McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
2. Solving Systems of Inequalities by Graphing
Continue to next subchapter

Exercise 55 Page 152

Begin by graphing the equations. First two equations represent horizontal lines and last two of them are in slope-intercept form.

(3/2,3), (7/2,7), (8,3), (10,7)

Practice makes perfect

Let's examine the given equations. y=3 y=7 y=2x y=2x-13 Equation I and Equation II represent horizontal lines that pass through the points (0,3) and (0,7), respectively. Let's graph the first two equations.

Next, we will graph the third equation. Notice that it is in slope-intercept form. Thus, we can immediately determine its slopes and y-intercepts. Equation III y= 2x+ 0 To graph it, we will first plot its y-intercept and use its slope to find a second point.

Now we have two points, we can draw a line.

The last equation is also in slope-intercept form, so it can be graphed proceeding in the same way. Equation IV y= 2x -13 Let's graph it!

The y-coordinates of the vertices are already known. We will find the x-coordinates of them by substituting y=3 and y=7 into Equation III and Equation IV. Let's begin by substituting y=3 into Equation III.
y=2x
3=2x
3/2=x
x=3/2
The coordinates of the first vertex is ( 32,3). The other vertices can be found in the same way.
Vertex y-coordinate Substitution x-coordinate Point
I 3 3=2x 3/2 (3/2,3)
II 7 7=2x 7/2 (7/2,7)
III 3 3=2x-13 8 (8,3)
IV 7 7=2x-13 10 (10,7)