McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
2. Solving Systems of Inequalities by Graphing
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Exercise 23 Page 149

Graph the given system and determine the vertices of the overlapping region.

Garph:

Vertices: (8,4), (6,- 4), (- 2,3)

Practice makes perfect

Our first step in finding the vertices is to graph the system and determine the overlapping region. Graphing a single inequality involves two main steps.

  1. Plotting the boundary line.
  2. Shading half of the plane to show the solution set.

For this exercise, we need to do this process for each of the inequalities in the system. 6y-24x≥- 168 & (I) 8y+7x>10 & (II) 20y-2x≤ 64 & (III) Let's begin!

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.
  • Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.

Let's find each of these key pieces of information for the inequalities in the system.

Information Inequality (I) Inequality (II) Inequality (III)
Given Inequality 6y-24x≥- 168 8y+7x>10 20y-2x≤64
Boundary Line Equation 6y-24x=- 168 8y+7x=10 20y-2x=64
Solid or Dashed? ≥ ⇒ Solid > ⇒ Dashed ≤ ⇒ Solid
y= mx+ b y= 4x+( - 28) y= -7/8x+ 5/4 y= 0.1x+ 3.2

Great! With all of this information, we can plot the boundary lines. Let's do it one at a time.

Inequality I

First, we will graph the line y=4x-28. Since it is difficult to plot its y-intercept we will use the x-intercept instead. To find the x-intercept we will substitute 0 for y and solve for x.
y=4x-28
Substitute 0 for y and simplify
0=4x-28
28=4x
7=x
x=7
The point (7,0) is the x-intercept. Let's plot this point and then we will use the slope to find another point. Finally, we will be able to draw the line.
To complete the graph, we will test a point that is not on the line and decide which region we should shade. Let's test the point ( 0, 0). If the point satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
6y-24x≥- 168
6( 0)-24( 0)? ≥- 168
Simplify
0+0? ≥- 168
0≥- 168
Since the point satisfies the inequality, we will shade the region that contains the point.

Inequality II

For the second inequality we will plot the y-intercept and use the slope to find another point. Let's plot the boundary line. Remember that this time it will be dashed.

Again, let's use ( 0, 0) as our test point.
8y+7x>10
8( 0)+7( 0)? >10
Simplify
0+0? >10
0≯ 10
This time we will shade the region opposite the test point.

Inequality III

Let's graph the third inequality. First, the boundary line.

Next let's determine which region to shade. We will use ( 0, 0), like before.
20y-2x≤ 64
20( 0)-2( 0)? ≤64
Simplify
0-0? ≤64
0≤ 64
Thus, we have to shade the region containing the point.

Combining the Inequality Graphs

Finally, we can draw the graphs of the inequalities on the same coordinate plane.

Now that we can see the overlapping region, let's highlight the vertices.

Looking at the graph, we can determine the vertices. (8,4), (6,- 4), (- 2,3)