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 6 Page 149

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

Graph:

Vertices: (3,5), (2,- 10), (- 4,4)

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. 3y≥- 7x-16 & (I) 7y≤ x+32 & (II) y≥ 15x-40 & (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 3y≥- 7x-16 7y≤x+32 y≥15x-40
Boundary Line Equation 3y=- 7x-16 7y=x+32 y=15x-40
Solid or Dashed? ≥ ⇒ Solid ≤ ⇒ Solid ≥ ⇒ Solid
y= mx+ b y= -7/3x+( -16/3) y= 1/7x+ 32/7 y= 15x+( - 40)
Great! With all of this information, we can plot the boundary lines. Let's do it one at a time.

Inequality I

To draw the first boundary line we will plot the y-intercept and then use the slope to find another point.

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.
3y≥- 7x-16
3( 0)? ≥- 7( 0)-16
Simplify
0? ≥0-16
0≥- 16
Since the point satisfies the inequality, we will shade the region that contains the point.

Inequality II

We will follow the same process for the second inequality. Let's change the plane dimensions at first, so that it's easier to plot the boundary line.

Now, we will graph this line on the plane with original dimensions. In order to transfer it, we will use the y-intercept and the lattice point, (3,5). Then we can sketch a straight line passing through these points.

Again, let's use ( 0, 0) as our test point.
7y≤ x+32
7( 0)? ≤ 0+32
Simplify
0? ≤0+32
0≤ 32
One more time we will have to shade the region containing the point.

Inequality III

One last time we have to follow the process for the third inequality. First, the boundary line.

Next let's determine which region to shade. We will use ( 0, 0), like before.
y≥ 15x-40
0? ≥15( 0)-40
Simplify
0? ≥0-40
0≥- 40
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 two of the vertices. (3,5) and (2,- 10) To find the third vertex, we will have to determine the point of intersection of the lines 3y=- 7x-16 and 7y=x+32. We can do it by solving the system of equations related to these lines. 3y=- 7x-16 & (I) 7y=x+32 & (II) Since there is no isolated variable, we will use the Elimination Method.
3y=- 7x-16 7y=x+32
Solve by elimination
3y=- 7x-16 49y=7x+224
3y+ 49y=- 7x-16+ 7x+224 49y=7x+224
52y=208 49y=7x+224
y=4 49y=7x+224
y=4 7y=x+32
y=4 7( 4)=x+32
y=4 28=x+32
y=4 - 4=x
y=4 x=- 4
We determined that the last vertex is (- 4,4).