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McGraw-Hill Ryerson
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McGraw-Hill Ryerson Pre-Calculus 11
1. This lesson is designed to help students conceptualize the main ideas of Chapter 9.
2. To view the lesson, go to Slide Show > View Show
(PowerPoint 2003).
3. To use the pen tool, view Slide Show, click on the icon in the lower left of your screen and select Ball Point Pen.
4. To reveal an answer, click on or follow the instructions on the slide. To reveal a hint, click on . To view the complete solution, click on the View Solution button. Navigate through the lesson using the and buttons.
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Chapter
9
The graph of the linear equation x
– y =
–2 is referred to as a boundary line. This line divides the Cartesian plane into two regions:
For one region, the condition x – y < –2 is true.
For the other region, the condition x – y > –2 is true.
Use the pen to label the conditions below to the corresponding parts of the graph on the Cartesian plane.
x
– y
=
–
2 x – y < – 2 x – y > – 2 x – y < – 2 x – y > – 2
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Chapter
9
The ordered pair ( x , y ) is a solution to a linear inequality if its coordinates satisfy the condition expressed by the inequality.
Which of the following ordered pairs ( x , y ) are solutions of the linear inequality x
– 4 y < 4?
Click on the ordered pairs to check your answer.


2,

3
2





3
2
,2




3
2


 
 
4,0
 x

4 y

4
Use the pen tool to graph the boundary line and plot the points on the graph. Then, shade the region that represents the inequality.
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Click here for the solution.

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Chapter
9
Match the inequality to the appropriate graph of a boundary line below.
Complete the graph of each inequality by shading the correct solution region.
Match Shade

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Chapter
9 a)
Use the pen tool to graph the following inequalities. Describe the steps required to graph the inequality.
Click here for the solution.

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Chapter
9
Match each inequality to its graph.
Then, click on the graph to check the answer.

1.
Chapter
9
(0, 3)
0
(2, -1)
Write an inequality that represents each graph.
2.
(2, 4)
0
(0, -2
)
3 x y 2
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2 x y 3

Chapter
9
Paul is hosting a barbecue and has decided to budget $48 to purchase meat. Hamburger costs $5 per kilogram and chicken costs $6.50 per kilogram.
Let h = kg of hamburger c = kg of chicken
Write an inequality to represent the number of kilograms of each that Paul may purchase.
Write the equation of the boundary line below and draw its graph.
Shade the solution region for the inequality.
Hamburger
Click here for the solution.
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Chapter
9 c
5 h

6.5
c

48 h
Hamburger
1. Can Paul buy 6 kg of hamburger and 4 kg chicken if he wants to stay within his set budget?
No
2. How many kilograms of chicken can Paul buy if he decides not to buy any hamburger?
7.38 kg
3. If Paul buys 3 kg of hamburger, what is the greatest number of kilograms of chicken he can buy?
5.08 kg
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Click here for the solution.

Chapter
9
Solve x
2
– x
– 12 > 0. Use the pen tool.
Solve the related equation to determine intervals of numbers that may be solutions of the inequality.
Plot the solutions on a number line creating the intervals for investigation.
Pick a number from each interval to test in the original inequality. If the number tested satisfies the inequality, then all of the numbers in that interval are solutions.
State the solution set.
-5 -4 -3 -2 -1 0 1 2 3 4 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
Click here for the solution.
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Chapter
9
Graph the corresponding quadratic function y = x 2 – x
– 12 to verify your solution from the previous page.
Solve x
2
– x – 12 > 0. Use the pen tool.
Click here for the solution.
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1.
Chapter
9
Use the pen tool to solve the related quadratic equation to obtain the boundary points for the intervals.
Solve x
2
– 3 x – 4 > 0.
Use the boundary points to mark off test intervals on the number line.
Determine the intervals when each of the factors is positive or negative.
x - 4 x + 1
( x - 4)( x + 1)
Determine the solution using the number line.
Click here for the solution.
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Chapter
9
1.
Use the pen tool to create a graph of the related function to confirm your solutions.
Solve x
2
– 3 x – 4 > 0.
Click here for the solution.
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Chapter
9
Choose the correct shaded region to complete the graph of the inequality.
Circle your choice using the pen tool.

