# 3.4 Solving Systems of Linear Inequalities ©2001 by R. Villar All Rights Reserved.

## Presentation on theme: "3.4 Solving Systems of Linear Inequalities ©2001 by R. Villar All Rights Reserved."— Presentation transcript:

Solving Systems of Linear Inequalities Recall that the graphs of linear inequalities are shaded regions. Solutions to systems of linear inequalities will be the intersection of shaded regions. Example: y –2/3 x + 2 m = 2 b = –2 m = –2 b = 2 1 3 Graph both inequalities on one x-y coordinate plane and find the intersection.

y –2/3 x + 2 m = 2 b = –2 m = –2 b = 2 1 3

y –2/3 x + 2 m = 2 b = –2 m = –2 b = 2 1 3

y –2/3 x + 2 m = 2 b = –2 m = –2 b = 2 1 3

y –2/3 x + 2 m = 2 b = –2 m = –2 b = 2 1 3 This region is the solution to the system.

Solve the system by graphing: –3x – y ≥ –4 y > 2x + 1 –y ≥ 3x – 4 –y ≥ 3x – 4 -1 -1 -1 y ≤ –3x + 4 m=–3 b=4 m=2 b=1 1 1 shade below shade above solid line dotted line

Solve the system by graphing: –3x – y ≥ –4 y > 2x + 1 –y ≥ 3x – 4 –y ≥ 3x – 4 -1 -1 -1 y ≤ –3x + 4 m=–3 b=4 m=2 b=1 1 1 shade below shade above solid line dotted line

Solve the system by graphing: x ≤ 3 y < –1 y < x – 2 x ≥ 0

This region is the solution to the system.

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