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SOLVING TWO VARIABLE LINEAR INEQUALITIES

SOLVING TWO VARIABLE LINEAR INEQUALITIES. INCLUDING ABSOLUTE VALUE INEQUALITIES. Summary of Inequality Signs. >. >. Shade above the line. <. <. Shade below the line. Continuous line. Dashed line. Graphing Linear Inequalities.

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SOLVING TWO VARIABLE LINEAR INEQUALITIES

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  1. SOLVING TWO VARIABLE LINEAR INEQUALITIES INCLUDING ABSOLUTE VALUE INEQUALITIES

  2. Summary of Inequality Signs > > Shade above the line < < Shade below the line Continuous line Dashed line

  3. Graphing Linear Inequalities The graph of a linear inequality is a region of the coordinate plane that is bounded by a line. This region represents the SOLUTION to the inequality.

  4. A linear inequality is an inequality in two variables whose graph is a region of the coordinate plane that is bounded by a line.

  5. Graph the following inequality:x > 2 Boundary is: x = 2 We shaded at the right of the line because x is more than 2. The line is dashed because it is not equal or less than x, so the line which is the boundary is not included in the solution.

  6. Graph the following inequality: y < 6 Boundary: y = 0x + 6 m= 0 y- intercept = (0,6) We shaded below the line because y is less than 6. The line is dashed because it is not equal or less than y, so the line which is the boundary is not included in the solution.

  7. Example

  8. 1. The boundary line is dashed.

  9. Substitute (0, 0) into the inequality to decide where to shade. • So the graph is shaded away from (0, 0).

  10. Graph the following inequality:4x + 2y < 10 Solve for y  y < -2x + 5 Boundary is: y = -2x + 5 m= -2 y- intercept = (0,5) We shaded below the line because y is less than the expression -2x + 5. The line is dashed because it is not equal or less than y, so the line which is the boundary is not included.

  11. Graph the following inequality:-9x + 3y< 3 Solve for y  y < 3x + 1 Boundary is: y = 3x + 1 m= 3 y- intercept = (0, 1) We shaded below the line because y is less than the expression 3x +1. The line is dashed because it is not equal or less than y, so the line which is the boundary is not included.

  12. Graph the following inequality: y – 2 > (x – 4) Solve for y  y > x – 3 Boundary is: y = x – 3 m = y-intercept = (0, -3) We shaded above because y is greater or equal than the expression and the line is continuous because the word equal in greater or equal indicates that the boundary is included in the solution.

  13. EXAMPLE

  14. EXAMPLE

  15. EXAMPLE

  16. EXAMPLE

  17. Problem, con’t

  18. Graph the following absolute value equation: y = |x| For x < 0 For x > 0 y = -x y = x Now let’s shift it two units up: y = |x| + 2 Now let’s shift it three units to the right: y = |x - 3| + 2 Now let’s graph it upside down y = – |x-3| + 2 Now let’s make it skinner y = – 6|x-3| + 2

  19. So, that’s how the different parameters in an absolute value equation affect our graph. Now let’s graph absolute value inequalities.

  20. Absolute Value Inequalities • Graph the absolute value function then shade above OR below Solid line…y<, y> Dashed line…y<, y> Shade above y>, y> Shade below…y<, y<

  21. Absolute Value Inequalities Graph y< |x – 2| + 3 y<|x– 2|+ 3 DASHED line Shade BELOW slope = 1 Vertex = (2, 3)

  22. Absolute Value Inequalities Graph y<|x– 2|+ 3 Vertex = (2, 3)

  23. Absolute Value Inequalities Graph y<|x– 2|+ 3 slope = 1

  24. Absolute Value Inequalities Graph y<|x– 2|+ 3 DASHED line Shade BELOW

  25. Absolute Value Inequalities Graph y<|x– 2|+ 3 Shade BELOW

  26. Absolute Value Inequalities Graph –y + 1 < -2|x + 2| -y< -2|x + 2| - 1 y> 2|x + 2| + 1 -y so CHANGE the direction of the inequality

  27. Absolute Value Inequalities y> 2|x + 2| + 1 Solid line Shade above Vertex = (-2, 1) Slope = 2

  28. Absolute Value Inequalities y> 2|x + 2| + 1

  29. Absolute Value Inequalities y> 2|x + 2| + 1

  30. Absolute Value Inequalities y> 2|x + 2| + 1

  31. Absolute Value Inequalities y> 2|x + 2| + 1

  32. Absolute Value Inequalities Write an equation for the graph below.

  33. Graph the following inequality:y > |x| Finding the boundary: For x < 0 For x > 0 y = -x y = x . There are two regions: Testing point (0,2) 2 > | 0| 2 > 0 true Therefore, the region where (0,2) lies is the solution region and we shade it.

  34. Graph the following inequality:y < |x+1|– 3 Finding the boundary: For x + 1 < 0 For x + 1 > 0 y = -(x+1) – 3 y = x+1 – 3 y = - x – 1 -3 y = - x – 4 y = x – 2 There are two regions: Testing point (0,0) 0 < | 0+1| – 3 0 < -2 false So the region where (0,0) lies is not in the solution region, therefore we shade the region below. .

  35. Steps: • Decide if the boundary graph is solid or dashed. • Graph the absolute value function as the boundary. • Use the point (0, 0), if it is not on the boundary graph, to decide how to shade.

  36. Graph y ≥ 2|x – 3| + 2

  37. Graph y ≥ 2|x – 3| + 2 1. The boundary graph is solid.

  38. y ≥ 2|x – 3| + 2 • 0 ≥ 2|0 – 3| + 2 • 0 ≥ 2|-3| + 2 • 0 ≥ 6 + 2 • 0 ≥ 8 False • So shade away from (0, 0).

  39. The boundary graph is dashed.

  40. Your Turn! 8. Graph 9. Graph

  41. 8. 9.

  42. Example 10

  43. Example 10 y ≤ |x+4| - 3 y ≥ 2x+5

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