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Published byPhoebe Walters Modified over 8 years ago
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Graphing a Two Variable Inequality
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Examining the Solutions of a Linear Equation Are the following points solutions to the equation y = -2x + 3 ? Justify each conclusion with both the graph and the equation? a. (-1,5) b1. (2,-1) b2. (0,0) Yes. No. The points on the line make the equation true.
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There are an infinite number of points that satisfy the inequality.
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Graphing an Inequality In order to find the points that satisfy an inequality statement: 1. Find the boundary 2. Test every region to find which one(s) satisfies the original statement
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Graphing a 2 Variable Inequality (0,0) Graphically represent the solutions to the following inequality: Find the Boundary Plot points for the equality Test Every Region (3,0) 0 > -3 0 > 1.5 True False Solid or Dashed? Pick a point in each region Substitute into Original Shade True Region(s)
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Graphing a 2 Variable Inequality (0,0) Graphically represent the solutions to the following inequality: Find the Boundary Plot points for the equality Test Every Region (-4,0) 0 ≤ 6 8 ≤ 6 True False Solid or Dashed? Pick a point in each region Substitute into Original Shade True Region(s)
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(0,0) Graphically represent the solutions to the following inequality: Find the Boundary Plot points for the equality Test Every Region (0,3) 0 > 2 3 > 2 False True Solid or Dashed? Pick a point in each region Substitute into Original Shade True Region(s) Open or Closed Dots? Graphing a 1 Variable Inequality in 2-Dimensions
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Graphing a 2 Variable Inequality (0,0) Graphically represent the solutions to the following inequality: Find the Boundary Plot points for the equality Test Every Region (0,5) 0 < 3 5 < 3 True False Solid or Dashed? Pick a point in each region Substitute into Original Shade True Region(s)
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Graphing a 2 Variable Inequality (0,0) Graphically represent the solutions to the following inequality: Find the Boundary Plot points for the equality Test Every Region (0,2) 0 < 1 2 < 1 True False Solid or Dashed? Pick a point in each region Substitute into Original Shade True Region(s)
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