Chapter
9 y
 
2 x
2
Match each inequality to its graph using the pen tool.
y
 
2 x 2 , y

2 x 2 , y
  x 2 y
  x 2 
4 x

6, y
 x 2 
6 x

10, y
 x 2 
6 x

5 y
 x 2 
6 x

5 y
 x 2 
6 x

10 y
  x 2 
4 x

6 y

2 x 2
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  x
2

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The following pages contain solutions for the previous questions.
Click here to return to the start

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Solutions
(-4, 0)
(0, 4)
0
(0, 0) (4, 0)
(0, -4)
Go back to the question.

Solutions
An example method for graphing an inequality would be:
1
• Slope of the line is . and the y -intercept is the point (0, 1).
• The inequality is less than. Therefore, the boundary line is a broken line.
• Use a test point (0, 0). The point makes the inequality true.
• Therefore, shade below the line.
• The x -intercept is the point (
–2, 0), the y -intercept is the point (0, –4).
• The inequality is greater than and equal to.
Therefore, the boundary line is a solid line.
• Use a test point (0, 0). The point makes the inequality true.
• Therefore, shade above the line.
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Go back to the question.

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Solutions c
Let h = kg of hamburger c = kg of chicken
Write an inequality to represent the number of kilograms of each that Paul may purchase.
Graph the boundary line for the inequality.
h
Hamburger
Go back to the question.

Solutions c
(0, 7.38)
(3, 5)
5 h

6.5
c

48
(6, 4) h
Hamburger
1. Can Paul buy 6 kg of hamburger and 4 kg chicken if he wants to stay within his set budget?
The point (6, 4) is not within the shaded region. Paul could not purchase 6 kg of hamburger and 4 kg of chicken.
2. How many kilograms of chicken can Paul buy if he decides not to buy any hamburger?
This is the point (0, 7.38). Buying no hamburger would be the y-intercept of the graph.
3. If Paul buys 3 kg of hamburger, what is the greatest whole number of kilograms of chicken he can buy?
This would be the point (3, 5). Paul could buy 5 kg of chicken.
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Go back to the question.

Solutions
Solve the related equation to determine intervals of numbers that may be solutions of the inequality.
Plot the solutions on a number line creating the intervals for investigation.
-5 -4 -3 -2 -1 0 1 2 3 4 5
Pick a number from each interval to test in the original inequality. If the number tested satisfies the inequality, then all of the numbers in that interval are solutions.
Test point -5:
( -5 )
2
- ( -5 ) - 12 > 0
True
Test point 0:
( 0 )
2
- (
False
0 ) - 12 > 0
Test point 5:
( 5 )
2
True
-5 -4 -3 -2 -1 0 1 2 3 4 5
- ( 5 ) - 12 > 0
State the solution set.
The solution set is { x | x < –3 or x > 4, x

R}.
Go back to the question.
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Solutions
The inequality may have been solved by examining the graph of the corresponding function, y = x 2 – x – 12. The quadratic inequality is greater than zero where the graph is above the x -axis.
Solve x
2
– x – 12 > 0 x < –3 or x > 4
Go back to the question.
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Solutions
1.
Solve the related quadratic equation to obtain the boundary points for the intervals.
Use the boundary points to mark off test intervals on the number line. x
–
4 x + 1
Determine the intervals when each of the factors is positive or negative.
( x
– 4)( x + 1) x
2
– 3 x – 4 = 0
( x – 4)( x + 1) = 0 x – 4 = 0 or x + 1 = 0 x = 4 x = – 1
+
–
–
–
+
–
+
+
+
Determine the solution using the number line.
Pen Tool x < – 1 or x > 4
Go back to the question.

Solutions
1.
A graph of the related function may be used to confirm your solutions.
Pen Tool x < – 1 or x > 4
Go back to the question.