane below the line should be shaded. The solution is the double-shaded region. 10x + 20y = 250 x + y = 10

  23. b. Select any three points in the double-shaded region. The points representing 12 paperback books and 6 hardcover books, 16 paperback books and 0 hardcover books, and 0 paperback books and 10 hardcover books are all within the double-shaded region

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