Lesson 3–3

Example 1  Intersecting Regions
Solve each system of inequalities.	
a. y > x
2x + y  7

solution of y > x 	 Regions 1 and 2
solution of 2x + y  7 	 Regions 1 and 3

The region that provides a solution of both inequalities is the solution of the system. Region 1 is the solution of the system.
	
b. x – y  5
| x | < 4

The inequality | x | < 4 can be written as x < 4 and x > –4.

Graph all of the inequalities on the same coordinate plane and shade the region or regions that are common to all. 

Example 2  Separate Regions
Solve the system of inequalities by graphing.	
3y + 2x  3
6y  –4x – 24

Graph both inequalities. The graphs do not overlap, so the solutions have no points in common. The solution set is Ø.

Example 3  Write and Use a System of Inequalities
Postage    The U.S. Postal Service allows packages up to 70 pounds with a combined length and girth not over 108 inches to be mailed under the classification of Priority Mail. Write and graph a system of inequalities that represents the range of weights and combined length and girth measures for Priority Mail.  Source: The World Almanac

Let w represent the weight of packages in pounds. The acceptable weights are 0 to 70 pounds. We can write two inequalities.

0  w and w  70
	
Let m represent the combined length and girth of a package. The acceptable measures can also be written as two inequalities.

0  m and m  108

Graph all of the inequalities. Any ordered pair in the intersection of the graphs is a solution of the system. In this case, a solution of the system of inequalities is a potential weight and girth combination for Priority Mail.

Example 4  Find Vertices
Find the coordinates of the vertices of the figure formed by y  6, x  –4, y  x – 2, and 
y + 2x  7.
	
Graph each inequality. The intersection of the graphs forms a quadrilateral. 

The coordinates (–4, 6) and (–4, –6) can be determined from the graph. To find the coordinates of the other two vertices, you need to solve two systems.

System 1
y = 6
y + 2x = 7

Substitute 6 for y into the second equation.
y + 2x = 7	Second equation
6 + 2x = 7	Replace y with 6.
      2x = 1	Subtract 6 from each side.
        x = 	Divide each side by 2.
The third vertex is.

System 2
y + 2x = 7
y = x – 2 

Rewrite the equations and subtract to eliminate y.

   y + 2x =   7
 (–)  y –    x = –2		
         3x  =  9	Subtract the equations.	
    x = 3	Divide each side by 3.	

Now find y by substituting 3 for x in the first equation.
y + 2x = 7	First equation	
y + 2(3) = 7	Substitute 3 for x.	
y + 6 = 7	Multiply.	
y = 1	Subtract 6 from each side.	

The fourth vertex is (3, 1). 

The vertices of the quadrilateral are (–4, 6), (–4, –6), , and (3, 1).