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

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 = 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 = This region is the solution to the system.

Solve the system by graphing: –3x – y ≥ –4 y > 2x + 1 –y ≥ 3x – 4 –y ≥ 3x – 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 – 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.