McGraw Hill Glencoe Algebra 1, 2012
MH
McGraw Hill Glencoe Algebra 1, 2012 View details
6. Graphing Inequalities in Two Variables
Continue to next subchapter

Exercise 43 Page 321

Start by drawing the boundary line. Then, decide which side of the boundary line to shade.

Ordered pairs that belong to the solution set: (2,-5), (5,3) and (7,5)

Practice makes perfect

Graphing an inequality involves two main steps.

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

Boundary Line

To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol with an equals sign. Inequality & Boundary Line 2x-2y ≥ 4 & 2x-2y = 4To draw this line, we will first rewrite the equation in slope-intercept form.
2x-2y=4
-2y=-2x+4
y=x-2
Now that the equation is in slope-intercept form, we can identify the slope m and y-intercept (0, b). y= 1x+( -2) We will plot the y-intercept (0, -2), then use the slope m= 1 to plot another point on the line. Connecting these points with a solid line will give us the boundary line of our inequality. Note that the boundary line is solid, not dashed, because the inequality is not strict.

Shading the Plane

To decide which side of the boundary line to shade, we will substitute a test point that is not on the boundary line into the given inequality. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point.
2x-2y ≥ 4
2( 0) - 2( 0) ? ≥ 4
0-0 ? ≥ 4
0 ≱ 4
Since the substitution of the test point did not create a true statement, we will shade the region that does not contain the point.

Determining the Solutions

Finally, we can check the given points to see which ones are contained in the solution set. We will do this by adding the list of points to our coordinate plane.

Points that lie within the shaded region are solutions to the inequality. Note that points that lie on the solid line are also solutions. This means that three of the given points are part of the solution set. (2,-5), (5,3) and (7,5)