McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
6. Graphing Inequalities in Two Variables
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Exercise 37 Page 320

Practice makes perfect
a Let x be the number of hot dogs and y the number of sodas the team sells. Note that the price of each hot dog is $1.00, and the price of each soda is $1.25. We can write an expression for the total earnings from these sales.

1 * x+ 1.25 * y ⇔ x+1.25y The cost of the new goals is $2000. Therefore, our expression should be greater than or equal to 2000. x+ 1.25y ≥ 2000

b To graph the inequality we will follow two steps.
  1. Graph the boundary line.
  2. Determine the region to shade.

Boundary Line

To obtain the boundary line, we replace the inequality sign with an equals sign. ccc Inequality & & Boundary Line [0.8em] x+1.25y ≥ 2000 & & x+1.25y = 2000 To graph the boundary line, we will first write in it slope-intercept form.
x+1.25y=2000
â–Ľ
Write in slope-intercept form
1.25y=- x+2000
5y=- 4x+8000
y=- 4x+8000/5
y=- 4x/5+8000/5
y=- 4x/5+1600
y=- 4/5x+1600
y= - 4/5x+ 1600
We will now use the slope - 45 and the y-intercept 1600 to graph the line. Note that the inequality is not strict, so the boundary line will be solid. Since the number of items cannot be a negative number, the domain and range contain only non-negative numbers.

Region

To determine the region we should shade, we will test a point. If substituting the point into the inequality produces a true statement, we will shade the half-plane that contains the point. Otherwise, we will shade the opposite region. For simplicity, we will test the point (0,0).
x+1.25y≥ 2000
0+1.25( 0)? ≥ 2000
0+0? ≥ 2000
0 ≱ 2000
Since the point did not produce a true statement, we will shade the region that does not contain (0,0).
c The answers may vary, since any point in the shaded region is a solution. Consider our graph.

For example, we may select the points (800,1200), (1000,1600), (1700,1400), (1400,800), and (500,1500).

d Finally, we will plot the points we mentioned in part C.