For this exercise, we need to do this process for each of the inequalities in the system.
y>2x+4 & (I) 2x-y≤ 4 & (II)
The system's solution set will be the intersection of the shaded regions in the graphs of Inequalities (I) and (II).
Boundary Lines
We can tell a lot of information about the boundary lines from the inequalities given in the system.
Exchanging the inequality symbols for equals signs gives us the boundary line equations.
Observing the inequality symbols tells us whether the inequalities are strict.
Let's find each of these key pieces of information for the inequalities in the system.
Information
Inequality (I)
Inequality (II)
Given Inequality
y >2x+4
2x-y ≤4
Boundary Line Equation
y =2x+4
2x-y =4
Solid or Dashed?
> ⇒ Dashed
≤ ⇒ Solid
y= mx+ b
y= 2x+ 4
y= 2x+( - 4)
Great! With all of this information, we can plot the boundary lines.
Shading the Solution Sets
Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.
We will use ( 0, 0) as a test point. To do so, we will substitute 0 for x and y in the given inequalities and simplify. If the substitution creates a true statement, we will shade the region that contains the point ( 0, 0). Otherwise, we shade the region which does not contain the point.
Information
Inequality (I)
Inequality (II)
Given Inequality
y>2x+4
2x-y≤ 4
Substitute ( 0, 0)
0? >2( 0)+4
2( 0)- 0? ≤4
Simplify
0≯ 4
0≤ 4
Shaded Region
does not contain (0,0)
contains (0,0)
For Inequality (I) we will shade the region opposite to the test point, or above the boundary line. For Inequality (II), however, we will shade the region containing our test point, or above the boundary line. The solution set of the system is the intersection of both shaded regions.
