# Graphing A System Linear Inequalities

## Presentation on theme: "Graphing A System Linear Inequalities"— Presentation transcript:

Graphing A System Linear Inequalities

Objectives Define the solution for a system of inequalities
Find the solution of a system of inequalities

Solving a system of Inequalities
Consider the system x + y ≥ -1 3 1 2 -1 -2 -3 -2x + y < 2

Solving a system of Inequalities
Consider the system x + y ≥ -1 y≥-x-1 3 1 2 -1 -2 -3 -2x + y < 2

Solving a system of Inequalities
Consider the system x + y ≥ -1 3 1 2 -1 -2 -3 -2x + y < 2 y<2x+2

3 1 2 -1 -2 -3 3 1 2 -1 -2 -3 x + y ≥ -1 -2x + y < 2

Solving a system of Inequalities
Consider the system x + y ≥ -1 3 1 2 -1 -2 -3 -2x + y < 2 SOLUTION: Lies where the two shaded regions intersect each other.

Solving a system of Inequalities
Consider the system -2x + 3y < -6 Graph 3 1 2 5x + 4y < 12 y < x - 2 2 3 3 2 4 -1 -2 1 -1 -2

Solving a system of Inequalities
Consider the system -2x + 3y < -6 3 1 2 5x + 4y < 12 Graph y < x + 3 5 4 3 2 4 -1 -2 1 -1 -2

Solving a system of Inequalities
Consider the system -2x + 3y < -6 3 1 2 5x + 4y < 12 Graph SOLUTION: (0,0) Lies where the two shaded regions intersect each other. 3 2 4 -1 -2 1 -1 -2 Hidden word: juice

Solving a system of Inequalities
Consider the system -2x + 3y < -6 3 1 2 5x + 4y < 12 Graph NOTE: (0,0) All order pairs in dark region are true in both inequalities. 3 2 4 -1 -2 1 -1 -2

Solving a system of Inequalities
Consider the system 10 8 12 6 4 2 -2 -4 -6 x - 4y ≤ 12 Graph 4y + x ≤ 12 TEST: (0,0) (0,0) (0) - 4(0) ≤ 12 0 - 0 ≤ 12 0 ≤ 12

Solving a system of Inequalities
Consider the system 10 8 12 6 4 2 -2 -4 -6 x - 4y ≤ 12 4y + x ≤ 12 Graph TEST: (0,0) (0,0) 4(0) + (0) ≤ 12 0 ≤ 12

Now you try When you have finished the work make sure to fill out a document. Find the hidden word with in the power point and write it next to your name.